Re: [R] How would I color points conditional on their value in a plot of a time series
> Eivind K Dovik > on Tue, 1 May 2018 23:18:51 +0200 writes: > You may also want to check this out: > plot(ttt, type = "p") > points(ttt, col = ifelse(ttt < 8, "black", "red")) > Eivind K. Dovik > Bergen, NO yes, indeed, or -- even nicer for a time series: using 'type = "c"' which many people don't know / have forgotten about: ttt <- ts(rpois(12, lambda = 8), start = c(2000, 1), freq = 4) plot (ttt, type = "c") points(ttt, col = ifelse(ttt < 8, "black", "red")) Martin Maechler > On Tue, 1 May 2018, Christopher W Ryan wrote: >> Excellent! Worked like a charm. Thanks. >> >> --Chris Ryan >> >> On Tue, May 1, 2018 at 4:33 PM, William Dunlap wrote: >> >>> The ts method for plot() is quirky. You can use the default method: >>> >>> plot(as.vector(time(ttt)), as.vector(ttt), type = "p", col=ifelse(ttt<8, >>> "black", "red")) >>> >>> >>> Bill Dunlap >>> TIBCO Software >>> wdunlap tibco.com >>> >>> On Tue, May 1, 2018 at 1:17 PM, Christopher W Ryan >>> wrote: >>> How would I color points conditional on their value in a plot of a time series. Something like this: ## demonstration data ttt <- ts(rpois(12, lambda = 8), start = c(2000, 1), freq = 4) ttt plot(ttt, type = "p") ## doesn't work--all points the same color plot(ttt, type = "p", col = ifelse(ttt < 8, "black", "red")) ## also doesn't work--all points the same color q <- as.numeric(ttt) q plot(ttt, type = "p", col = ifelse(q < 8, "black", "red")) ## works OK with a simple, non-time-series scatterplot, as in sss <- data.frame(x = rpois(12, lambda = 8), y = rnorm(12, mean = 100, sd = 25)) with(sss, plot(y ~ x, col = ifelse(y > 100, "black", "red"))) ## but I am missing something about time series. Thanks. --Chris Ryan Broome County Health Department and Binghamton University Binghamton, NY [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posti ng-guide.html and provide commented, minimal, self-contained, reproducible code. >>> >>> >> >> [[alternative HTML version deleted]] >> >> __ >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Merging dataframes
Thanks - Peter, Eivind, Rui Sorry, I perhaps could not explain it properly in the first go. Trying to simplify it here with an example - Say I have two dataframes as below that are NOT equally-sized data frames (i.e., number of columns are different in each table): Table_A: Email Name Phone a...@gmail.com John Chan 0909 b...@yahoo.com Tim Ma89089 .. Table_B: Email Name SexPhone a...@gmail.comJohn ChanM 0909 k...@hotmail.com Rosy Kim F 7779 . Now, I have used - merge (Table_A, Table_B, by="Email", all = FALSE)) - to find only the rows that match from these data frames - based on Email as primary key. Further, I am also interested (using "Email" as the common key) which rows from Table_A did not match with Table_B. I am not sure how to do this here. Thanks and regards, Chintanu On Tue, May 1, 2018 at 8:48 PM, Rui Barradas wrote: > Hello, > > Is it something like this that you want? > > x <- data.frame(a = c(1:3, 5, 5:10), b = c(1:7, 7, 9:10)) > y <- data.frame(a = 1:10, b = 1:10) > > which(x != y, arr.ind = TRUE) > > > Hope this helps, > > Rui Barradas > > > On 5/1/2018 11:35 AM, Chintanu wrote: > >> Hi, >> >> >> May I please ask how I do the following in R. Sorry - this may be trivial, >> but I am struggling here for this. >> >> >> >> For two dataframes (A and B), I wish to identify (based on a primary >> key-column present in both A & B) - >> >> 1. Which records (rows) of A did not match with B, and >> >> >> >> 2. Which records of B did not match with A ? >> >> >> >> I came across a setdt function while browsing, but when I tried it, it >> says >> - Could not find function "setdt". >> >> >> >> Overall, if there is any way of doing it (preferably in some simplified >> way), please advise. >> >> >> Many thanks in advance. >> >> >> regards, >> >> Tito >> >> [[alternative HTML version deleted]] >> >> __ >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posti >> ng-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> >> [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Merging dataframes
Hi,
I'll coded your example into R code:
Table_A <- c('a...@gmail.com', 'John Chan', '0909')
Table_A <- rbind(Table_A, c('b...@yahoo.com', 'Tim Ma', '89089'))
colnames(Table_A) <- c('Email', 'Name', 'Phone')
Table_A
Table_B <- c('a...@gmail.com', 'John Chan', 'M', '0909')
Table_B <- rbind(Table_B, c('k...@hotmail.com', 'Rosy Kim', 'F', '7779'))
colnames(Table_B) <- c('Email', 'Name', 'Sex', 'Phone')
Table_B
Did you have a look at this one?
Table_C <- merge (Table_A, Table_B, by="Email", all = TRUE)
Table_C[is.na(Table_C$Name.y),]
Table_C[is.na(Table_C$Name.x),]
Table_C contains all data from Table_A and Table_B. The key.x is NA if
the row comes from Table_B and key.y is NA if the row comes from Table_A.
Best, Robin
On 05/02/2018 11:38 AM, Chintanu wrote:
> Thanks - Peter, Eivind, Rui
>
>
> Sorry, I perhaps could not explain it properly in the first go.
>
> Trying to simplify it here with an example - Say I have two dataframes as
> below that are NOT equally-sized data frames (i.e., number of columns are
> different in each table):
>
>
>
> Table_A:
>
> Email Name Phone
>
> a...@gmail.com John Chan 0909
>
> b...@yahoo.com Tim Ma89089
>
> ..
>
>
>
> Table_B:
>
> Email Name SexPhone
>
> a...@gmail.comJohn ChanM 0909
>
> k...@hotmail.com Rosy Kim F 7779
>
> .
>
>
>
> Now, I have used -
>
> merge (Table_A, Table_B, by="Email", all = FALSE))
>
>
>
> - to find only the rows that match from these data frames - based on Email
> as primary key.
>
>
>
> Further, I am also interested (using "Email" as the common key) which rows
> from Table_A did not match with Table_B.
>
> I am not sure how to do this here.
>
> Thanks and regards,
> Chintanu
>
>
>
> On Tue, May 1, 2018 at 8:48 PM, Rui Barradas wrote:
>
>> Hello,
>>
>> Is it something like this that you want?
>>
>> x <- data.frame(a = c(1:3, 5, 5:10), b = c(1:7, 7, 9:10))
>> y <- data.frame(a = 1:10, b = 1:10)
>>
>> which(x != y, arr.ind = TRUE)
>>
>>
>> Hope this helps,
>>
>> Rui Barradas
>>
>>
>> On 5/1/2018 11:35 AM, Chintanu wrote:
>>
>>> Hi,
>>>
>>>
>>> May I please ask how I do the following in R. Sorry - this may be trivial,
>>> but I am struggling here for this.
>>>
>>>
>>>
>>> For two dataframes (A and B), I wish to identify (based on a primary
>>> key-column present in both A & B) -
>>>
>>> 1. Which records (rows) of A did not match with B, and
>>>
>>>
>>>
>>> 2. Which records of B did not match with A ?
>>>
>>>
>>>
>>> I came across a setdt function while browsing, but when I tried it, it
>>> says
>>> - Could not find function "setdt".
>>>
>>>
>>>
>>> Overall, if there is any way of doing it (preferably in some simplified
>>> way), please advise.
>>>
>>>
>>> Many thanks in advance.
>>>
>>>
>>> regards,
>>>
>>> Tito
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> __
>>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide http://www.R-project.org/posti
>>> ng-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Dr. Robin Haunschild
Max Planck Institute
for Solid State Research
Heisenbergstr. 1
D-70569 Stuttgart (Germany)
phone: +49 (0) 711-689-1285
fax: +49 (0) 711-689-1292
email: r.haunsch...@fkf.mpg.de
http://www.fkf.mpg.de/ivs
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Re: [R] Merging dataframes
Have a look at anti_join() from the dplyr package. It does exactly
what you want. Here is an example based on the code of Robin
Table_A <- as.data.frame(Table_A, stringsAsFactors = FALSE)That is
Table_B <- as.data.frame(Table_B, stringsAsFactors = FALSE)
library(dplyr)
anti_join(Table_A, Table_B, by = "Email")
anti_join(Table_B, Table_A, by = "Email")
Best regards,
ir. Thierry Onkelinx
Statisticus / Statistician
Vlaamse Overheid / Government of Flanders
INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE
AND FOREST
Team Biometrie & Kwaliteitszorg / Team Biometrics & Quality Assurance
thierry.onkel...@inbo.be
Havenlaan 88 bus 73, 1000 Brussel
www.inbo.be
///
To call in the statistician after the experiment is done may be no
more than asking him to perform a post-mortem examination: he may be
able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data. ~ Roger Brinner
The combination of some data and an aching desire for an answer does
not ensure that a reasonable answer can be extracted from a given body
of data. ~ John Tukey
///
2018-05-02 13:23 GMT+02:00 Dr. Robin Haunschild :
> Hi,
>
> I'll coded your example into R code:
>
> Table_A <- c('a...@gmail.com', 'John Chan', '0909')
> Table_A <- rbind(Table_A, c('b...@yahoo.com', 'Tim Ma', '89089'))
> colnames(Table_A) <- c('Email', 'Name', 'Phone')
> Table_A
>
> Table_B <- c('a...@gmail.com', 'John Chan', 'M', '0909')
> Table_B <- rbind(Table_B, c('k...@hotmail.com', 'Rosy Kim', 'F', '7779'))
> colnames(Table_B) <- c('Email', 'Name', 'Sex', 'Phone')
> Table_B
>
> Did you have a look at this one?
> Table_C <- merge (Table_A, Table_B, by="Email", all = TRUE)
> Table_C[is.na(Table_C$Name.y),]
> Table_C[is.na(Table_C$Name.x),]
>
> Table_C contains all data from Table_A and Table_B. The key.x is NA if
> the row comes from Table_B and key.y is NA if the row comes from Table_A.
>
> Best, Robin
>
>
> On 05/02/2018 11:38 AM, Chintanu wrote:
>> Thanks - Peter, Eivind, Rui
>>
>>
>> Sorry, I perhaps could not explain it properly in the first go.
>>
>> Trying to simplify it here with an example - Say I have two dataframes as
>> below that are NOT equally-sized data frames (i.e., number of columns are
>> different in each table):
>>
>>
>>
>> Table_A:
>>
>> Email Name Phone
>>
>> a...@gmail.com John Chan 0909
>>
>> b...@yahoo.com Tim Ma89089
>>
>> ..
>>
>>
>>
>> Table_B:
>>
>> Email Name SexPhone
>>
>> a...@gmail.comJohn ChanM 0909
>>
>> k...@hotmail.com Rosy Kim F 7779
>>
>> .
>>
>>
>>
>> Now, I have used -
>>
>> merge (Table_A, Table_B, by="Email", all = FALSE))
>>
>>
>>
>> - to find only the rows that match from these data frames - based on Email
>> as primary key.
>>
>>
>>
>> Further, I am also interested (using "Email" as the common key) which rows
>> from Table_A did not match with Table_B.
>>
>> I am not sure how to do this here.
>>
>> Thanks and regards,
>> Chintanu
>>
>>
>>
>> On Tue, May 1, 2018 at 8:48 PM, Rui Barradas wrote:
>>
>>> Hello,
>>>
>>> Is it something like this that you want?
>>>
>>> x <- data.frame(a = c(1:3, 5, 5:10), b = c(1:7, 7, 9:10))
>>> y <- data.frame(a = 1:10, b = 1:10)
>>>
>>> which(x != y, arr.ind = TRUE)
>>>
>>>
>>> Hope this helps,
>>>
>>> Rui Barradas
>>>
>>>
>>> On 5/1/2018 11:35 AM, Chintanu wrote:
>>>
Hi,
May I please ask how I do the following in R. Sorry - this may be trivial,
but I am struggling here for this.
For two dataframes (A and B), I wish to identify (based on a primary
key-column present in both A & B) -
1. Which records (rows) of A did not match with B, and
2. Which records of B did not match with A ?
I came across a setdt function while browsing, but when I tried it, it
says
- Could not find function "setdt".
Overall, if there is any way of doing it (preferably in some simplified
way), please advise.
Many thanks in advance.
regards,
Tito
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posti
ng-guide.html
and provide commented, minimal, self-contained, reproducible code.
>>
>> [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.et
[R] Converting a list to a data frame
I suspect this is pretty easy, but I'm having trouble figuring it out.
Basically, I have a list of data frames such as the following example:
list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
I would like to turn this into data frame where the list elements are
essentially rbind'ed together and the element name becomes a new
variable. For example, I would like to turn the list above into a data
frame that looks like this:
data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))
Appreciate any pointers.
Kevin
--
Kevin E. Thorpe
Head of Biostatistics, Applied Health Research Centre (AHRC)
Li Ka Shing Knowledge Institute of St. Michael's Hospital
Assistant Professor, Dalla Lana School of Public Health
University of Toronto
email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Converting a list to a data frame
Hi Kevin,
There is probably a better way, but it can be done in two steps like this
temp <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
temp <- lapply(names(temp), function(n, temp) {
temp[[n]]$type <- n
return(temp[[n]])
}, temp = temp)
do.call(rbind, temp)
On Wed, May 2, 2018 at 1:11 PM, Kevin E. Thorpe
wrote:
> I suspect this is pretty easy, but I'm having trouble figuring it out.
> Basically, I have a list of data frames such as the following example:
>
> list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
>
> I would like to turn this into data frame where the list elements are
> essentially rbind'ed together and the element name becomes a new variable.
> For example, I would like to turn the list above into a data frame that
> looks like this:
>
> data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))
>
> Appreciate any pointers.
>
> Kevin
>
> --
> Kevin E. Thorpe
> Head of Biostatistics, Applied Health Research Centre (AHRC)
> Li Ka Shing Knowledge Institute of St. Michael's Hospital
> Assistant Professor, Dalla Lana School of Public Health
> University of Toronto
> email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posti
> ng-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Converting a list to a data frame
> x1 <- do.call(rbind, c(x, list(make.row.names=FALSE)))
> x2 <- cbind(type=rep(names(x), vapply(x, nrow, 0)), x1)
> str(x2)
'data.frame': 4 obs. of 3 variables:
$ type: Factor w/ 2 levels "A","B": 1 1 2 2
$ x : int 1 2 5 6
$ y : int 3 4 7 8
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, May 2, 2018 at 10:11 AM, Kevin E. Thorpe
wrote:
> I suspect this is pretty easy, but I'm having trouble figuring it out.
> Basically, I have a list of data frames such as the following example:
>
> list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
>
> I would like to turn this into data frame where the list elements are
> essentially rbind'ed together and the element name becomes a new variable.
> For example, I would like to turn the list above into a data frame that
> looks like this:
>
> data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))
>
> Appreciate any pointers.
>
> Kevin
>
> --
> Kevin E. Thorpe
> Head of Biostatistics, Applied Health Research Centre (AHRC)
> Li Ka Shing Knowledge Institute of St. Michael's Hospital
> Assistant Professor, Dalla Lana School of Public Health
> University of Toronto
> email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posti
> ng-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Converting a list to a data frame
On Wed, 2 May 2018, Kevin E. Thorpe wrote:
I suspect this is pretty easy, but I'm having trouble figuring it out.
Basically, I have a list of data frames such as the following example:
list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
I would like to turn this into data frame where the list elements are
essentially rbind'ed together and the element name becomes a new variable.
For example, I would like to turn the list above into a data frame that looks
like this:
data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))
Appreciate any pointers.
Kevin
Hi, Kevin.
Here's code that will generate your desired data frame.
# List as provided
thelist <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
thelist
# Creating the type-vector
type <- c()
for(i in 1:length(thelist)){
type <- c(type, rep(names(thelist)[i], sapply(thelist, nrow)[i]))
}
# Creating the data frame
df <- data.frame(type, do.call(rbind.data.frame, c(thelist, make.row.names
= FALSE)))
df
Kind regards,
Eivind K. Dovik
Bergen, NO
--
Kevin E. Thorpe
Head of Biostatistics, Applied Health Research Centre (AHRC)
Li Ka Shing Knowledge Institute of St. Michael's Hospital
Assistant Professor, Dalla Lana School of Public Health
University of Toronto
email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016
__
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Re: [R] Converting a list to a data frame
Another approach:
library(tidyr)
L <- list( A = data.frame( x=1:2, y=3:4 )
, B = data.frame( x=5:6, y=7:8 )
)
D <- data.frame( Type = names( L )
, stringsAsFactors = FALSE
)
D$data <- L
unnest(D, data)
#> Type x y
#> 1A 1 3
#> 2A 2 4
#> 3B 5 7
#> 4B 6 8
On Wed, 2 May 2018, Eivind K. Dovik wrote:
On Wed, 2 May 2018, Kevin E. Thorpe wrote:
I suspect this is pretty easy, but I'm having trouble figuring it out.
Basically, I have a list of data frames such as the following example:
list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
I would like to turn this into data frame where the list elements are
essentially rbind'ed together and the element name becomes a new variable.
For example, I would like to turn the list above into a data frame that
looks like this:
data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))
Appreciate any pointers.
Kevin
Hi, Kevin.
Here's code that will generate your desired data frame.
# List as provided
thelist <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
thelist
# Creating the type-vector
type <- c()
for(i in 1:length(thelist)){
type <- c(type, rep(names(thelist)[i], sapply(thelist, nrow)[i]))
}
# Creating the data frame
df <- data.frame(type, do.call(rbind.data.frame, c(thelist, make.row.names =
FALSE)))
df
Kind regards,
Eivind K. Dovik
Bergen, NO
--
Kevin E. Thorpe
Head of Biostatistics, Applied Health Research Centre (AHRC)
Li Ka Shing Knowledge Institute of St. Michael's Hospital
Assistant Professor, Dalla Lana School of Public Health
University of Toronto
email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016
__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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[R] using apply
Hi
I have 3 dataframes, a,b,c with 0/1 values...i have to check a condition
for dataframe a and b and then input the rows ids to datframe c . In the if
condition, I AND the 2 rows of from a and b and then see if the result is
equal to one of them.
I have done this using a for loop, however, it takes a long time to execute
with larger dataset..Can you help me do it using apply function so that i
can do it faster?
a
V1.x V2.x V3.x V1.y V2.y V3.y
1110101
2101101
3111101
4000101
b
V1 V2 V3 V4 V5 V6
1 1 0 1 1 0 0
2 1 0 1 0 0 0
c
xy
1 21
2 22
3 31
4 32
for(i in 1:nrow(a)){
for(j in 1:nrow(b)){
if(all((a[i,]&b[j,])==b[j,]))
{ c[nrow(c)+1, ]<-c(paste(i,j)
}
}
}
Thanks,
Neha
[[alternative HTML version deleted]]
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Re: [R] Converting a list to a data frame
On 05/02/2018 07:11 PM, Kevin E. Thorpe wrote:
I suspect this is pretty easy, but I'm having trouble figuring it out.
Basically, I have a list of data frames such as the following example:
list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
I would like to turn this into data frame where the list elements are
essentially rbind'ed together and the element name becomes a new
variable. For example, I would like to turn the list above into a data
frame that looks like this:
data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))
Appreciate any pointers.
Hi Kevin,
data.table::rbindlist does exactly what you want in a very efficient way:
library(data.table)
dat <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
rbindlist(dat, idcol = "type")
Regards,
Denes
Kevin
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Re: [R] using apply
Hi Neha,
Perhaps merge() from base or join from dplyr is what you are looking for.
data. table could also be interesting.
Hth
Ulrik
On Wed, 2 May 2018, 21:28 Neha Aggarwal, wrote:
> Hi
>
> I have 3 dataframes, a,b,c with 0/1 values...i have to check a condition
> for dataframe a and b and then input the rows ids to datframe c . In the if
> condition, I AND the 2 rows of from a and b and then see if the result is
> equal to one of them.
> I have done this using a for loop, however, it takes a long time to execute
> with larger dataset..Can you help me do it using apply function so that i
> can do it faster?
>
> a
> V1.x V2.x V3.x V1.y V2.y V3.y
> 1110101
> 2101101
> 3111101
> 4000101
>
> b
> V1 V2 V3 V4 V5 V6
> 1 1 0 1 1 0 0
> 2 1 0 1 0 0 0
>
> c
> xy
> 1 21
> 2 22
> 3 31
> 4 32
>
> for(i in 1:nrow(a)){
> for(j in 1:nrow(b)){
> if(all((a[i,]&b[j,])==b[j,]))
> { c[nrow(c)+1, ]<-c(paste(i,j)
> }
> }
> }
>
>
> Thanks,
> Neha
>
> [[alternative HTML version deleted]]
>
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> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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Re: [R] Converting a list to a data frame
Or add the type column first and then rbind:
x <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
x2 <- do.call(rbind, lapply(names(x), function(z)
data.frame(type=z, dat[[z]])))
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77843-4352
-Original Message-
From: R-help On Behalf Of William Dunlap via
R-help
Sent: Wednesday, May 2, 2018 12:28 PM
To: Kevin E. Thorpe
Cc: R Help Mailing List
Subject: Re: [R] Converting a list to a data frame
> x1 <- do.call(rbind, c(x, list(make.row.names=FALSE)))
> x2 <- cbind(type=rep(names(x), vapply(x, nrow, 0)), x1)
> str(x2)
'data.frame': 4 obs. of 3 variables:
$ type: Factor w/ 2 levels "A","B": 1 1 2 2
$ x : int 1 2 5 6
$ y : int 3 4 7 8
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, May 2, 2018 at 10:11 AM, Kevin E. Thorpe
wrote:
> I suspect this is pretty easy, but I'm having trouble figuring it out.
> Basically, I have a list of data frames such as the following example:
>
> list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
>
> I would like to turn this into data frame where the list elements are
> essentially rbind'ed together and the element name becomes a new variable.
> For example, I would like to turn the list above into a data frame
> that looks like this:
>
> data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))
>
> Appreciate any pointers.
>
> Kevin
>
> --
> Kevin E. Thorpe
> Head of Biostatistics, Applied Health Research Centre (AHRC) Li Ka
> Shing Knowledge Institute of St. Michael's Hospital Assistant
> Professor, Dalla Lana School of Public Health University of Toronto
> email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posti
> ng-guide.html and provide commented, minimal, self-contained,
> reproducible code.
>
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Re: [R] using apply
This might be faster. It uses apply() which is just another way of constructing
a loop so it is not necessarily faster. We can vectorize the logical
comparisons, but all() is not vectorized. Use dput() to share your data. That
makes it easier to replicate your example:
a <- structure(list(V1. = c(1, 1, 1, 0), V2.x = c(1, 0, 1, 0), V3.x = c(0,
1, 1, 0), V1.y = c(1, 1, 1, 1), V2.y = c(0, 0, 0, 0), V3.y = c(1, 1,
1, 1)), .Names = c("V1.x", "V2.x", "V3.x", "V1.y", "V2.y", "V3.y"),
row.names = c(NA, -4L), class = "data.frame")
b <- structure(list(V1 = c(1, 1), V2 = c(0, 0), V3 = c(1, 1), V4 = c(1,
0), V5 = c(0, 0), V6 = c(0, 0)), .Names = c("V1", "V2", "V3",
"V4", "V5", "V6"), row.names = c(NA, -2L), class = "data.frame")
# Generate the row indices that we need so we can vectorize the logical
operation:
idxa <- rep(1:4, each=2)
idxb <- rep(1:2, 4)
ab <- (a[idxa, ] & b[idxb, ]) == b[idxb, ]
c <- cbind(idxa, idxb)[apply(ab, 1, all), ]
c
# idxa idxb
# [1,]21
# [2,]22
# [3,]31
# [4,]32
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77843-4352
-Original Message-
From: R-help On Behalf Of Ulrik Stervbo
Sent: Wednesday, May 2, 2018 3:49 PM
To: Neha Aggarwal
Cc: r-help@r-project.org
Subject: Re: [R] using apply
Hi Neha,
Perhaps merge() from base or join from dplyr is what you are looking for.
data. table could also be interesting.
Hth
Ulrik
On Wed, 2 May 2018, 21:28 Neha Aggarwal, wrote:
> Hi
>
> I have 3 dataframes, a,b,c with 0/1 values...i have to check a
> condition for dataframe a and b and then input the rows ids to
> datframe c . In the if condition, I AND the 2 rows of from a and b and
> then see if the result is equal to one of them.
> I have done this using a for loop, however, it takes a long time to
> execute with larger dataset..Can you help me do it using apply
> function so that i can do it faster?
>
> a
> V1.x V2.x V3.x V1.y V2.y V3.y
> 1110101
> 2101101
> 3111101
> 4000101
>
> b
> V1 V2 V3 V4 V5 V6
> 1 1 0 1 1 0 0
> 2 1 0 1 0 0 0
>
> c
> xy
> 1 21
> 2 22
> 3 31
> 4 32
>
> for(i in 1:nrow(a)){
> for(j in 1:nrow(b)){
> if(all((a[i,]&b[j,])==b[j,]))
> { c[nrow(c)+1, ]<-c(paste(i,j)
> }
> }
> }
>
>
> Thanks,
> Neha
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
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Re: [R] Converting a list to a data frame
Typo: dat[[z]] should be x[[z]]:
x2 <- do.call(rbind, lapply(names(x), function(z)
data.frame(type=z, x[[z]])))
x2
type x y
1A 1 3
2A 2 4
3B 5 7
4B 6 8
David C
-Original Message-
From: R-help On Behalf Of David L Carlson
Sent: Wednesday, May 2, 2018 3:51 PM
To: William Dunlap ; Kevin E. Thorpe
Cc: r-help mailing list
Subject: Re: [R] Converting a list to a data frame
Or add the type column first and then rbind:
x <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
x2 <- do.call(rbind, lapply(names(x), function(z)
data.frame(type=z, dat[[z]])))
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77843-4352
-Original Message-
From: R-help On Behalf Of William Dunlap via
R-help
Sent: Wednesday, May 2, 2018 12:28 PM
To: Kevin E. Thorpe
Cc: R Help Mailing List
Subject: Re: [R] Converting a list to a data frame
> x1 <- do.call(rbind, c(x, list(make.row.names=FALSE)))
> x2 <- cbind(type=rep(names(x), vapply(x, nrow, 0)), x1)
> str(x2)
'data.frame': 4 obs. of 3 variables:
$ type: Factor w/ 2 levels "A","B": 1 1 2 2
$ x : int 1 2 5 6
$ y : int 3 4 7 8
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, May 2, 2018 at 10:11 AM, Kevin E. Thorpe
wrote:
> I suspect this is pretty easy, but I'm having trouble figuring it out.
> Basically, I have a list of data frames such as the following example:
>
> list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
>
> I would like to turn this into data frame where the list elements are
> essentially rbind'ed together and the element name becomes a new variable.
> For example, I would like to turn the list above into a data frame
> that looks like this:
>
> data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))
>
> Appreciate any pointers.
>
> Kevin
>
> --
> Kevin E. Thorpe
> Head of Biostatistics, Applied Health Research Centre (AHRC) Li Ka
> Shing Knowledge Institute of St. Michael's Hospital Assistant
> Professor, Dalla Lana School of Public Health University of Toronto
> email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posti
> ng-guide.html and provide commented, minimal, self-contained,
> reproducible code.
>
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[R] Calling the curve function with a character object converted into an expression
Hi,
Down a cascade of function calls, I want to use the curve function with an
expression that is a variable. For various reason, this variable must be a
character object and cannot be an expression as required by the curve function.
How do I convert my variable into a expression that is accepted by curve?
Thanks in advance for your help.
## The following attempts do not work
myf <- '1+x^2'
curve(myf, from = 0, to = 10) # obviously !
curve(parse(text=myf), from = 0, to = 10) # not sure why this does not work
## This works but does not feel elegant:
eval(parse(text=sprintf('curve(%s, from = 0, to = 10)', myf)))
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[R] The stages of standard function evaluation
Dear R Help folks --
I have been trying to put together a list of the steps or stages of R
function evaluation, with particular focus on those that have "standard" or
"nonstandard" forms. This is both for my own edification and also because I
am thinking of joining the world of R bloggers and have been trying to put
together some draft posting that might be useful. I seem to have an
affirmative genius for finding incorrect interpretations of R's evaluation
rules; I'm trying to make that an asset.
I am hoping that you can tell me:
1. Is this list complete, or are there additional stages I am missing?
2. Have I inserted one or more imaginary stages?
3. Are the terms I use below to name each stage appropriate, or are
there other terms more widely used or recognizable?
4. Is the order correct?
I begin each name with “Standard,” to express my belief that each of these
things has a usual or default form, but also that (unless I am mistaken)
almost none of them exist only in a single form true of all R functions. (I
have marked with an asterisk a few evaluation steps that I think may always
be followed).
It is my ultimate goal (which I do not feel at all close to accomplishing)
to determine a way to mechanically test for “standardness” along each of
these dimensions, so that each function could be assigned a logical vector
showing the ways that it is and is not standard. One thing I think is
conceptually or procedurally difficult about this project is that I think
“standardness” should be determined by what a function does, rather than by
how it does it, so that a primitive function that takes unevaluated
arguments positionally could still have standard matching, scoping, etc.,
by internal emulation. A related goal is to identify which evaluation steps
most often use an alternative form, and perhaps determine if there is more
than one such alternative. Finally, an easier short-term goal is simply to
find instances of one or more function with standard and non-standard
evaluation for each evaluation step.
For the most part below I am treating the evaluation of closures as the
standard from which “nonstandard” is defined. However, I do not assume that
other kinds of functions are automatically nonstandard on any particular
dimension below. Most of this comes from the R Language Definition, but
there are numerous places where I am by no means certain that my
interpretation is correct. I have highlighted some of these below with a
“??”.
I look forward to learning from you.
Warmest regards,
J. Andrew Hoerner
** Standard function recognition:* recognizing some or all of a string code
as a function. (Part of code line parsing)
*Standard environment construction:* construction of the execution
environment, and of pointers to the calling and enclosing environments.
*Standard function identification:* Get the name of the function, if any
** Standard f**unction scoping*: Search the current environment and then up
the chain of enclosing environments until you find the first binding
environment, an environment where the name of the function is bound, i.e.
linked to a pointer to the function definition. The binding environment is
usually (but not always) the same as the defining environment (i.e. the
enclosing environment when the function is defined. Note that function
lookup, unlike function argument lookup, necessarily starts from the
calling environment, because a function does not know what it is – its
formals, body, and environments – until it is found. Named functions are
always found by scoping. R never learns "where" they are -- they have to be
looked up each time. For this reason, anonymous functions must be used in
place, and called by a function that takes a function as an argument, or by
(function(formals){body})(actual args)
*Standard f**unction **retrieval**:*
load (??) the function, i.e. transfer the list (??) of formals and defaults
and the list (??) of expressions that constitute the function body into the
execution environment
Note that the function body is parsed at the time the function is created
(true?? Or is it parsed every time the function is called?)
*Standard argument matching*:
assignment of expressions and default arguments to formals via the
usually-stated matching rulesIf matched positionall at call time, the name
is scoped like an actual argument.. Note that giving an argument the same
name as a formal when calling the function will only match it to that
formal if matched positionally or by tag, not by name.
*Standard a**rgument parsing:*
Converts argument expressions into abstract syntax trees. “Standard”
argument parsing and promise construction take place before the arguments
are passed into the body.
*Standard p**romise construction*:
Assigning each name (including function names) in an AST to its binding the
calling environment (if ordinary) or the execution environment (if default)
(Am I right that the action here for calls and for names is essenti
Re: [R] Calling the curve function with a character object converted into an expression
Sebastian:
This is somewhat arcane, perhaps even a bug (correction on this
welcomed). The problem is that the "expr" argument to curve() must be
an actual expression, not a call to parse that evaluates to an
expression. If you look at the code of curve() you'll see why
(substitute() does not evaluate expr in the code). Another simple
workaround other than sticking in the eval() call is this:
myf <- function(x)NUL
body(myf)<- parse(text = "{1+x^2}") ## note the additional "{ }"
for safety, though not necessary here
## this idiom will continue to work for any text.
curve(myf, from = 0, to = 10) ## myf is now the name of a function
that executes the parsed text.
*** I would appreciate any wiser R programmers correcting any
misunderstanding or error in my explanation ***
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, May 2, 2018 at 8:11 PM, Sebastien Bihorel
wrote:
>
> Hi,
>
> Down a cascade of function calls, I want to use the curve function with an
> expression that is a variable. For various reason, this variable must be a
> character object and cannot be an expression as required by the curve
> function. How do I convert my variable into a expression that is accepted by
> curve?
>
> Thanks in advance for your help.
>
> ## The following attempts do not work
>
> myf <- '1+x^2'
> curve(myf, from = 0, to = 10) # obviously !
> curve(parse(text=myf), from = 0, to = 10) # not sure why this does not work
>
> ## This works but does not feel elegant:
> eval(parse(text=sprintf('curve(%s, from = 0, to = 10)', myf)))
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
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Re: [R] Converting a list to a data frame
This is very nice to learn about, Denis, but it seems only fair to point out
that the result of rbindlist is not a data frame. You can convert it to a data
frame easily, but the copy and indexing semantics of data tables are quite
different than data tables, which could be a real headache for someone not
prepared for those differences. (To learn more, read the data tables vignette.)
Tibbles (as produced by unnest in my previous response) are not data tables
either, but they behave much more like data frames than data tables do.
On May 2, 2018 1:30:37 PM MDT, "Tóth Dénes" wrote:
>
>
>On 05/02/2018 07:11 PM, Kevin E. Thorpe wrote:
>> I suspect this is pretty easy, but I'm having trouble figuring it
>out.
>> Basically, I have a list of data frames such as the following
>example:
>>
>> list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
>>
>> I would like to turn this into data frame where the list elements
>are
>> essentially rbind'ed together and the element name becomes a new
>> variable. For example, I would like to turn the list above into a
>data
>> frame that looks like this:
>>
>> data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))
>>
>> Appreciate any pointers.
>
>Hi Kevin,
>
>data.table::rbindlist does exactly what you want in a very efficient
>way:
>
>library(data.table)
>dat <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
>rbindlist(dat, idcol = "type")
>
>Regards,
>Denes
>
>
>>
>> Kevin
>>
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
--
Sent from my phone. Please excuse my brevity.
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Re: [R] Calling the curve function with a character object converted into an expression
Typo: should be NULL not NUL of course
An alternative approach closer to your original attempt is to use
do.call() to explicitly evaluate the expr argument:
w <- "1 + x^2"
do.call(curve, list(expr = parse(text = w), ylab ="y"))
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, May 2, 2018 at 10:35 PM, Bert Gunter wrote:
> Sebastian:
>
> This is somewhat arcane, perhaps even a bug (correction on this
> welcomed). The problem is that the "expr" argument to curve() must be
> an actual expression, not a call to parse that evaluates to an
> expression. If you look at the code of curve() you'll see why
> (substitute() does not evaluate expr in the code). Another simple
> workaround other than sticking in the eval() call is this:
>
> myf <- function(x)NUL
> body(myf)<- parse(text = "{1+x^2}") ## note the additional "{ }"
> for safety, though not necessary here
> ## this idiom will continue to work for any text.
>
> curve(myf, from = 0, to = 10) ## myf is now the name of a function
> that executes the parsed text.
>
> *** I would appreciate any wiser R programmers correcting any
> misunderstanding or error in my explanation ***
>
> Cheers,
> Bert
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Wed, May 2, 2018 at 8:11 PM, Sebastien Bihorel
> wrote:
>>
>> Hi,
>>
>> Down a cascade of function calls, I want to use the curve function with an
>> expression that is a variable. For various reason, this variable must be a
>> character object and cannot be an expression as required by the curve
>> function. How do I convert my variable into a expression that is accepted by
>> curve?
>>
>> Thanks in advance for your help.
>>
>> ## The following attempts do not work
>>
>> myf <- '1+x^2'
>> curve(myf, from = 0, to = 10) # obviously !
>> curve(parse(text=myf), from = 0, to = 10) # not sure why this does not work
>>
>> ## This works but does not feel elegant:
>> eval(parse(text=sprintf('curve(%s, from = 0, to = 10)', myf)))
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
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