Hi Andrew,
thanks a lot, that fully explains it.
Sorry for the HTML text. For the record I put the original code again
below.
Best regards
Hilmar
On 27.10.22 18:34, Andrew Simmons wrote:
> $ does not evaluate its second argument, it does something like
> as.character(substitute(name)).
>
> Yo
Your message is garbled. Please send plain text to the mailing list.
On October 27, 2022 2:31:47 AM PDT, Hilmar Berger wrote:
>Dear all,
>
>I'm a little bit surprised by the behavior of the $ operator when used
>in lapply - any indication what might be wrong is appreciated.
>
>> xx = list(A=list(
$ does not evaluate its second argument, it does something like
as.character(substitute(name)).
You should be using
lapply(list, function(x) x$a)
or
lapply(list, `[[`, "a")
On Thu, Oct 27, 2022, 12:29 Hilmar Berger wrote:
> Dear all,
>
> I'm a little bit surprised by the behavior of the $ o
Thans Richard, it works well. --- Kai
On Monday, December 13, 2021, 04:00:33 AM PST, Richard O'Keefe
wrote:
You want to DELETE rows satisfying the condition P & Q.The subset() function
requires an expression saying whatyou want to RETAIN, so you need subset(PD,
!(P & Q)).
test <- subs
You want to DELETE rows satisfying the condition P & Q.
The subset() function requires an expression saying what
you want to RETAIN, so you need subset(PD, !(P & Q)).
test <- subset(PD, !(Class == "1st" & Survived == "No"))
By de Morgan's laws, !(P & Q) is the same as (!P) | (!Q)
so you could als
Use one ampersand, not two.
And post plain text.
On December 12, 2021 8:30:11 PM PST, Kai Yang via R-help
wrote:
>Hi R team,I want to delete records from a data frame if Class = '1st' and
>Survived = 'No'. I wrote the code below, test <- subset(PD, Class != '1st' &&
>Survived != 'No')
>but th
I assume that prim, etc. are columns of your data frame, mydata. Ergo, the
error message "prim not found" as 'prim' etc. does not exist in the Global
environment.
exclude <- with(mydata, prim == -9, etc. ) should get what you want to
evaluate your second subset statement if I have understood corre
Thanks. YES the second call to subset is there, trying to use my failed
definition of "exclude". Read on..
On 2021/10/16 上午 09:35, Jeff Newmiller wrote:
I don't see a "second one". Looks like you forgot the subset function call?
On October 15, 2021 6:23:56 PM PDT, Steven Yen wrote:
The follo
I don't see a "second one". Looks like you forgot the subset function call?
On October 15, 2021 6:23:56 PM PDT, Steven Yen wrote:
>The following "subset command works. I was hoping the second would as
>well but it does not.
>
>My definition of exclude is rejected.
>
>Help please? Thanks.
>
> > m
-Original Message-
> From: R-help On Behalf Of Bert Gunter
> Sent: Tuesday, January 14, 2020 8:10 AM
> To: ani jaya
> Cc: r-help
> Subject: Re: [R] Subset a data frame with specific date
>
> That's fine, but do note that the which() function is wholly unnecessary
in
Thank you Bert.
And yes another topic to study.
On Tue, Jan 14, 2020 at 4:10 PM Bert Gunter wrote:
>
> That's fine, but do note that the which() function is wholly unnecessary in
> your last line as R allows logical indexing. Perhaps another topic you need
> to study.
>
> -- Bert
>
>
>
> On Mon
That's fine, but do note that the which() function is wholly unnecessary in
your last line as R allows logical indexing. Perhaps another topic you need
to study.
-- Bert
On Mon, Jan 13, 2020 at 10:56 PM ani jaya wrote:
> Dear Jeff and Bert,
>
> Thank you very much for your correction and expl
Dear Jeff and Bert,
Thank you for your correction and explanation.
Yes, I need more study regarding date format and
sorry for HTML mail.
I was able to subset data that I want.
mjo30<-read.table("rmm.txt", header=FALSE, skip=4234, nrows=10957)
mjo30$V8<-NULL
names(mjo30)<-c("year","month","day",
Dear Jeff and Bert,
Thank you very much for your correction and explanation.
And yes, I need to study about date format more.
Sorry for HTML mail, don't realize.
I was able to subset the data that I want.
mjo30<-read.table("rmm.txt", header=FALSE, skip=4234, nrows=10957)
mjo30$V8<-NULL
names(mjo
Inline.
Bert Gunter
On Mon, Jan 13, 2020 at 8:54 PM ani jaya wrote:
> Good morning R-Help,
>
> I have a dataframe with 7 columns and 1+ rows. I want to subset/extract
> those data frame with specific date (not in order). Here the head of my
> data frame:
>
> head(mjo30)
> year month d
The dput function is for re-creating an R object in another R workspace, so it
uses fundamental base types to define objects. A Date is really the number of
days since a specific date (typically 1970-01-01) that get converted to look
like dates whenever you display or print them, so what you are
Look at the help docs and examples for textcat and sapply:
print(as.character(data$x[sapply(data$x, textcat)=="english"]))
Although textcat defaults classify "This book is amazing" as dutch, so
you may want to read the help for textcat and change the profile db
("p") or "method".
On 19/11/20
Do you also want lines 38 and 39 (in addition to 40:44), or do I
misunderstand your problem?
When you deal with runs of data, think of the rle (run-length encoding)
function. E.g. here is
a barely tested function to find runs of a given minimum length and a given
difference between
successive val
Hi Jim,
thank's it is working with the given example,
but whats the difference when using
testdata=data.frame(TIME=c("17:11:20", "17:11:21", "17:11:22",
"17:11:23", "17:11:24", "17:11:25", "17:11:26", "17:11:27", "17:11:28",
"17:21:43",
"17:22:16", "17:22:19", "18:04:48"
Bugger! It's
eval(parse(text=paste0("kkdf[c(",paste(starts,ends,sep=":",collapse=","),"),]")))
What a mess!
Jim
On Fri, Sep 28, 2018 at 8:35 AM Jim Lemon wrote:
>
> Hi Knut,
> As Bert said, you can start with diff and work from there. I can
> easily get the text for the subset, but despite fool
Hi Knut,
As Bert said, you can start with diff and work from there. I can
easily get the text for the subset, but despite fooling around with
"parse", "eval" and "expression", I couldn't get it to work:
# use a bigger subset to test whether multiple runs can be extracted
kkdf<-subset(airquality,Te
1. I assume the values are integers, not floats/numerics (which woud make
it more complicated).
2. Strategy: Take differences (e.g. see ?diff) and look for >3 1's in a
row.
I don't have time to work out details, but perhaps that helps.
Cheers,
Bert
Bert Gunter
"The trouble with having an open
Dear David,
Subsetting works but the 'date' information is lost in the new file.
Thanks, Mike. I was not aware of the bug but will work on learning
about (getZ) and (setZ). Thanks again!
Sincerely,
Milu
On Tue, Jun 19, 2018 at 7:32 AM, Michael Sumner wrote:
>
>
> On Mon, 18 Jun 2018, 22:09 D
On Mon, 18 Jun 2018, 22:09 David Winsemius, wrote:
>
>
> > On Jun 18, 2018, at 7:21 AM, Miluji Sb wrote:
> >
> > Dear all,
> >
> > I have a rasterbrick with the date/time information provided which I
> would
> > like to subset by year.
> >
> > However, when I use the following code for sub-setti
> On Jun 18, 2018, at 7:21 AM, Miluji Sb wrote:
>
> Dear all,
>
> I have a rasterbrick with the date/time information provided which I would
> like to subset by year.
>
> However, when I use the following code for sub-setting;
>
> new_brick <- subset(original, which(getZ( original ) >= as.D
You realize, do you not, that in fact there are no numbers in your "list"
(actually a vector).
It looks like you would do well to spend some time with an R tutorial or
two before posting further to this list. We can help, but cannot substitute
for the basic knowledge that you would gain from doin
Super, thanks Boris. Top notch :-)
On Mon, Sep 25, 2017 at 1:05 PM, Boris Steipe
wrote:
> Always via logical expressions. In this case you can use the logical
> expression
>
> myDF$b != "0"
>
> to give you a vector of TRUE/FALSE
>
>
>
> B.
>
>
> > On Sep 25, 2017, at 8:00 AM, Shane Carey wrote
Always via logical expressions. In this case you can use the logical expression
myDF$b != "0"
to give you a vector of TRUE/FALSE
B.
> On Sep 25, 2017, at 8:00 AM, Shane Carey wrote:
>
> This is super, really helpfull. Sorry, one final question, lets say I wanted
> to remove 0's rather t
This is super, really helpfull. Sorry, one final question, lets say I
wanted to remove 0's rather than NAs , what would it be?
Thanks
On Mon, Sep 25, 2017 at 12:41 PM, Boris Steipe
wrote:
> myDF <- data.frame(a = c("<0.1", NA, 0.3, 5, "Nil"),
>b = c("<0.1", 1, 0.3, 5, "Nil")
myDF <- data.frame(a = c("<0.1", NA, 0.3, 5, "Nil"),
b = c("<0.1", 1, 0.3, 5, "Nil"),
stringsAsFactors = FALSE)
# you can subset the b-column in several ways
myDF[ , 2]
myDF[ , "b"]
myDF$b
# using the column, you make a logical vector
! is.na(as.numeric(myDF
Hi,
Lets say this was a dataframe where I had two columns
a <- c("<0.1", NA, 0.3, 5, "Nil")
b <- c("<0.1", 1, 0.3, 5, "Nil")
And I just want to remove the rows from the dataframe where there were NAs
in the b column, what is the syntax for doing that?
Thanks in advance
On Fri, Sep 22, 2017 at
Super,
Thanks
On Fri, Sep 22, 2017 at 4:57 PM, Boris Steipe
wrote:
> > a <- c("<0.1", NA, 0.3, 5, "Nil")
> > a
> [1] "<0.1" NA "0.3" "5""Nil"
>
> > b <- as.numeric(a)
> Warning message:
> NAs introduced by coercion
> > b
> [1] NA NA 0.3 5.0 NA
>
> > b[! is.na(b)]
> [1] 0.3 5.0
>
>
>
> a <- c("<0.1", NA, 0.3, 5, "Nil")
> a
[1] "<0.1" NA "0.3" "5""Nil"
> b <- as.numeric(a)
Warning message:
NAs introduced by coercion
> b
[1] NA NA 0.3 5.0 NA
> b[! is.na(b)]
[1] 0.3 5.0
B.
> On Sep 22, 2017, at 11:48 AM, Shane Carey wrote:
>
> Hi,
>
> How do I extract just n
Hi Elise,
One of the quirks of POSIXt time values is that they are lists. This
should give you the plot:
plot(Soil_Temp~as.numeric(DateTime),eldf,xaxt="n",xlab="DateTime")
and this the x axis:
axis.POSIXct(1,eldf$DateTime)
If you want a different format for the date values on the axis, look
at
Hi Elise,
If I create a CSV file like your example and read it into a data frame:
eldf<-read.csv("el.csv")
Then convert the first field to POSIXt dates:
eldf$DateTime<-strptime(eldf$DateTime,"%Y-%m-%d %H:%M:%S")
class(eldf$DateTime)
[1] "POSIXlt" "POSIXt"
I can subset the file like this:
time_
Hi Elise,.
I would ask:
class(data$DateTime)
and see if it returns:
"POSIXct" "POSIXt"
Jim
On Sat, Jan 21, 2017 at 3:02 AM, Elise LIKILIKI
wrote:
> Hello,
>
> I have a dataset containing Date Time, Air Temperature, PPFD, Sol
> Temperature...
> The first data are false so I would like to extr
How are we supposed to help you if you don’t read the Posting Guide and don’t
provide any information about the classes of columns in `data`.?
— David.
> On Jan 20, 2017, at 10:02 AM, Elise LIKILIKI wrote:
>
> Hello,
>
> I have a dataset containing Date Time, Air Temperature, PPFD, Sol
> Te
Ashta,
## I may have misunderstood your question and if so I apologize.
## I had to remove the extra line after "45" before
## the ",sep=" to use your code.
## You could have used dput(dat) to send a more reliable (robust) version.
dat <- structure(list(ID = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L,
For the data you provide, it's simply:
summary(subset(dat, x1 == "x" & x2 == "z")$y)
Note that x1 and x2 are factors in your example.
We also don't know what you want to do if there are more than one
combination of that per ID, or if there ID values with no matching
rows.
Sarah
On Fri, Oct 14,
>If you want to drop levels, use droplevels() either on the factor or on
the >subset of your data frame. Example:
>droplevels(f[1]) #One element, only one level
Calling factor() on a factor, as the OP did, also drops any unused levels,
as the examples showed.
> str(factor(factor(letters)[11:13]))
> You did not change df$quant - you made a new object called 'subdf'
> containing a column called 'quant' that had only one level. Changing subdf
> has
> no effect on df.
Also, subsetting a factor _intentionally_ does not change the number of levels.
Example:
f <- factor(sample(letters[1:3], 30
You did not change df$quant - you made a new object called 'subdf'
containing a column called 'quant' that had only one level. Changing
subdf has no effect on df.
> df <- data.frame(quant=factor(letters))
> str(df)
'data.frame': 26 obs. of 1 variable:
$ quant: Factor w/ 26 levels "a","b","c",
Hello,
Don't use subset, use indexing.
subdf <- df[df$quant %in% "VeryFast", ]
By the way, instead of %in% you can use ==, since you're interested in
just one value of quant.
Hope this helps,
Rui Barradas
Citando ch.elahe via R-help :
> Hi all,
> I have the following df and I want to know
This is quite a different question. I suggest you start a new post with a
new subject line for this. And I suggest you include code for an example
plot that you want to use.
Otherwise, you might look here for some ideas on how to control colors in a
scatter plot using base r, http://stackoverflo
Thanks Jean, Does anyone know how to set these [hast1] and [hast2] as the
colors of a plot?
On Friday, April 22, 2016 7:39 AM, "Adams, Jean" wrote:
You can use the grepl() function to give you logicals for each criterion, then
combine them as needed. For example:
# example version of Co
You can use the grepl() function to give you logicals for each criterion,
then combine them as needed. For example:
# example version of Command
Command <- paste0("_localize_", c("PD","t2","t1_seq", "abc", "xyz",
"PD_t1"))
hasPD <- grepl("PD", Command, fixed=TRUE)
hast1 <- grepl("t1", Command, f
You may investigate a solution based on regular expressions.
Some tutorials to help:
http://www.regular-expressions.info/rlanguage.html
http://www.endmemo.com/program/R/grep.php
http://biostat.mc.vanderbilt.edu/wiki/pub/Main/SvetlanaEdenRFiles/regExprTalk.pdf
https://rstudio-pubs-static.s3.ama
R's subscripting operators do not "guess" the value of a missing
argument: a missing k'th subscript means seq_len(dim(x)[k]).
I bet that you use syntax like x[,1] (the entire first column of x)
all the time and that you don't want this syntax to go away.
Some languages use a placeholder like '.' o
Thanks Bill,
This is more clear.
In any case, I find very inappropriate that a programming language tries to
guess the value of a missing argument. It is unfair towards code developers
and it promotes the production of bugged piece of software.
I hope R will revise its policies sooner or later.
The "missingness" of an argument gets passed down through nested function
calls. E.g.,
fOuter <- function(x) c(outerMissing=missing(x), innerMissing=fInner(x))
fInner <- function(x) missing(x)
fInner()
#[1] TRUE
fOuter()
#outerMissing innerMissing
# TRUE TRUE
It is only
Hi Petr,
Thank you for your answer.
I'm not sure how the empty index reflects what I'm showing in my example.
If my function was
emptySubset <- function(vec) vec[]
I would then agree that this was the case. But I think it's different: I'm
specifically telling my function that it should have two
Hi
Help page for ?"[" says
An empty index selects all values: this is most often used to replace all the
entries but keep the attributes.
and actually you function construction works with empty index
> x<-c(1,2,5)
> letters[x]
[1] "a" "b" "e"
> letters[]
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i
t;V046", "V047", "V048", "V049", "V050", "V051", "V052",
"V053", "V054", "V055", "V056", "V057"), class = "factor"), linecode =
structure(c(1L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
m1 <- c()
> x <- length(plot.id)
> for (i in (1:x)) {
> m1 <- as.numeric(strsplit(as.character(dataset$ranges2use),
> ",")[[i]])
> }
> m2
> }
>
> I am not sure where I am making a mistake.
> Thanks.
> Nilesh
&g
lit(as.character(dataset$ranges2use),
",")[[i]])
}
m2
}
I am not sure where I am making a mistake.
Thanks.
Nilesh
-Original Message-
From: Michael Dewey [mailto:li...@dewey.myzen.co.uk]
Sent: Monday, November 23, 2015 12:11 PM
To: DIGHE, NILESH [AG/2362]; r-help@r-pro
er 23, 2015 10:17 AM
To: DIGHE, NILESH [AG/2362]; r-help@r-project.org
Subject: Re: [R] subset data using a vector
length(strsplit(as.character(mydata$ranges2use), ","))
was that what you expected? I think not.
On 23/11/2015 16:05, DIGHE, NILESH [AG/2362] wrote:
Dear R users,
NILESH [AG/2362]; r-help@r-project.org
Subject: Re: [R] subset data using a vector
length(strsplit(as.character(mydata$ranges2use), ","))
was that what you expected? I think not.
On 23/11/2015 16:05, DIGHE, NILESH [AG/2362] wrote:
> Dear R users,
> I like to split
length(strsplit(as.character(mydata$ranges2use), ","))
was that what you expected? I think not.
On 23/11/2015 16:05, DIGHE, NILESH [AG/2362] wrote:
Dear R users,
I like to split my data by a vector created by using variable "ranges".
This vector will have the current range (r
On Tue, 30 Jun 2015, Rolf Turner wrote:
If you want a pointer to the correct syntax for subset(), try
help("subset")!!!
The syntax of your "extstream" function is totally screwed up, convoluted and
over-complicated. Note that even if you had your "subset" argument specified
correctly, the re
If you want a pointer to the correct syntax for subset(), try
help("subset")!!!
The syntax of your "extstream" function is totally screwed up,
convoluted and over-complicated. Note that even if you had your "subset"
argument specified correctly, the return() call will give you only the
resu
On Mon, 29 Jun 2015, David Winsemius wrote:
No. A pointer to the correct use of "[" is needed.
Thanks, David. This puts me on the the right path.
Much appreciated,
Rich
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https:/
On Tue, 30 Jun 2015, Steve Taylor wrote:
Using return() within a for loop makes no sense: only the first one will be
returned.
Steve,
Mea culpa. Didn't catch that.
How about:
alldf.B = subset(alldf, stream=='B') # etc...
I used to do each stream manually, like the above, and want to
On Jun 29, 2015, at 5:03 PM, Rich Shepard wrote:
> Moving from interactive use of R to scripts and functions and have bumped
> into what I believe is a problem with variable names. Did not see a solution
> in the two R programming books I have or from my Web searches. Inexperience
> with ess-tra
Well, your code is, ah, too incorrect to convey what you want out of this
effort. If I were to guess based on your description, you want all of the data,
not a subset. An example data frame containing what you hope to extract might
be helpful.
However, extracting subsets is rarely done for just
Using return() within a for loop makes no sense: only the first one will be
returned.
How about:
alldf.B = subset(alldf, stream=='B') # etc...
Also, have a look at unique(alldf$stream) or levels(alldf$stream) if you want
to use a for loop on each unique value.
cheers,
Steve
-Original
, 2, 2, 2, 2, 2, 2, 2, 2, 2), Wgt =
>> c(0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.004357455208
83, 0.0043574552083,
> 0.0043574552083), SPCLORatingValue = c(14L, 15L, 15L,
> 12L, 15L, 12L, 13L, 15L, 14L, 15L, 14L)), .Names = c("WgtBand",
> "Wgt", "SPCLORatingValue"), row.names = 12:22, class = "data.frame"),
> V10 = stru
2, 2, 2, 2, 2,
2, 2, 2), Wgt = c(0.0043574552083, 0.00435745520833333,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083
), SPCLORatingVal
Can you show a small self-contained example of you data and expected
results?
I tried to make one and your expression returned a single number in a 1 by
1 matrix.
library(doBy)
Generation<-list(
data.frame(Wgt=c(1,2,4), SPCLORatingValue=c(10,11,12)),
data.frame(Wgt=c(8,16), SPCLORatingValue=
On 20/05/2015 7:13 PM, Vin Cheng wrote:
> Hi,
>
> I'm trying to group rows in a dataframe with SPCLORatingValue factor >16 and
> summing the Wgt's that correspond to this condition. There are 100
> dataframes in a list.
>
> Some of the dataframes won't have any rows that have this conditio
> -Original Message-
> A consulting client has a large data set with a binary response
> (negative) and two factors (ctry and member) which have many levels, but
> many occur with very small frequencies. It is far too sparse with a model
> like
> glm(negative ~ ctry+member, family=binom
These two commands will compute the cell frequencies and then sort them:
e <- as.data.frame(xtabs(~ctry+member, Dataset))
f <- e[order(e$Freq, decreasing=TRUE),]
Then draw your subset
g <- head(f, 10)
or
g <- f[cumsum(f$Freq)/sum(f$Freq) >.8,]
Finally merge the sample with the original data a
... nd nevermind, figured it out (from the final example on the
Extract.data.frame page):
`[.MyClass` <- function(x, i, ...) {
NextMethod("[")
mostattributes(RV) <- attribute(x)
RV
}
cheers,
-m
On Wed, Nov 12, 2014 at 11:02 PM, Murat Tasan wrote:
> And as a follow-up, I impleme
And as a follow-up, I implemented a barebones as.data.frame.MyClass(...).
It works when dealing with non-subsetted data frames, but fails upon a
subset(...) call:
> as.data.frame.MyClass <- function(x, ...) as.data.frame.vector(x, ...)
This works for a single column, e.g.:
> str(data.frame(MyCla
On Sep 4, 2014, at 2:58 PM, Kuma Raj wrote:
> This post has NOT been accepted by the mailing list yet.
Well, it has now. Were you earlier posting from Nabble? (Not an efficient
strategy.)
> I would like to subset a column based on the contents of a column with
> specific character. In the samp
try this:
> x <- structure(list(date = structure(c(15765, 15766, 15767, 15768,
+ 15769, 15770, 15771, 15772, 15773, 15780, 15781, 15782, 15788,
+ 15789, 15790, 15791, 15792, 15795), class = "Date"), mon = c("Mrz",
+ "Mrz", "Mrz", "Mrz", "Mrz", "Mrz", "Mrz", "Mrz", "Mrz", "Mrz",
+ "Mrz", "Mrz", "Mr
Hi,
In your example the "wea", only showed "dw". Suppose the data is like this:
dat1 <- structure(list(date = structure(c(15765, 15766, 15767, 15768,
15769, 15770, 15771, 15772, 15773, 15780, 15781, 15782, 15788,
15789, 15790, 15791, 15792, 15795, 15796, 15797, 15798, 15799,
15800, 15801, 1
Hi,
#if 'dat` is the dataset
May be this helps.
lst1 <- setNames(split(dat, cumsum(c(TRUE,diff(dat$date)!=1))),LETTERS[1:4])
A.K.
On Tuesday, May 20, 2014 12:17 PM, Christoph Schlächter
wrote:
Dear all,
I have a subset of a data frame with 3 columns and a few rows. The columns
are “date” [%
Dear,
This is a very efficient way. I have checked some values and it is what is
was looking for.
I want to find the days called dry-weather days. It depends but in my case
it is a dry day if on this day the precipitation "N" is equal or less than
0.3 mm and if on the day before the day with N <=
nal Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of arun
> Sent: 28. april 2014 13:58
> To: R. Help
> Subject: Re: [R] subset of obersevation depending on multiple conditions
>
> Hi,
>
> The conditions are not very c
Hi
Here is a very simple way.
mydat <- read.table(text="
DateN
2010-01-010
2010-01-021.9
2010-01-030
2010-01-040
2010-01-051.6
2010-01-060
2010-01-070.3
2010-01-080
2010-01-091.1
2010-01-101.7
2010-01-112.6
2010-01-120
2010-01-130
2010-01-1
Hi,
The conditions are not very clear. Based on the rows you wanted to pick, may
be this helps,
#It is better to dput() the example.
dat <- structure(list(Date = structure(c(14610, 14611, 14612, 14613,
14614, 14615, 14616, 14617, 14618, 14619, 14620, 14621, 14622,
14623, 14624, 14625, 14
Hi Christoph,
I'm not sure I understand your conditions. It is an "AND" or an "OR",
i.e. must both conditions be met to subset or any one of them?
From your explanation, I would think you really mean AND, but then I
don't understand why you would select both lines 3 and 4.
I would just say:
d
Hi,
It is not mentioned whether your dataset is a matrix of data.frame. Also,
please use ?dput() to show the dataset. I get similar errors with matrix.
MOPrice <-
data.frame(Date=c("2013-12-31","2013-12-31","2013-12-31","2013-11-28"),stringsAsFactors=FALSE)
subset(MOPrice, as.Date(Date,"%Y-%m-
On Apr 2, 2014, at 3:35 PM, jcrosbie wrote:
> I'm getting this error: "Error in MOPrice$Date : $ operator is invalid for
> atomic vectors"
>
> The cost is: subset(MOPrice,
> as.Date(MOPrice$Date,"%Y-%m-%d")==as.Date("2013-11-28","%Y-%m-%d"))
>
> The date column looks like:
> "2013-12-31" "2013
So are the names of the columns in the dataset x, y, and z? or are
they area, concentration, and year? you seem to be mixing these
together? If you provide a minimal reproducible example (provide some
data with dput, or the commands to generate random data, or use a
built in dataset) then it make
Hi,
Try:
set.seed(49)
dat1 <- data.frame(year=
rep(2010:2013,c(10,8,9,13)),x=sample(1e4,40,replace=TRUE),y=sample(40,40,replace=TRUE))
plot(x~y,data=dat1,subset=year > 2012)
#or
with(subset(dat1,year > 2012),plot(y,x))
A.K.
Hi R people
This might take me the whole day to figure out, instea
Hi,
Try:
x[x$Species%in%names(sp)[1:2],]
A.K.
On Saturday, February 8, 2014 10:21 PM, Yuanzhi Li
wrote:
Hi, everyone
I met a small problem when I want take a subset from a data frame. The
data frame(x) looks like the followings(10 species with 3 measured
traits):
Species trait1 tra
Hello,
First of all use ?dput to post a data example.
dput(head(x, 20)) # paste the output of this in a post
Now, without a reproducible example it's difficult to say but maybe ?%in%
x[x$Species %in% names(sp)[1:2],]
Hope this helps,
Rui Barradas
Em 09-02-2014 00:03, Yuanzhi Li escreveu:
Hi,
Try ?split()
If `dat1` is the dataset:
lst1 <- split(dat1,dat1$ID)
lst1$an1
# ID V1 mean SD SE
#1 an1 5 72.21719 22.27118 9.092172
#2 an1 6 100.0 NA NA
lst1$an2
# ID V1 mean SD SE
#3 an2 5 79.27999 25.08938 10.2427
#4 an2 6 100.0
subset.data.frame() does not have an na.rm argument!
-pd
On 23 Jan 2014, at 00:58 , Jeff Johnson wrote:
> I have a dataset "mydf" with a field EMAIL_ADDRESS. When importing, I
> specified:
> mydf <- read.csv(file = extract, header = TRUE, stringsAsFactors = FALSE,
> na.strings=c("NA",""))
>
>
I don't think na.rm is a valid at parameter for the subset function. I would
normally use the is.na function to logically test for NA values. I also don't
know where your VALID_EMAIL variable is coming from.
a <- subset(mydf, !is.na(EMAIL_ADDRESS))
The na.strings argument to read.csv and friend
Hi Veepsirtt,
May be this helps:
dat1 <- structure(list(V2 = c(11109.75, 11135.15, 11105.85, 11099.75,
11055.55, 11063.45, 11045.65, 11065, 11061.2, 11070.25, 11069.3,
11076, 11081.85, 11086.4, 11086.7, 11065.6, 11071.25, 11073.15,
11077.8, 11067.7, 11061.1, 11065.9, 11069.1, 11063.3, 11070.45,
#300 1 3 10
#301 1 3 2
#437 1 3 9
#672 1 3 8
A.K.
- Original Message -
From: Rui Barradas
To: Noah Silverman
Cc: "R-help@r-project.org"
Sent: Friday, July 5, 2013 3:51 PM
Subject: Re: [R] Subset and order
Hello,
If time is one of the problems, precompute an ordered index
David Carlson tamu.edu> writes:
>
> It may be that single and efficient are opposing goals. Two steps
> lets you create the subset and then just order each query.
> Alternatively, if the data do not change often, create an ordered
> version and query that.
>
I don't know the data.table pack
[mailto:r-help-boun...@r-project.org] On Behalf Of Noah Silverman
Sent: Friday, July 5, 2013 2:47 PM
To: Rui Barradas
Cc: R-help@r-project.org
Subject: Re: [R] Subset and order
That would work, but is painfully slow. It forces a new sort of the
data with every query. I have 200,000 rows and need
Hello,
If time is one of the problems, precompute an ordered index, and use it
every time you want the df sorted. But that would mean you can't do it
in a single operation.
iord <- order(x$a)
subset(x[iord, ], b == 3)
Rui Barradas
Em 05-07-2013 20:47, Noah Silverman escreveu:
That would w
That would work, but is painfully slow. It forces a new sort of the data with
every query. I have 200,000 rows and need almost a hundred queries.
Thanks,
-N
On Jul 5, 2013, at 12:43 PM, Rui Barradas wrote:
> Hello,
>
> Maybe like this?
>
> subset(x[order(x$a), ], b == 3)
>
>
> Hope thi
Hello,
Maybe like this?
subset(x[order(x$a), ], b == 3)
Hope this helps,
Rui Barradas
Em 05-07-2013 20:33, Noah Silverman escreveu:
Hello,
I have a data frame with several columns.
I'd like to select some subset *and* order by another field at the same time.
Example:
a b c
1
On Jul 1, 2013, at 9:39 PM, Ben Bolker wrote:
> Philip A. Viton osu.edu> writes:
>
>> suppose "state" is a variable in a dataframe containing abbreviations
>> of the US states, as a factor. What I'd like to do is to include
>> dummy variables for a few of the states, (say, CA and MA) among th
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