> x <- runif(10)
> y <- runif(10)
> cov(cbind(x,y))
x y
x 0.1205034 0.02642830
y 0.0264283 0.09945432
I understand wanting to calculate covariance matrices.
What I DON'T understand is wanting to do it using apply().
(And that's what looked like a homework problem, it's so artifi
On Fri, 04 Oct 2024 11:16:45 -0700
Jeff Newmiller via R-help wrote:
> Even if this is not a homework question, it smells like one. If you
> read the Posting Guide it warns you that homework is off-topic, so
> when you impose an arbitrary constraint like "must use specific
> unrelated function" we
Even if this is not a homework question, it smells like one. If you read the
Posting Guide it warns you that homework is off-topic, so when you impose an
arbitrary constraint like "must use specific unrelated function" we feel like
you are either cheating or wasting our time, and it is up to you
Pardon me!!!
What makes you think this is a homework question? You are not obligated
to respond if the question is not intelligent enough for you.
I did the following: two ways to calculate a covariance matrix but
wonder how I might replicate the results with "apply". I am not too
comfortable
Hello,
You don't need apply, covariance calculations are so frequent that R or
any other statistics package already has pre-programmed functions.
This time with two vectors x and y.
set.seed(123)
n <- 3
x <- rnorm(n)
y <- rnorm(n)
# the two main diagonal values
var(x)
#> [1] 1.300025
var(y)
Why must the answer use apply? It feels like there are elements of the problem
that are not explained.
-Original Message-
From: R-help On Behalf Of Ben Bolker
Sent: Friday, October 4, 2024 8:45 AM
To: r-help@r-project.org
Subject: Re: [R] apply
[External Email]
It's still ha
It's still hard to figure out what you want. If you have two vectors
you can compute their (2x2) covariance matrix using cov(cbind(x,y)).
If you want to compute all pairwise squared differences between elements
of x and y you could use outer(x, y, "-")^2.
Can you explain a little bit more
В Fri, 4 Oct 2024 20:28:01 +0800
Steven Yen пишет:
> Suppose I have two vectors, x and y. Is there a way
> to do the covariance matrix with “apply”.
There is no covariance matrix for just two samples (vectors) 'x' and
'y'. You can only get one covariance value for these.
If you had a pair of v
OK. Thanks to all. Suppose I have two vectors, x and y. Is there a way
to do the covariance matrix with “apply”. The matrix I need really
contains the deviation products divided by the degrees of freedom (n-1).
That is, the elements
(1,1), (1,2),...,(1,n)
(2,1), (2,2),, (2,n)
(n,1)
Hello,
This doesn't make sense, if you have only one vector you can estimate
its variance with
var(x)
but there is no covariance, the joint variance of two rv's. "co" or
joint with what if you have only x?
Note that the variance of x[1] or any other vector element is zero, it's
only one va
В Fri, 4 Oct 2024 19:14:30 +0800
Steven Yen пишет:
> I have a vector:
> set.seed(123) > n<-3 > x<-rnorm(n); x [1] -0.56047565 -0.23017749
> 1.55870831
> var(x[1]) cov(x[1],x[2])
Are you sure you don't have a matrix? If you type var(x[1]) or
cov(x[1],x[2]) into R, you can see that all these are
Hello
I have a vector:
set.seed(123) > n<-3 > x<-rnorm(n); x [1] -0.56047565 -0.23017749
1.55870831 I like to create a matrix with elements containing variances
and covariances of x. That is var(x[1]) cov(x[1],x[2]) cov(x[1],x[3])
cov(x[2],x[1]) var(x[2]) cov(x[2],x[3]) cov(x[3],x[1]) cov(x[3]
Hello,
If you have a numeric matrix or data.frame, try something like
cov(mtcars)
Hope this helps,
Rui Barradas
Às 10:15 de 04/10/2024, Steven Yen escreveu:
On 10/4/2024 5:13 PM, Steven Yen wrote:
Pardon me!!!
What makes you think this is a homework question? You are not
obligated to res
On 04.10.2024 11:13, Steven Yen wrote:
Pardon me!!!
What makes you think this is a homework question? You are not obligated
Otherwise you called cov()
Best,
Uwe Ligges
to respond if the question is not intelligent enough for you.
I did the following: two ways to calculate a covariance
On 10/4/2024 5:13 PM, Steven Yen wrote:
> Pardon me!!!
>
> What makes you think this is a homework question? You are not
> obligated to respond if the question is not intelligent enough for you.
>
> I did the following: two ways to calculate a covariance matrix but
> wonder how I might replicate
Homework questions are not answered on this list.
Best,
Uwe Ligges
On 04.10.2024 10:32, Steven Yen wrote:
The following line calculates standard deviations of a column vector:
se<-apply(dd,1,sd)
How can I calculate the covariance matrix using apply? Thanks.
The following line calculates standard deviations of a column vector:
se<-apply(dd,1,sd)
How can I calculate the covariance matrix using apply? Thanks.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/l
On Tue, 19 Sep 2023 17:14:58 +0200
arnaud gaboury wrote:
> non_empty_df <- function(l) {
> lapply(l, function(df) df[sapply(df, function(df) nrow(df) !=0)])
> }
> If I test this way: non_empty_df(my.list[1]) it does the job. It will
> return the data.frame from the first list of my_list with
I have a list of 9 lists called my.list. Each one of these 9 lists is
itself a list of 6 data.frames. Most of these data.frames have 0 rows and 0
columns.
I want to return all data.frames from the list with row numbers different
from 0.
I first created the following function:
non_empty_df <- funct
Define
mean0na <- function(x) {
x[ 0 == x ] <- NA
mean( x, na.rm = TRUE )
}
and then use that instead of mean.
On August 10, 2023 7:25:08 PM PDT, ani jaya wrote:
>Hello,
>
>I try to calculate the mean of an array with a condition. My array is
>B=c(181,101,420) in dimensions. And I want to f
Hello,
I try to calculate the mean of an array with a condition. My array is
B=c(181,101,420) in dimensions. And I want to find a specific member
in 3rd dimension (time monthly) based on a condition of another data
frame, A =c(420). My final array would be C=c(181,101,12) which 12 is
a monthly mea
Bert provided an excellent answer to your question.
FYI here is a different approach to do the calculation.
It uses data.frame rather than matrix. A data frame is a list of its columns.
Here the function supplied to sapply operates on each column of the data.frame.
> m <- as.data.frame(t(matrix(1:
from ?apply:
"If each call to FUN returns a vector of length n, and simplify is
TRUE, then apply returns an array of dimension c(n, dim(X)[MARGIN]) ."
For margin = 1 (cumsum over rows), each call to cumsum return a vector
of length 2. Hence the array returned will be of dimension c(2,
c(5,2)[1]) =
Hi there,
I try to calculate the cumsum of row and column of a matrix as follows.
> m <- matrix(1:10, ncol = 2)
> m
[,1] [,2]
[1,] 1 6
[2,] 2 7
[3,] 3 8
[4,] 4 9
[5,] 5 10
> apply(m, 1, cumsum)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 3 4 5
[2,]
Thanks. I have a clarification and a follow-up question. I should have
asked this in the original post, and I should have provided a better
example for the FUN argument, I apologize.
For use in an example, here is a "silly" example of a function that
requires arguments such as x and y to be "separ
Thank you, it works!
On Mon, Aug 9, 2021 at 3:26 PM Andrew Simmons wrote:
>
> Hello,
>
>
> There are two convenient ways to access a column in a data.frame using `$`
> and `[[`. Using `df` from your first email, we would do something like
>
> df <- data.frame(VAR = 1:3, VAL = c("value is blue",
Hello,
There are two convenient ways to access a column in a data.frame using `$`
and `[[`. Using `df` from your first email, we would do something like
df <- data.frame(VAR = 1:3, VAL = c("value is blue", "Value is red",
"empty"))
df$VAL
df[["VAL"]]
The two convenient ways to update / / replac
I wanted to remove possible white spaces before or after the string.
Actually, it worked, I used `gsub("[:blank:]*val[:blank:]*", "",
df$VAL, ignore.case=TRUE)`. I don't know why in the example there were
extra columns -- they did not came out in the real case.
Thank you, I think the case is closed
Hi Luigi,
You want to get rid of certain strings in the "VAL" column. You are
assigning to:
df[df$VAL]
Error in `[.data.frame`(df, df$VAL) : undefined columns selected
when I think you should be assigning to:
df$VAL
What do you want to remove other than "[V|v]alue is" ?
JIim
On Mon, Aug 9, 20
Sorry, silly question, gsub works already with regex. But still, if I
add `[[:blank:]]` still I don't get rid of all instances. And I am
keeping obtaining extra columns
```
> df[df$VAL] = gsub("[[:blank:]Value is]", "", df$VAL, ignore.case=TRUE)
> df[df$VAL] = gsub("[[:blank:]Value is]", "", df$VAL
Thank you, that is much appreciated. But on the real data, the
substitution works only on few instances. Is there a way to introduce
regex into this?
Cheers
Luigi
On Mon, Aug 9, 2021 at 11:01 AM Jim Lemon wrote:
>
> Hi Luigi,
> Ah, now I see:
>
> df$VAL<-gsub("Value is","",df$VAL,ignore.case=TRU
Hi Luigi,
Ah, now I see:
df$VAL<-gsub("Value is","",df$VAL,ignore.case=TRUE)
df
VAR VAL
1 1 blue
2 2 red
3 3 empty
Jim
On Mon, Aug 9, 2021 at 6:43 PM Luigi Marongiu wrote:
>
> Hello,
> I have a dataframe where I would like to change the string of certain
> rows, essentially I am lo
Hello,
I have a dataframe where I would like to change the string of certain
rows, essentially I am looking to remove some useless text from the
variables.
I tried with:
```
> df = data.frame(VAR = 1:3, VAL = c("value is blue", "Value is red", "empty"))
> df[df$VAL] = gsub("value is ", "", df$VAL,
ing
>
> -Original Message-
> From: Jim Lemon [mailto:drjimle...@gmail.com]
> Sent: Thursday, June 17, 2021 8:29 PM
> To: Yuan Chun Ding
> Cc: r-help@r-project.org
> Subject: Re: [R] apply a color range (kind of like a heat map) to the
> values in each cell of a data frame
Ding
Cc: r-help@r-project.org
Subject: Re: [R] apply a color range (kind of like a heat map) to the values in
each cell of a data frame
Hi Ding,
There are a number of "value to color" functions in various packages.
One is "color.scale" in the plotrix package:
lib
Hi Ding,
There are a number of "value to color" functions in various packages.
One is "color.scale" in the plotrix package:
library(plotrix)
s1 <-c(0.085,0.086,0.139,0.129,0.235,0.177,0.000,0.126,0.271,0.000,0.083,0.163)
s2 <-c(0.000,0.093,0.000,0.080,0.072,0.388,0.138,0.107,0.000,0.000,0.474,0.00
Dear R users,
I have a numeric table with 140 rows and 30 columns, here I only made partial
table, test1, as an example. I want to apply a blue color range to the value
in each cell of the data frame test1.
I found some R code using DT library. However, I only can see the colored table
at m
-help [mailto:r-help-boun...@r-project.org] On Behalf Of David L Carlson
Sent: Friday, May 12, 2017 10:48 AM
To: Ismail SEZEN ; Micha Silver
Cc: R-help@r-project.org
Subject: Re: [R] apply and cor()
Actually, r is a vector, not an index value. You need
apply(compare_data, 1, function(r) cor(r, t(test
-project.org
Subject: Re: [R] apply and cor()
Actually, r is a vector, not an index value. You need
apply(compare_data, 1, function(r) cor(r, t(test_data)))
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-
to:r-help-boun...@r-project.org] On Behalf Of Ismail SEZEN
Sent: Friday, May 12, 2017 10:11 AM
To: Micha Silver
Cc: R-help@r-project.org
Subject: Re: [R] apply and cor()
> On 12 May 2017, at 17:57, Micha Silver wrote:
>
> I have two data.frames, one with a single row of 31 columns, a
> On 12 May 2017, at 17:57, Micha Silver wrote:
>
> I have two data.frames, one with a single row of 31 columns, and the second
> with 269 rows and the same 31 columns.
> > dim(compare_data)
> [1] 269 31
> > dim(test_data)
> [1] 1 31
>
> I want to apply cor() between the one row of 'test_dat
I have two data.frames, one with a single row of 31 columns, and the
second with 269 rows and the same 31 columns.
> dim(compare_data)
[1] 269 31
> dim(test_data)
[1] 1 31
I want to apply cor() between the one row of 'test_data', and each row
of the 'compare_data' .
I tried 'apply' but I get
these functions could result in
object suitable for plm
Cheers
Petr
From: laura roncaglia [mailto:roncaglia.la...@gmail.com]
Sent: Friday, September 16, 2016 9:36 AM
To: PIKAL Petr
Subject: Re: [R] apply weight to a data frame
Thank you for your answer.
The data frame contains social and economic
r-help-boun...@r-project.org] On Behalf Of laura
> roncaglia
> Sent: Thursday, September 15, 2016 6:40 PM
> To: r-help@r-project.org
> Subject: [R] apply weight to a data frame
>
> I am a beginner user of R.
>
> I am writing the master thesis using a data frame from a national s
I am a beginner user of R.
I am writing the master thesis using a data frame from a national survey.
The data frame contains several variables, one of which contains the survey
weights.
I need to apply the survey weights to the data frame, in order to use the
data frame with the plm package (I ne
> I would like to define an arbitrary function of an arbitrary number of
> variables,
> for example, for 2 variables:
>
> func2 <- function(time, temp) time + temp
>
> I'd like to keep variable names that have a meaning in the problem (time and
> temperature above).
Not quite enough information
c2,
>> as.list(a)))
>>
>> although this does work:
>>
>> do.call(func2, as.list(c(10, 121)))
>>
>> And, this also works:
>>
>> apply(expand.grid(temps,times), 1, function(a) do.call("+", as.list(a)))
>>
>> There
= c(1:5), this doesn't quite seem to work:
>>
>> apply(expand.grid(temps,times), 1, function(a) do.call(func2,
>> as.list(a)))
>>
>> although this does work:
>>
>> do.call(func2, as.list(c(10, 121)))
>>
>> And, this also works:
>>
>
iginal Message-
From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us]
Sent: Friday, September 09, 2016 5:39 PM
To: Steve Kennedy; r-help@r-project.org
Subject: Re: [R] Apply a multi-variable function to a vector
Your architecture has a bad smell to me. For one thing you are mixing
different units
Your architecture has a bad smell to me. For one thing you are mixing different
units in the same vector but should be putting multiple instances of the same
variable into one vector. Lists of vectors (data frames) are typically used
when multiple variables need to be grouped.
Another problem i
Try do.call(), as in
> func2 <- function(time, temp) paste(time, temp)
> func2(121, 10)
[1] "121 10"
> do.call(func2, as.list(c(121,10)))
[1] "121 10"
> do.call(func2, list(121,10))
[1] "121 10"
>
> func2(121, time=10:12)
[1] "10 121" "11 121" "12 121"
> do.call(func2, list(121,time=10:12))
[1] "1
Hello,
I would like to define an arbitrary function of an arbitrary number of
variables, for example, for 2 variables:
func2 <- function(time, temp) time + temp
I'd like to keep variable names that have a meaning in the problem (time
and temperature above).
If I have a vector of values
Hello,
I would like to define an arbitrary function of an arbitrary number of
variables, for example, for 2 variables:
func2 <- function(time, temp) time + temp
I'd like to keep variable names that have a meaning in the problem (time and
temperature above).
If I have a vector of values for th
Thanks Jim and others (and sorry Jim - an early version of this slipped
into your inbox :))
Apologies for not giving some concrete code - I was trying to explain in
words.
What I need to do is to fit a simple linear model to successive sections
of a long matrix.
So far, the best solution I have
Hi John,
With due respect to the other respondents, here is something that might help:
# get a vector of values
foo<-rnorm(100)
# get a vector of increasing indices (aka your "recent" values)
bar<-sort(sample(1:100,40))
# write a function to "clump" the adjacent index values
clump_adj_int<-functio
Hopefully Bert and William won't be offended if I more or less summarize:
Are you assuming a loop will take ages, or have you actually tested it? I
wouldn't assume a loop will take ages, or that it will take much longer
than apply().
What's wrong with
apply( X[ {logical expression } , ] , 1, f
John:
1. Please read and follow the posting guide. In particular, provide a
small reproducible example so that we know what your data and looping
code look like.
2. apply-type commands are *not* vectorized; they are disguised loops
that may or may not offer any speedup over explicit loops.
3. A
>It is easy in a loop but that will take ages. Is there any vectorised
>apply-like solution to this?
If you showed the loop that takes ages, along with small inputs for
it (and an indication of how to expand those small inputs to big ones),
someone might be able to show you some code that does the
Folks
Is there any way to get the row index into apply as a variable?
I want a function to do some sums on a small subset of some very long
vectors, rolling through the whole vectors.
apply(X,1,function {do something}, other arguments)
seems to be the way to do it.
The subset I want is the mos
;David L Carlson" , "r-help"
Inviato: Venerdì, 13 maggio 2016 19:22:21
Oggetto: Re: [R] apply formula over columns by subset of rows in a dataframe
(to get a new dataframe)
ave() encapsulates the split/lapply/unsplit stuff so
transform(mydf, v1.mod = ave(v1, blocks, FUN=mynorm
ave() encapsulates the split/lapply/unsplit stuff so
transform(mydf, v1.mod = ave(v1, blocks, FUN=mynorm))
also gives what you got above.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, May 13, 2016 at 7:44 AM, Massimo Bressan <
massimo.bres...@arpa.veneto.it> wrote:
> yes, thanks
>
> yo
yes, thanks
you pointed me in the right direction: split/unplist was the trick
I completely left behind that possibility!
here the final version
mynorm <- function(x) {(x - min(x, na.rm=TRUE))/(max(x, na.rm=TRUE) - min(x,
na.rm=TRUE))}
mydf<-data.frame(blocks=rep(c("a","b","c"
ocks)
> str(mydf2)
'data.frame': 15 obs. of 4 variables:
$ blocks: Factor w/ 3 levels "a","b","c": 1 1 1 1 1 2 2 2 2 2 ...
$ v1: num 19 15 17 22 16 12 24 25 22 18 ...
$ v2: num 35 31 35 31 39 31 19 35 32 38 ...
$ v1mod : num 0.571 0 0.286 1 0.143 ...
-----
hi
I need to apply a user defined formula over some selected columns of a
dataframe by subsetting group of rows (blocks) and get back a new dataframe
I’ve been managed to get the the calculations right but I’m not satisfied at
all by the form of the results
please refer to my reproducible e
.117015
>
> --
> David L. Carlson
> Department of Anthropology
> Texas A&M University
>
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Jim Lemon
> Sent: Saturday, April 2, 2016 6:14 PM
> To: Pedro Mar
2016 6:14 PM
To: Pedro Mardones
Cc: r-help mailing list
Subject: Re: [R] apply mean function to a subset of data
Hi Pedro,
This may not be much of an improvement, but it was a challenge.
selvec<-as.vector(matrix(c(nsel,unlist(by(toy$diam,toy$group,length))-nsel),
ncol=2,byrow=TRUE))
TFv
Hi Pedro,
This may not be much of an improvement, but it was a challenge.
selvec<-as.vector(matrix(c(nsel,unlist(by(toy$diam,toy$group,length))-nsel),
ncol=2,byrow=TRUE))
TFvec<-rep(c(TRUE,FALSE),length.out=length(selvec))
toynsel<-rep(TFvec,selvec)
by(toy[toynsel,]$diam,toy[toynsel,]$group,mean)
Your toy code does not reproduce what you describe: mean(toy$diam[1:nsel[i]])
both times selects from elements of group 1. YOu probably want to subset like
toy$diam[toy$group == i]. Also, if there is any real inefficiency here, it is
_not_ because you are executing a for-loop for two iterations.
Dear all;
This must have a rather simple answer but haven't been able to figure it
out: I have a data frame with say 2 groups (group 1 & 2). I want to select
from group 1 say "n" rows and calculate the mean; then select "m" rows from
group 2 and calculate the mean as well. So far I've been using a
The calculation appears to be sum(a)/(sum(a)+sum(b)).
library(dplyr)
library(tidyr)
result <- ( this_df
%>% gather( group, truth, -c(a,b) )
%>% group_by( group, truth )
%>% summarise( calc = sum(a)/(sum(a)+sum(b)) )
%>% as.data.frame
)
-
Hello all,
I've been banging my head over what must be a simple solution. I would
like to apply a function across columns of a dataframe for rows grouped
across different columns. These groups are not exclusive. See below
for an example. Happy to use dplyr, data.table, or whatever. Any
g
On Oct 6, 2015, at 10:42 AM, Adrienne Wootten wrote:
> R-Helpers,
>
> I've seen some similar threads about this question online, but not quite
> what I'm looking for. I apologize in advance if someone's already answered
> this and I just can't find it online.
>
> Say that I have an array like
Bill,
Thanks a bunch that works great!
A
On Tue, Oct 6, 2015 at 2:56 PM, William Dunlap wrote:
> Since the model matrix, cbind(1,time) is the same for all your
> response variables,
> you can calculate this on one call to lm, but you have to rearrange the
> response
> values so that each x,y s
Since the model matrix, cbind(1,time) is the same for all your
response variables,
you can calculate this on one call to lm, but you have to rearrange the response
values so that each x,y set is in one column. I think the following
function does it:
f <- function (time, y)
{
stopifnot(length(
Almost forgot that function lmfunc is this:
lmfunc = function(valist,input){
fitted.values(lm(valist~input))
}
A
On Tue, Oct 6, 2015 at 2:41 PM, Adrienne Wootten wrote:
> FYI I did try something like this:
>
> test = apply(test3,c(1,2),lmfunc,input=t)
>
> but that gives me an array that is
FYI I did try something like this:
test = apply(test3,c(1,2),lmfunc,input=t)
but that gives me an array that is 10 rows by 5 columns by 5 slices, and I
need it to keep the same dimensions as test3 (5x5x10)
A
On Tue, Oct 6, 2015 at 1:42 PM, Adrienne Wootten wrote:
> R-Helpers,
>
> I've seen so
R-Helpers,
I've seen some similar threads about this question online, but not quite
what I'm looking for. I apologize in advance if someone's already answered
this and I just can't find it online.
Say that I have an array like test3 in the little example code I have below:
test1 = array(rep(1:1
Hi R Colleagues,
I have a small R script that relies on two for-loops to pull data from a
database, make some edits to the data returned from the query, then inserts
the updated data back into the database. The script works just fine, no
problems, except that I am striving to get away from loops,
ON Canada
> -Original Message-
> From: shah_su...@hotmail.co.uk
> Sent: Fri, 24 Jul 2015 04:08:08 -0700 (PDT)
> To: r-help@r-project.org
> Subject: [R] apply kendall tau to a split data set
>
> $`19179222`
>Unique.IDStart.YearL1.Risk.Cat
$`19179222`
Unique.IDStart.YearL1.Risk.Category
Gross.amount.sum
17 19179222 2013 Execution, Delivery & Process Management
161212.1
18 19179222 2015 Execution, Delivery & Process Management
110880.0
$`25182498`
Uniqu
Thank you for your help.
Your explanations have been very useful.
Stefano
Da: Rui Barradas [ruipbarra...@sapo.pt]
Inviato: venerdì 22 maggio 2015 20.26
A: Stefano Sofia; r-help@r-project.org
Oggetto: Re: [R] apply a function to a list of data frames
Hello,
You should change your function to accept only one argument, the
data.frames, and then use lapply (not sapply).
Something like the following.
calc <- function(dat)
{
bias_dmo_max <- round(mean((dat$dmo_12-dat$Eonestep_12), na.rm=TRUE),
digits=2)
rmse_dmo_max <- round(sqrt(mean((dat$d
Where is your code? I see no invocation of sapply.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Fri, May 22, 2015 at 11:02 AM, Stefano Sofia
wrote:
> D
Dear R-users,
given a list of dataframes (like below reported), for each month I need to
apply a function (called calc).
The result should be written in a new list of data frames, one row for each
month.
I have been trying to use sapply, with no success.
Could somebody help me in this?
$df1
da
Many thanks, Petr.You solved my problem.AT.
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-proje
ject.org] On Behalf Of Zilefac
> Elvis via R-help
> Sent: Wednesday, February 18, 2015 4:46 AM
> To: R. Help
> Subject: [R] Apply t-test on list in R
>
> I have a list object in R with dataframe names as:
>
>
> "pav_DJF_histo.csv"&qu
I have a list object in R with dataframe names as:
"pav_DJF_histo.csv""pav_DJF_rcp26_2040s.csv"
"pav_DJF_rcp26_2080s.csv"
"pav_DJF_rcp45_2040s.csv""pav_DJF_rcp45_2080s.csv"
"pav_DJF_rcp85_2040s.csv" "pav_DJF_rcp85_2080s.csv"
"pav_JJA_histo.csv"
"pav_JJA_rcp26_2040s.csv" "pav_JJ
You don't need to do these operations in pieces so the mapply is unnecessary.
Neither dplyr nor data.table can go faster than (assuming your data frame is
called DF):
DF$dtm <- as.POSIXct( as.numeric( DF$ts ), tz="GMT", origin="1970-01-01" )
They can in this case save you from having to retype
I am little lost between all the possibilities to apply a function to
a data.frame or data.table.
Here is mine:
structure(list(name = c("poisonivy", "poisonivy", "poisonivy",
"poisonivy", "poisonivy", "poisonivy", "poisonivy", "poisonivy",
"cruzecontrol", "agreenmamba", "agreenmamba", "vairis", "
Hello,
Please cc the list, the odds of getting more and better answers are greater.
And you should tell us from what package do the function plot.correlog
comes. library(what)?
As for your question, assuming you want to save your plots as PNG files,
you could do something like the following.
> Le 24 oct. 2014 à 09:23, Sarah a écrit :
>
> Thank you very much, it helped a lot!
>
> I just have another question know. I want to make plot for every species. I
> just add the function « plot correlog » to the previous function and I have
> now the following script:
>
> ddeg.correlog.li
Hello,
Yes, you can use lapply. Maybe something like the following. Note that
the result is a list with one member per species. (Untested).
ddeg.correlog.list <- lapply(9:11, function(p)
correlog(plant[plant[,p]=="1", 2], plant[plant[,p]=="1", 3],
plant[plant[,p]=="1", 4]))
Hope this help
Hello List,
I have a database which consist of 912 plots. For each plot, I have the
presence/absence information of 260 species of plants and also 5 different
environmental variables (ddeg, mind, srad, slp, topo).
The dataframe looks like this:
Plot_NumberX Y ddeg mi
Hello,
Have you tried
mapply(f, list_df, list_par, MoreArgs = list(z = fix), SIMPLIFY = FALSE)
?
Hope this helps,
Rui Barradas
Em 14-10-2014 19:42, Carlos Nasher escreveu:
Hi R helpers,
I'm struggling how to apply a function to multiple lists. My function uses
a dataframe, a list of param
Hi R helpers,
I'm struggling how to apply a function to multiple lists. My function uses
a dataframe, a list of parameters and a fixed value as arguments. Now I
want to apply that function to several dataframes contained in list, with
several lists of parameters (also contained in a list) and the
On 02-10-2014, at 11:01, r...@openmailbox.org wrote:
> Subscribers,
>
> What is the correct syntax to apply the 'if else' conditional statement to
> vector objects?
>
> Example:
>
> vectorx<-c(50,50,20,70)
> vectory<-c(50,50,20,20)
> vectorz<-function () {
> if (vectorx>vectory)
>
ct.org] On Behalf Of r...@openmailbox.org
> Sent: Thursday, October 02, 2014 11:02 AM
> To: r-help@r-project.org
> Subject: [R] apply if else statement to vector
>
> Subscribers,
>
> What is the correct syntax to apply the 'if else' conditional statement
> to
Subscribers,
What is the correct syntax to apply the 'if else' conditional statement
to vector objects?
Example:
vectorx<-c(50,50,20,70)
vectory<-c(50,50,20,20)
vectorz<-function () {
if (vectorx>vectory)
vectorx
else vectorx<-0
}
vectorz()
Warning message:
In if (vec
Hi
> -Original Message-
> From: r...@openmailbox.org [mailto:r...@openmailbox.org]
> Sent: Thursday, September 18, 2014 4:35 PM
> To: PIKAL Petr
> Cc: r-help@r-project.org
> Subject: RE: [R] apply block of if statements with menu function
>
> On 2014-09-16
On 2014-09-16 12:35, PIKAL Petr wrote:
So if result of menu is 0 (you did not choose anything) you can either
stay with 0, then switch does not return anything or add 1 and let
evaluate something meaningful specified in second and following
positions of switch command.
Thanks for your explana
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