HI,
May be this helps:
set.seed(5)
dat1<-as.data.frame(matrix(sample(1:100,100,replace=TRUE),ncol=20))
dat1[,8:ncol(dat1)]
A.K.
- Original Message -
From: eliza botto
To: "r-help@r-project.org"
Cc:
Sent: Wednesday, January 2, 2013 8:59 AM
Subject: [R] column selec
ject.org [mailto:r-help-boun...@r-project.org] Namens
eliza botto
Verzonden: woensdag 2 januari 2013 15:00
Aan: r-help@r-project.org
Onderwerp: [R] column selection
Dear R users,
sorry for a very basic question. i wanted to ask that if your column are too
much in number and you want to se
Hmmm... May be
q[, 117:ncol(q)]
And, by the way, do not use "q" as the name for your data.frame() or
matrix() as it is a reserved name. See ?q and
require(fortunes)
fortune('dog')
for reasons why.
HTH,
Jorge.-
On Thu, Jan 3, 2013 at 12:59 AM, eliza botto <> wrote:
>
> Dear R users,
> sorry
On 02-01-2013, at 14:59, eliza botto wrote:
>
> Dear R users,
> sorry for a very basic question. i wanted to ask that if your column are too
> much in number and you want to select the remaining columns, starting from
> column number 117. one way is to use usual command
>> q[ ,(117:2300)].
Dear R users,
sorry for a very basic question. i wanted to ask that if your column are too
much in number and you want to select the remaining columns, starting from
column number 117. one way is to use usual command
>q[ ,(117:2300)].
is there a way by which i can select the remaining columns
If the columns of all elements of the list are in the same order, then
you can collapse it first and then extract.
out <- do.call("rbind", SPECSHOR_tx_Asfc)
out[ , "Asfc.median"]
Regards, Adai
Ivan Calandra wrote:
Hi everybody!
I have a (stupid) question but I cannot find a way to do
Thanks, I didn't think about the help for such a character.
Ivan
Le 1/22/2010 18:50, baptiste auguie a écrit :
?"[" should give you enough information. In short, "[" is an operator
to extract elements, you can think of it as a function with special
semantics. For a simple vector,
v = c("one", "
?"[" should give you enough information. In short, "[" is an operator
to extract elements, you can think of it as a function with special
semantics. For a simple vector,
v = c("one", "two")
v[2] selects the second element of the vector, and is equivalent to,
`[`(v, 2)
# "two"
as you can see fro
Thanks Baptiste, it does help.
However, I don't really understand what "[" means. Could you please tell
me more about it? I didn't find anything helpful on that in the help.
Thanks in advance
Ivan
Le 1/22/2010 17:19, baptiste auguie a écrit :
Hi,
Try this,
a = replicate(3, data.frame(x=1
Hi,
Try this,
a = replicate(3, data.frame(x=1:10, y=rnorm(10)), simplify=FALSE)
lapply(a, "[", "y")
HTH,
baptiste
2010/1/22 Ivan Calandra :
> Hi everybody!
>
> I have a (stupid) question but I cannot find a way to do it!
>
> I have a list like:
>> SPECSHOR_tx_Asfc
> $cotau
> SPECSHOR Asfc.m
Hi everybody!
I have a (stupid) question but I cannot find a way to do it!
I have a list like:
> SPECSHOR_tx_Asfc
$cotau
SPECSHOR Asfc.median
38cotau381.0247
39cotau154.6280
40cotau303.3219
41cotau351.2933
42cotau156.5327
$eqgre
SPECSHOR Asfc.median
Not really, I tried without select = - c(MEASUREM, SEL_FACET, SEL_MEAS)
and indeed the mean was not computed, but it still appeared in the data,
which I didn't want.
Thanks a lot for your help
Ivan
Gabor Grothendieck a écrit :
> It looks ok except you have both specified the wanted factors and
Hi
If I really wanted aggregate all numerics by all non numerics this is how
I would do it
my.numerics <- which(sapply(zeta, is.numeric))
my.factor <- which(sapply(zeta, is.factor))
aggregate(zeta[, my.numerics], zeta[, my.factor], mean)
Regards
Petr
r-help-boun...@r-project.org napsal dne 1
It looks ok except you have both specified the wanted factors and
removed the undesired factors from the data frame. You only need to
do one of these as in the example I gave, not both, so the solution
could be simpler.
On Mon, Jan 18, 2010 at 11:19 AM, Ivan Calandra
wrote:
> Hi!
>
> It looks li
Hi!
It looks like it works perfectly.
However, since I cannot check whether I get the good result or not, can
you please let me know if you see any mistakes?
Here is the code:
ssfamean <- summaryBy(.~SPECSHOR+BONE+TO_POS+FACETTE+SHEARFAC+ENA_BA,
data = subset(ssfa, select = - c(MEASUREM, SEL_FA
Try summaryBy in the doBy package. e.g. using the built-in CO2
summarize each numeric variable by each factor except for the factors
Plant and Type:
library(doBy)
summaryBy(. ~ ., data = subset(CO2, select = - c(Plant, Type)))
On Mon, Jan 18, 2010 at 9:53 AM, Ivan Calandra
wrote:
> Hi everybody
On Mon, Jan 18, 2010 at 10:33 AM, Ivan Calandra <
ivan.calan...@uni-hamburg.de> wrote:
> I didn't understand from the help what really does the function rowMeans
> but it looks like it doesn't take into account the categorical variables (I
> want to calculate the means when the values of all categ
I didn't understand from the help what really does the function rowMeans
but it looks like it doesn't take into account the categorical variables
(I want to calculate the means when the values of all categorical
variables are the same, second part of aggregate). Moreover, ssfa_num
contains only
On Mon, Jan 18, 2010 at 10:17 AM, Ivan Calandra <
ivan.calan...@uni-hamburg.de> wrote:
> Thanks for your answer, but it doesn't work...
>
> Here is what I get:
> > ssfamean <- aggregate(ssfa[[10:24]],ssfa[c("SPECSHOR", "BONE", "TO_POS",
> "FACETTE", "SHEARFAC", "ENA_BA")],mean)
> Error in .subset2
Thanks for your answer, but it doesn't work...
Here is what I get:
> ssfamean <- aggregate(ssfa[[10:24]],ssfa[c("SPECSHOR", "BONE",
"TO_POS", "FACETTE", "SHEARFAC", "ENA_BA")],mean)
Error in .subset2(x, i, exact = exact) :
recursive indexing failed at level 2
I think I've tried everything tha
On Mon, Jan 18, 2010 at 9:53 AM, Ivan Calandra wrote:
> Hi everybody!
>
> I'm working on R today so I have a lot of questions (you may have
> noticed that it's the 3rd email today). I'm new on R, so please excuse
> the "spam"!
>
> I have a dataset "ssfa" with many rows and the column names are:
>
Hi everybody!
I'm working on R today so I have a lot of questions (you may have
noticed that it's the 3rd email today). I'm new on R, so please excuse
the "spam"!
I have a dataset "ssfa" with many rows and the column names are:
> names(ssfa)
[1] "SPECSHOR" "BONE" "TO_POS""MEASUREM"
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