Hi John,
I was thinking that you created df1 in a way that set the 'year'
column as a factor when this is not what you wanted to do.
The data.frame() function takes an argument stringsAsFactors which
controls this behavior.
For R versions 3.6.3 or earlier, the default setting is
stringsAsFactors=TR
Sure John,
df1<-df1[order(as.character(df1$year),decreasing=TRUE),]
Jim
On Tue, Nov 10, 2020 at 8:05 PM John wrote:
> Thanks Jim. Can we do descending order?
>
> Jim Lemon 於 2020年11月10日 週二 下午4:56寫道：
>
>> Hi John,
>>
>> df1<-sapply(df1,as.character)
>>
>> Should do what you ask. The error mess
Thanks Jim. Can we do descending order?
Jim Lemon 於 2020年11月10日 週二 下午4:56寫道：
> Hi John,
>
> df1<-sapply(df1,as.character)
>
> Should do what you ask. The error message probably means that you should
> do this:
>
> df1<-df1[order(as.character(df1$year)),]
>
> as "year" is the name of the first co
Hi John,
df1<-sapply(df1,as.character)
Should do what you ask. The error message probably means that you should do
this:
df1<-df1[order(as.character(df1$year)),]
as "year" is the name of the first column in df1, not a separate object.
Jim
On Tue, Nov 10, 2020 at 6:57 PM John wrote:
> Hi,
>
Hi,
I would like to sort the following simple dataframe by "year"
(characters), but the factor structure prevents me from doing so. How can I
remove the factor structure? Thanks!
> df1
year country
4 2007 Asia; survey
5 2010 8 countries in E/SE Asia
6 2015
Hi Pedro,
I see you use dplyr and ggplot2. Are you looking for something like
this:
```
library(ggplot2)
library(dplyr)
test_data <- data.frame(
year = c(rep("2018", 10), rep("2019", 8), rep("2020", 6)),
value = sample(c(1:100), 24)
)
test_data <- test_data %>%
group_by(year) %>%
mut
Hi all,
I am trying to draw a plot with cumsum values but each "line" has different
lengths
Ilibrary(dplyr)
library(tibble)
library(lubridate)
library(PerformanceAnalytics)
library(quantmod)
library(ggplot2)
getSymbols('TSLA')
I want to create the variables:
a<-cumsum(dailyReturn(TSLA, subset
Hi Thomas,
Jeff is correct that this can be handled via merge, e.g.
df3 <- merge( df2, df1, by="Serial", all=FALSE )
This operation is called an "inner join", and you could use other tools,
such as the dplyr package to accomplish the same thing
df3 <- dplyr::inner_join( df2, df1, by="Serial" )
H
> On Jan 8, 2020, at 6:52 AM, Thomas Subia wrote:
>
> Colleagues,
>
> I have two data frames which look like this.
>
> Data frame 1
>
> Serial Pre.HolePre.flowPre.Date
> 1 30361 0.2419-Nov-19
> 2 30362
"merge" is generally the base R answer to this question, and there are
equivalent functions in various contributed packages.
However, it is necessary to identify which columns in each table uniquely
identify each row ("primary key"). If your Serial 3036 shows up 10 times in the
first table and
Colleagues,
I have two data frames which look like this.
Data frame 1
Serial Pre.HolePre.flowPre.Date
1 30361 0.2419-Nov-19
2 30362 0.212 19-Nov-19
3 30363
But it's also a convenience feature. Note that $E returned null
because there was an ambiguity. By the time you got to $Ex the column
you were referencing was unambiguous and you didn't have to type out
the whole thing. Useful if you have very long column names, for
example imported from a spreadsh
Hello Yannick,
That behavior is documented in the help for subsetting ( ?'$' ):
Both ‘[[’ and ‘$’ select a single element of the list. The main
difference is that ‘$’ does not allow computed indices, whereas
‘[[’ does. ‘x$name’ is equivalent to ‘x[["name", exact =
FALSE]]’.
Hello all
I noticed today that you can access dataframe columns by using incomplete
names. This is a really unexpected behavior which led to some unexpected errors
and I was wondering whether it's a bug or not and whether it should be changed
in the future.
Here's a working example using the pre
, while there
are certain cols to each U denoted as 1s in each URow in dataframe2.
I have to find relationships between Rs and Us. So i start with each U row
in U dataframe (lets say U1 row) and try to find all the rows in R
dataframe, which are subset of U1 row.
I cant find a way to compare rows to
Good one, did not even notice that...!
--
Dr. Ivan Calandra
TraCEr, laboratory for Traceology and Controlled Experiments
MONREPOS Archaeological Research Centre and
Museum for Human Behavioural Evolution
Schloss Monrepos
56567 Neuwied, Germany
+49 (0) 2631 9772-243
https://www.researchgate.net/pr
class("dat") is different from class(dat), which is what you actually want.
On 17-11-17, P. Roberto Bakker wrote:
> Hi everybody,
>
> Question: why are my dataframe and numeric variables a character?
>
> I read an excel file via readxl but my dataframe is a character, and
> numeric variables, eg
Hi Roberto,
This often happens when there are some non-numeric characters. You would
have to check it.
Without more information, e.g. dput(dat), you will have to find by yourself.
HTH,
Ivan
--
Dr. Ivan Calandra
TraCEr, laboratory for Traceology and Controlled Experiments
MONREPOS Archaeologic
Hi everybody,
Question: why are my dataframe and numeric variables a character?
I read an excel file via readxl but my dataframe is a character, and
numeric variables, eg "yi", are also a character.
My excelfile is in English numeric
Sometimes the dataframe was indeed a dataframe, but I do not kn
rrected these things by hand in the example data.
>>>>>
>>>>> HTH
>>>>> Ulrik
>>>>>
>>>>> On Wed, 30 Aug 2017 at 10:16 Hemant Sain
>>>>> wrote:
>>>>>
>>>>>> Hey PIKAL,
>&
;>>>> my company so that i can share the original data file, Actually I'm
>>>>> working
>>>>> on a market basket analysis task but not able to convert my existing
>>>>> data
>>>>> table to appropriate format so that i can apply Apriori
; R,
>>>> and this is very important me to get it done because I'm an intern and
>>>> if i
>>>> won't get it done they will not going to hire me as a full-time
>>>> employee.
>>>> i tried everything by myself but not able to
ave mentioned .
>>>
>>> Thanks
>>>
>>> On 30 August 2017 at 12:49, PIKAL Petr wrote:
>>>
>>> > Hi
>>> >
>>> > It seems to me like homework, there is no homework policy on this help
>>> > list.
>>> &g
uger does not match sugar (I wonder that sugar is dairy product)
> >
> > and you mix uppercase and lowercase letters which could be also
> > problematic, when matching words.
> >
> > Cheers
> > Petr
> >
> > > -Original Message-
> > > Fro
; and you mix uppercase and lowercase letters which could be also
> problematic, when matching words.
>
> Cheers
> Petr
>
> > -Original Message-
> > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Hemant
> Sain
> > Sent: Wednesday, August
> Sent: Wednesday, August 30, 2017 8:28 AM
> To: r-help@r-project.org
> Subject: [R] Dataframe Manipulation
>
> i want to do a market basket analysis and I’m trying to create a dataset for
> that
> i have two tables, one table contains daily transaction of products in which
>
i want to do a market basket analysis and I’m trying to create a dataset
for that i have two tables,
one table contains daily transaction of products in which each row of table
shows item purchased by the customer,
The second table contains parent group under those products are fallen, for
example
The apply function operates on arrays, so your data frame is being converted to
an array (matrix) before doing its thing. So use lapply or one of its
variants.
--
Sent from my phone. Please excuse my brevity.
On December 18, 2016 9:07:52 AM PST, "Cleber N.Borges via R-help"
wrote:
>Why colu
Read ?apply and you shall be be enlightened.
--Ista
On Dec 18, 2016 12:09 PM, "Cleber N.Borges via R-help"
wrote:
> Why columns classes are function dependents?
> Like this example:
>
> > for( i in 1:5 ) print( class( iris[,i] ) )
> [1] "numeric"
> [1] "numeric"
> [1] "numeric"
> [1] "numeric"
Why columns classes are function dependents?
Like this example:
> for( i in 1:5 ) print( class( iris[,i] ) )
[1] "numeric"
[1] "numeric"
[1] "numeric"
[1] "numeric"
[1] "factor"
>
> apply( iris, 2, class )
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
"character" "character" "cha
(10),ID=i))
instead.
Cheers
Petr
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Troels
Ring
Sent: Friday, December 04, 2015 5:03 PM
To: r-help@r-project.org
Subject: [R] dataframe rbind
Dear friends - I have a very simple question - I generate a num
On 04 Dec 2015, at 17:03 , Troels Ring wrote:
> Dear friends - I have a very simple question -
> I generate a number of dataframes with identical names and want to combine
> them into one large dataframe with the same names -
> here is an example
>
> DF <- data.frame(a=rnorm(10),b=runif(10),ID
Try reading and following the Help file, ?rbind.data.frame.
You are inventing your own syntax, not using R's.
Incidentally, growing the frames as you do is generally a bad idea.
Search r-help archives for why.
Cheers,
Bert
Bert Gunter
"Data is not information. Information is not knowledge. And
rom: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Troels Ring
Sent: Friday, December 4, 2015 10:03 AM
To: r-help@r-project.org
Subject: [R] dataframe rbind
Dear friends - I have a very simple question -
I generate a number of dataframes with identical names and want to
combine them int
boun...@r-project.org] On Behalf Of Troels
> Ring
> Sent: Friday, December 04, 2015 5:03 PM
> To: r-help@r-project.org
> Subject: [R] dataframe rbind
>
> Dear friends - I have a very simple question - I generate a number of
> dataframes with identical names and want to combine t
Dear friends - I have a very simple question -
I generate a number of dataframes with identical names and want to
combine them into one large dataframe with the same names -
here is an example
DF <- data.frame(a=rnorm(10),b=runif(10),ID=0)
for (i in 1:10){
DF <- DF+rbind(data.frame(a=rnorm(10),
I am trying to analyse time-series .netcdf (3D lat,long and time domain)
climate data. I want to apply the SPEI package (calculation of standardized
precipitation evapotranspiration index) on it. But unable to arrange my data
in the required data frame. As I am a beginner in R, it will be very much
Hi
Please do not use html formating in your post. It does not bring any advantage.
See inline.
From: Verena Weinbir [mailto:vwein...@gmail.com]
Sent: Thursday, May 29, 2014 3:33 PM
To: PIKAL Petr
Subject: Re: [R] Dataframe: Average cells of two rows and replace them with one
row
Hey,
Thank you
-help
Subject: Re: [R] Dataframe: Average cells of two rows and replace them with one
row
Hello,
thank you for your reply.
Actually, the whole rows would have to be averaged anyways - my mistake :-)
Besides the first column "name" there is one other string (chr) variable "Test&q
nal Message-
> > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> > project.org] On Behalf Of Verena Weinbir
> > Sent: Wednesday, May 28, 2014 2:00 PM
> > To: arun
> > Cc: r-help
> > Subject: Re: [R] Dataframe: Average cells of two rows and repl
41.
Show us at least structure of your data frame.
?str
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Verena Weinbir
> Sent: Wednesday, May 28, 2014 2:00 PM
> To: arun
> Cc: r-help
> Su
Hey guys,
thank you very much for your help. Since I am a R-newbie I am still
checking out how your code works and how I could adapt it to my dataframe,
which has 124 rows and 41 columns/variables. The first column would be
"name", the last ones, 40 and 41, contain the cells I want to average fo
Hi,
You can also try:
dat <- read.table(text="Name C1 C2 C3
1 A 3 3 5
2 B 2 7 4
3 C 4 3 3
4 C 4 4 6
5 D 5 5 3",sep="",header=TRUE,stringsAsFactors=FALSE)
library(plyr)
ddply(dat,.(Name),numcolwise(mean,na.rm=TRUE))
A.K.
On Tuesday, May 27, 2014 4:08 PM, Verena We
Hello,
Try the following.
dat <- read.table(text = "
Name C1 C2 C3
1 A 3 3 5
2 B 2 7 4
3 C 4 3 3
4 C 4 4 6
5 D 5 5 3
", header = TRUE)
str(dat)
aggregate(dat[, -1], list(dat$Name), mean)
Hope this helps,
Rui Barradas
Em 27-05-2014 21:06, Verena Weinbir escreve
Look at the aggregate function. As long as you have a column like
Name that indicates which rows should be averaged together it will
work (technically it will average the other rows as well, but since
the average of 1 number is that number you will not see a difference).
On Tue, May 27, 2014 at 2
Hello,
I have a big dataframe, and want to average two specific cells of two
specific rows and then replace those two rows with one row which contains
the averaged cells. Example (row 3 and 4: Cells2 and Cells3 averaged and
replaced)
NameC1 C2 C3
1 A 3 3 5
2 B 2 7 4
3 C 4 3
Hi,
It is better to show example data using ?dput().
dat <- structure(list(row.names = 1:4, XYZ = c("sample", "sample2",
"sample3", "sample4"), `000_001` = c("sample", "Au5", "C", "C"
), `000_002` = c("sample", "Au32", "C", "Au4"), `000_003` = c("sample",
"Au5", "A", "AC")), .Names = c("row.nam
Hello,
thank you for accepting me into the list.
I have the following dataframe:
row.names X Y Z 000_001 000_002 000_003
1sample samplesample sample
2sample2 Au5 Au32 Au5
Thanks, your solution using ave() works perfectly.
/johannes
-Ursprüngliche Nachricht-
Von: Bert Gunter
An: Johannes Radinger
Cc: R help
Gesendet: Mittwoch, 26. März 2014 16:45:43 GMT+00:00
Betreff: Re: [R] dataframe calculations based on certain values of a column
I believe this
dplyr's group_by and mutate can create those columns for you:
var1 <- c("a","b","c","a","b","c","a","b","c")
var2 <- c("X","X","X","Y","Y","Y","Z","Z","Z")
var3 <- c(1,2,2,5,2,6,7,4,4)
df <- data.frame(var1,var2,var3)
dt <- tbl_df(df)
dt %.%
group_by(var2) %.%
mutate(
div = var3[var1 =
On 26-03-2014, at 17:09, Johannes Radinger wrote:
> Hi,
>
> I have data in a dataframe in following structure
> var1 <- c("a","b","c","a","b","c","a","b","c")
> var2 <- c("X","X","X","Y","Y","Y","Z","Z","Z")
> var3 <- c(1,2,2,5,2,6,7,4,4)
> df <- data.frame(var1,var2,var3)
>
> Now I'd like to
I believe this will generalize. But check carefully!
Using your example (Excellent!), use ave():
with(df,ave(seq_along(var1),var2,FUN=function(i)
var3[i]/var3[i][var1[i]=="c"]))
[1] 0.500 1.000 1.000 0.833 0.333 1.000 1.750
[8] 1.000 1.000
This is kind of a l
Hi,
I have data in a dataframe in following structure
var1 <- c("a","b","c","a","b","c","a","b","c")
var2 <- c("X","X","X","Y","Y","Y","Z","Z","Z")
var3 <- c(1,2,2,5,2,6,7,4,4)
df <- data.frame(var1,var2,var3)
Now I'd like to calculate relative values of var3. This values
should be relative to th
Hi,
dat1 <- read.table(text="X1
1,1
1,2
1,3
2,1
2,2
2,3
3,1
3,2
3,3",sep="",header=TRUE)
dat2 <- setNames(as.data.frame(matrix(dat1[,1],3,3)),paste0("X",1:3))
str(dat2)
#'data.frame': 3 obs. of 3 variables:
# $ X1: Factor w/ 3 levels "1,1","1,2","1,3": 1 2 3
# $ X2: Factor w/ 3 levels "2,1",
Hi,
The question is not clear.
set.seed(45)
spe <- data.frame(Col1=sample(c(1:10,letters[1:2]),400,replace=TRUE))
mat1 <- matrix(spe[,1],nrow=20,ncol=20)
#If you want to replace the non-numeric values with NAs,
mat2 <- matrix(as.numeric(as.character(spe[,1])),nrow=20,ncol=20)
A.K.
Hell
Hi,
Try:
d[match(unique(d$fac),d$fac),]
A.K.
On Friday, December 13, 2013 4:17 PM, Gang Chen wrote:
Suppose I have a dataframe defined as
L3 <- LETTERS[1:3]
(d <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace
= TRUE)))
x y fac
1 1 1 C
2 1 2 A
3 1 3
-project.org] On
> Behalf
> Of Gang Chen
> Sent: Friday, December 13, 2013 1:35 PM
> To: arun
> Cc: R help
> Subject: Re: [R] dataframe manipulation
>
> Perfect! Thanks a lot, A.K!
>
>
> On Fri, Dec 13, 2013 at 4:21 PM, arun wrote:
>
> >
> >
Another neat solution! Thanks a lot, Sarah!
On Fri, Dec 13, 2013 at 4:35 PM, Sarah Goslee wrote:
> What about:
>
> lapply(levels(d$fac), function(x)head(d[d$fac == x,], 1))
>
>
> Thanks for the reproducible example. If you put set.seed(123) before
> the call to sample, then everyone who tries it
What about:
lapply(levels(d$fac), function(x)head(d[d$fac == x,], 1))
Thanks for the reproducible example. If you put set.seed(123) before
the call to sample, then everyone who tries it will get the same data
frame d.
Sarah
On Fri, Dec 13, 2013 at 4:15 PM, Gang Chen wrote:
> Suppose I have a
Perfect! Thanks a lot, A.K!
On Fri, Dec 13, 2013 at 4:21 PM, arun wrote:
>
>
> Hi,
> Try:
> d[match(unique(d$fac),d$fac),]
> A.K.
>
>
> On Friday, December 13, 2013 4:17 PM, Gang Chen
> wrote:
> Suppose I have a dataframe defined as
>
> L3 <- LETTERS[1:3]
> (d <- data.frame(cbind(x
Suppose I have a dataframe defined as
L3 <- LETTERS[1:3]
(d <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace
= TRUE)))
x y fac
1 1 1 C
2 1 2 A
3 1 3 B
4 1 4 C
5 1 5 B
6 1 6 B
7 1 7 A
8 1 8 A
9 1 9 B
10 1 10 A
I want to extract
t;",x[-length(x)]));
Nexttripstore<- ave(x2$Store,x2$PANID,FUN=function(x) c(x[-1],""))
}
)
What am i doing wrong?
- Original Message -
From: arun
To: R help
Cc:
Sent: Monday, September 2, 2013 11:29 PM
Subject: Re: R dataframe and looping help
HI Satish,
colnam
;- lapply(seq_len(ncol(dat2)),function(i)
{
x1<-cbind(Output[,c(1:3)],dat2[,i]);
colnames(Output)[4]<- colnames(dat2)[i];
x2<-x1[x1[,4]!=0,];
previoustripstore<-ave(x2$store,x2$CUSTID,FUN=function(x) c(NA,x[-length(x)]));
Nexttripstore<- ave(x2$store,x2$CUSTID,FUN=function(x) c
HI,
You may try this:
dat1<- read.table(text="
CustID TripDate Store Bread Butter Milk Eggs
1 2-Jan-12 a 2 0 2 1
1 6-Jan-12 c 0 3 3 0
1 9-Jan-12 a 3 3 0 0
1 31-Mar-13 a 3 0 0 0
2 31-Aug-12 a 0 3 3 0
2 24-Sep-12 a 3 3 0 0
2 25-Sep-12 b 3 0 0 0
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat2<- d
#this should also work
within(X,a<- ifelse(b,c,a))
# a b c
#1 2 TRUE 2
#2 2 TRUE 2
#3 1 FALSE 2
#4 1 FALSE 2
#5 1 FALSE 2
#6 2 TRUE 2
A.K.
- Original Message -
From: Pascal Oettli
To: fgrelier
Cc: r-help@r-project.org
Sent: Tuesday, May 14, 2013 4:47 AM
Subject: Re:
ine-
De : arun [mailto:smartpink...@yahoo.com]
Envoyé : mardi 14 mai 2013 15:19
À : Pascal Oettli
Cc : R help; fgrelier
Objet : Re: [R] Dataframe and conditions
#this should also work
within(X,a<- ifelse(b,c,a))
# a b c
#1 2 TRUE 2
#2 2 TRUE 2
#3 1 FALSE 2
#4 1 FALSE 2
#5 1 FALSE
Hello,
One approach is using "ifelse":
> X <- data.frame(a=c(1,1,1,1,1,1),
b=c(TRUE,TRUE,FALSE,FALSE,FALSE,TRUE), c=c(2,2,2,2,2,2))
> X
a b c
1 1 TRUE 2
2 1 TRUE 2
3 1 FALSE 2
4 1 FALSE 2
5 1 FALSE 2
6 1 TRUE 2
>
> X <- within(X, a <- ifelse(b==TRUE, c, a))
> X
a b c
1 2 TRUE 2
Hello,
Try the following.
X$a[X$b] <- X$c[X$b]
Hope this helps,
Rui Barradas
Em 14-05-2013 09:06, fgrelier escreveu:
I have in a dataframe X : 3 Variables X$a , X$b, X$c
I would like to replace in X the values of X$a by the values of X$c but
only when X$b=="TRUE"
I have tried to put in
I have in a dataframe X : 3 Variables X$a , X$b, X$c
I would like to replace in X the values of X$a by the values of X$c but
only when X$b=="TRUE"
I have tried to put in place a loop but as I have a lot of rows it is very
very long to run.
Thanks for your help
--
View this message in cont
Fantastic, thanks alot for that!
Take care.
Adam.
Date: Sat, 30 Mar 2013 01:14:49 -0700
From: ml-node+s789695n466289...@n4.nabble.com
To: english.fel...@hotmail.com
Subject: Re: Dataframe manipulation
Hi Adam,
I hope this is what you wanted:
dat1<- read.csv("example.csv",sep="\t",s
Hi Adam,
I hope this is what you wanted:
dat1<- read.csv("example.csv",sep="\t",stringsAsFactors=FALSE)
str(dat1)
#'data.frame': 102 obs. of 5 variables:
# $ species : chr "B. barbastrellus" "E. nilssonii" "H. savii" "M. alcathoe"
...
# $ period : chr "dusk" "dusk" "dusk" "dusk" ...
# $
omJSON(url)
> >>
> >> str(dat) # List with 3 components
> >>
> >> str(dat[[1]][[1]])
> >> dat[[1]][[1]]
> >>
> >> str(dat[[1]][[2]])
> >> unlist(dat[[1]][[2]])
> >>
> >> str(dat[[1]][[3]])
> >> dat[[1]][[3
amp;te=";<http://apistat.istat.it/?q=getdatajson&dataset=DCIS_POPSTRBIL&dim=1,0,0,0&lang=1&tr=&te=>
>> dat <- fromJSON(url)
>>
>> str(dat) # List with 3 components
>>
>> str(dat[[1]][[1]])
>> dat[[1]][[1]]
>>
>> s
][[1]])
> dat[[1]][[1]]
>
> str(dat[[1]][[2]])
> unlist(dat[[1]][[2]])
>
> str(dat[[1]][[3]])
> dat[[1]][[3]]
>
>
> Hope this helps,
>
> Rui Barradas
> Em 27-12-2012 11:55, sp.duccio escreveu:
>
> Hello to everybody,
>
> I need to convert a json datas
escreveu:
> Hello to everybody,
>
> I need to convert a json dataset in an R dataframe.
> I suppose that I'd need to use rjson or rjsonio package.
> The json dataset is:
> http://apistat.istat.it/?q=getdatajson&dataset=DCIS_POPSTRBIL&dim=1,0,0,0&lang=1&tr=&am
Hello to everybody,
I need to convert a json dataset in an R dataframe.
I suppose that I'd need to use rjson or rjsonio package.
The json dataset is:
http://apistat.istat.it/?q=getdatajson&dataset=DCIS_POPSTRBIL&dim=1,0,0,0&lang=1&tr=&te=
It would be nice if someon
nal Message-
> From: kokila.krish...@quantilez.com
> Sent: Thu, 8 Nov 2012 02:18:22 -0800 (PST)
> To: r-help@r-project.org
> Subject: [R] Dataframe,Matrix,Table
>
> Hi for all...
>
> I am having one Dataset. Mode of the dataset is dataframe. I need to add
&g
Hi for all...
I am having one Dataset. Mode of the dataset is dataframe. I need to add
headerline,row and column line.
How to do this...Help me to do this
Thanks
Kokila
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Sent f
Thank you Jorge & Michael.
I was being stupid - its the only explanation!
The line I had been executing was
mtcars[rownames=="Valiant"] # missing rownames argument
but the line I quoted in my post was
mtcars[rownames(mtcars) != "Valiant",] # How could I write the correct line in
the mailin
Hi Ajay,
Like Jorge, I can't seem to reproduce the behavior you are worried about.
mtcars[rownames(mtcars) != "Valiant",]
returns a 31x11 data.frame as expected.
When you say it "fails," what error message / result are you seeing?
Michael
On Fri, Feb 17, 2012 at 3:27 AM, Jorge I Velez wrote
Hi Ajay,
In the first case, you need "==" instead of "=" :
R> mtcars[ rownames(mtcars) == "Valiant", ]
mpg cyl disp hp drat wt qsec vs am gear carb
Valiant 18.1 6 225 105 2.76 3.46 20.22 1 031
For the second case,
R> mtcars[rownames(mtcars) != "Valiant",]
will do it.
See als
data(mtcars)
mtcars[rownames(mtcars)!="Valiant",] # fails
mtcars[list(rownames(mtcars))!="Valiant",] # runs but I am not getting the
expected result
With the latter statement, I expected all rows except the one where the name is
"Valiant".
I must have got something simple wrong; what is it?
This works but I do not know if there is a better way
tmp = df[df$myvalue<2000,]
ind = match(tmp$myvalue, df$myvalue)
res = df$DateTime[ind]
solution = list(ind[1], res[1])
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Thank you Jorge and Florent for your responses.
Now, I 'd like to get the date *(and its index) *where myvalue < 2000 for
the first time.
I expect for a result like (index, date) = (3, 2012-01-07 )
This way does not work:
ind = match(df$myvalue <2000, df$myvalue)
res = df$DateTime[ind]
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Thank you Florent, it really helps.
Regards,
Jorge.-
On Thu, Jan 19, 2012 at 7:28 PM, Florent D. <> wrote:
> Another possibility:
>
> df$Date[match(1800, df$myvalue)]
>
> match() stops at the first value encountered so it may be a bit faster
> than a full subset(), depending on your table size.
Another possibility:
df$Date[match(1800, df$myvalue)]
match() stops at the first value encountered so it may be a bit faster
than a full subset(), depending on your table size.
Another difference: this approach would return NA if there was no
match, subset(...)[1, ] would trigger an error. Depend
Hi ikuzar,
Try
subset(df, myvalue == 1800, select = Date)[1, ]
See ?subset for more information.
HTH,
Jorge.-
On Thu, Jan 19, 2012 at 6:42 PM, ikuzar <> wrote:
> Hi,
> I 'd like to select the Date where myvalue =1800 appears the* first time*.
>
> For instance:
> df =data.frame(date, myvalue,
Hi,
I 'd like to select the Date where myvalue =1800 appears the* first time*.
For instance:
df =data.frame(date, myvalue, ...)
...
Datemyvalue
2012-01-052500
2012-01-06 2450
*2012-01-07 1800*
2012-01-08
t; Datum: Thu, 24 Nov 2011 09:12:57 -0500
> Von: Gabor Grothendieck
> An: Johannes Radinger
> CC: r-help@r-project.org
> Betreff: Re: [R] dataframe indexing by number of cases per group
> On Thu, Nov 24, 2011 at 7:02 AM, Johannes Radinger
> wrote:
> > Hello,
> >
&g
On Thu, Nov 24, 2011 at 7:02 AM, Johannes Radinger wrote:
> Hello,
>
> assume we have following dataframe:
>
> group <-c(rep("A",5),rep("B",6),rep("C",4))
> x <- c(runif(5,1,5),runif(6,1,10),runif(4,2,15))
> df <- data.frame(group,x)
>
> Now I want to select all cases (rows) for those groups
> whi
A very similar question was asked a couple of days ago - see the
thread titled "Removing rows in dataframe w'o duplicated values" - in
particular, the responses by Dimitris Rizopoulos and David Winsemius.
The adaptation to this problem is
df[ave(as.numeric(df$group), as.numeric(df$group), FUN = le
Hello,
assume we have following dataframe:
group <-c(rep("A",5),rep("B",6),rep("C",4))
x <- c(runif(5,1,5),runif(6,1,10),runif(4,2,15))
df <- data.frame(group,x)
Now I want to select all cases (rows) for those groups
which have more or equal 5 cases (so I want to select
all cases of group A and
On 26 May 2011, at 08:02, Vijayan Padmanabhan wrote:
> I have a requirement for which I am seeking help.
Best to just ask, compactly. This is a very straightforward question: best to
read on how to use R: You are just set 1 column of a dataframe to a value based
on the others, applying this to
Dear RGroup
I have a requirement for which I am seeking help.
I am looking at automating the last column calculation through R when
having the data of the other columns as a dataframe, In excel I can do
using the formula function as given below, however, hereagain for the
number of observations
sDaply2(X, X$ID)
# list is not a timeSeries object
str(cbind(t(res)))
res <- as.timeSeries(cbind(t(res)))
-Original Message-
From: h.wick...@gmail.com [mailto:h.wick...@gmail.com] On Behalf Of Hadley
Wickham
Sent: 14 March 2011 15:07
To: Daniele Amberti
Cc: r-help@r-project.org
Subjec
> res <- daply(X, "ID", buildTimeSeriesFromDataFrame2, .parallel = FALSE)
> return(res)
> }
> # tsDaply2 .parallel = FALSE work but list discart timeSeries class
>
> # bind after ts creation
> res <- tsDaply2(X, X$ID)
> # list is not a timeSeries object
&g
ot;, buildTimeSeriesFromDataFrame2, .parallel = FALSE)
return(res)
}
# tsDaply2 .parallel = FALSE work but list discart timeSeries class
# bind after ts creation
res <- tsDaply2(X, X$ID)
# list is not a timeSeries object
str(cbind(t(res)))
res <- as.timeSeries(cbind(t(res)))
stopW
Well, I'd start by removing all explicit use of environments, which
makes you code very hard to follow.
Hadley
On Monday, March 14, 2011, Daniele Amberti wrote:
> I found that plyr:::daply is more efficient than base:::by (am I doing
> something wrong?), below updated code for comparison (I als
I found that plyr:::daply is more efficient than base:::by (am I doing
something wrong?), below updated code for comparison (I also fixed a couple
things).
Function daply from plyr package has also a .parallel argument and I wonder if
creating timeseries objects in parallel and then combining th
I’m wondering which is the most efficient (time, than memory usage) way to
obtain a multivariate time series object from a data frame (the easiest data
structure to get data from a database trough RODBC).
I have a starting point using timeSeries or xts library (these libraries can
handle time zo
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