On 05 Jan 2015, at 00:21 , Pete Brecknock peter.breckn...@bp.com wrote:
n - c(1,2,3,4,5)
lambda - c(0.1,0.8,1.2,2.2,4.2)
mapply(function(x,y) rpois(x,y), n, lambda)
Yes. I'd throw in a SIMPLIFY=FALSE to avoid getting results in a different
format if n is constant (then again, sapply()
thank you for your answer.Yes,that sounds right.I thought the same thing
but the problem is how can i generalize the command for every vector of
numbers not only for the specific example?not only for c(1,2),c(0.1,0.8).
2015-01-04 0:45 GMT+00:00 Pete Brecknock [via R]
dimnik wrote
thank you for your answer.Yes,that sounds right.I thought the same thing
but the problem is how can i generalize the command for every vector of
numbers not only for the specific example?not only for c(1,2),c(0.1,0.8).
2015-01-04 0:45 GMT+00:00 Pete Brecknock [via R]
dimnik wrote
i want to find a functionthattakes in two vectors of numbers
thathave
the same
length.The output should be a listof vectors, where each vector
is a
sequence of
randomly generated Poisson variableswhere the
I can read the documentation, I see why it happens, but who in their right
mind would design a function this way?
I think you're possibly starting from the wrong perspective, or at least it
might be useful to look at it from a different perspective.
In many cases, such as simulations, lapply
Can I follow-up with what I've learned about my own myopia regarding
sapply()?
First, I appreciate all the feedback. After thinking about it for a
while I realized R designers have often chosen to accommodate
interactive usage, and in that context, sapply() returning different
types makes
As you ignored the posting guide and posted in HTML, your below
didn't get through. So one can only guess that it has something to do
with (see ?sapply)
Simplification in sapply is only attempted if X has length greater
than zero and if the return values from all elements of X are all of
the same
Hey thanks for the helpful snark, Bert.
To everyone else, I apologize for neglecting to actually include the
examples.
a - function(i) { list(1) }
b - function(i) { list(1,2) }
ll - sapply(seq(3), a, simplfy=list)
mm - sapply(seq(3), b)
class(ll)
class(mm)
class(ll)
[1] list
class(mm)
[1]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of chris warth
Sent: Friday, January 31, 2014 2:22 PM
To: r-help@r-project.org
Subject: Re: [R] sapply returning list instead of matrix
Hey thanks for the helpful snark, Bert
Pot, meet kettle. You claim to be able to read documentation, yet you don't
reference knowledge gained or clarity lost from such activity in your question.
I think this is a case of inertia of history that we all have to live with at
this point. If you thoroughly read the documentation for
colnames(dd)
#[1] col1 colb
null_vector- colnames(dd)
sapply(null_vector,makeNull,dd)
# col1 colb
#[1,] NA 4
#[2,] 2 NA
#[3,] 3 2
#[4,] 4 NA
#[5,] 1 4
#[6,] NA 5
#[7,] 1 6
A.K.
I am trying to make a column value in a dataframe = NA if there is a 0
or
On 16-05-2013, at 17:31, Patel, Shreena s.pate...@lancaster.ac.uk wrote:
Dear R User,
I'm trying to perform a grid-search for the ML estimator of the Box-Cox
parameter for a linear mixed model. However using sapply to perform the grid
search returns an error message. Here's a small
On Wed, Aug 8, 2012 at 10:37 AM, alijk1989 [via R]
ml-node+s789695n4639622...@n4.nabble.com wrote:
Hi Michael,
Thanks for your response. Here is a simple example of what I am trying to
do:
w=rep(0.02,10)
Q=rep(0.02,10)
rho=matrix(0.5,nrow=10,ncol=10)
m=10
LGD=0.45
M1=sum(sapply(1:m,
Hi,
I have made some progress speeding up my code. This is what I have at the
moment:
M1=sum(sapply(1:m, function(k){sum(sapply(1:m,function(j){w[k]*w[j]*LGD^2
(pmnorm(c(qnorm(Q[k]),qnorm(Q[j])),c(0,0),equicorr(2,rho[k,j]))-Q[k]*Q[j])}))}))
I tried setting up a function as so:
f1 -
On Wed, Aug 8, 2012 at 9:17 AM, alijk1989 [via R]
ml-node+s789695n4639595...@n4.nabble.com wrote:
Hi,
I have made some progress speeding up my code. This is what I have at the
moment:
M1=sum(sapply(1:m, function(k){sum(sapply(1:m,function(j){w[k]*w[j]*LGD^2
Hi Michael,
Thanks for your response. Here is a simple example of what I am trying to
do:
w=rep(0.02,10)
Q=rep(0.02,10)
rho=matrix(0.5,nrow=10,ncol=10)
m=10
LGD=0.45
M1=sum(sapply(1:m,
Dominic,
It's great that you provided some example data, but a much smaller data
frame would have sufficed. For example, 10 randomly selected rows from
your data ...
LF - structure(list(Serra.da.Foladoira = c(27.335652173913,
25.4632608695652,
24.464652173913, 22.550652173913,
On Aug 6, 2012, at 7:34 AM, Dominic Roye wrote:
Hello everyone,
I have a dataset with 5 colums (4 colums with thresholds of weather
stations and one with month - data of 5 years). Now I would like to
calculate the average for each month.
I tried this unsuccessfully:
lf.med -
Thanks again for the help looks like this will be useful for what I'm doing.
Is there any way to use combn to return combinations of values with
themselves:
e.g.
combn(1:3,2)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]111 2 2 3
[2,]1 232 3 3
Take a look at ?expand.grid
Michael
On Aug 4, 2012, at 5:03 PM, alijk1989 [via R]
ml-node+s789695n463919...@n4.nabble.com wrote:
Thanks again for the help looks like this will be useful for what I'm doing.
Is there any way to use combn to return combinations of values with
themselves:
HI,
You can also try this:
d-1:25
A-sample(combn(20:30,2))
B-sample(combn(20:30,2))
lapply(d,function(x) matrix(c(1,A[x],B[x],1),2,2))
[[1]]
[,1] [,2]
[1,]1 23
[2,] 271
[[2]]
[,1] [,2]
[1,]1 21
[2,] 211
[[3]]
[,1] [,2]
[1,]1 29
[2,] 231
[[4]]
Go back and reread the section about the scoping of variables and that
functions do not have side effects; they only return values. You are
changinga local copy of df1 within the function which is returning
the changed values to df3.
On Wed, Jul 11, 2012 at 7:36 PM, Charles Stangor
Le vendredi 03 février 2012 à 18:51 +, William Dunlap a écrit :
Instead of colSums(t(aMatrix)), why not the more
direct rowSums(aMatrix)?
Because I felt it was more didactic. The question was about counting
occurrences per column, so using rowSums() could be a little confusing
without an
3-02-2012, 08:37 (-0800); Filoche escriu:
Hi every one.
I'm learning how to use sapply (and other function of this family).
Here's what I'm trying to do.
I have a vector of lets say 5 elements. I also have a matrix of nX5. I would
like to know how many element by column are inferior to
Le vendredi 03 février 2012 à 18:27 +0100, Ernest Adrogué a écrit :
3-02-2012, 08:37 (-0800); Filoche escriu:
Hi every one.
I'm learning how to use sapply (and other function of this family).
Here's what I'm trying to do.
I have a vector of lets say 5 elements. I also have a
-help-boun...@r-project.org] On
Behalf Of Milan Bouchet-
Valat
Sent: Friday, February 03, 2012 10:17 AM
To: Ernest Adrogué
Cc: r-help@r-project.org
Subject: Re: [R] sapply help
Le vendredi 03 février 2012 à 18:27 +0100, Ernest Adrogué a écrit :
3-02-2012, 08:37 (-0800); Filoche escriu
Thank you sire.
You explained it very well. This give ma a good point to start using sapply
more frequently.
Cordially,
Phil
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You are right that the problem is that DummyFunc isn't vectorized. R
looks for a single logical value in an if statement but x0 gives
it a whole vector's worth -- as the warning indicates, it only uses
the first and pushes the whole vector through the loop in the
return(-x) branch, which explains
Dear Alex,
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alex Zhang
Sent: December-27-11 2:14 PM
To: r-help@r-project.org
Subject: [R] sapply Call Returning the condition has length 1
Error
Dear all,
Happy new
Subject: RE: [R] sapply Call Returning the condition has length 1 Error
Dear Alex,
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alex Zhang
Sent: December-27-11 2:14 PM
To: r-help@r-project.org
Subject: [R] sapply Call
...@mcmaster.ca
To: 'Alex Zhang' alex.zh...@ymail.com
Cc: r-help@r-project.org
Sent: Tuesday, December 27, 2011 3:10 PM
Subject: RE: [R] sapply Call Returning the condition has length 1 Error
Dear Alex,
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r
Dear Alex,
-Original Message-
From: Alex Zhang [mailto:alex.zh...@ymail.com]
Sent: December-27-11 3:34 PM
To: John Fox
Cc: r-help@r-project.org
Subject: Re: [R] sapply Call Returning the condition has length 1
Error
John,
Thanks for the pointers.
The DummyFunc is just
...@mcmaster.ca
To: 'Alex Zhang' alex.zh...@ymail.com
Cc: r-help@r-project.org
Sent: Tuesday, December 27, 2011 4:06 PM
Subject: RE: [R] sapply Call Returning the condition has length 1 Error
Dear Alex,
-Original Message-
From: Alex Zhang [mailto:alex.zh...@ymail.com]
Sent: December-27
!�
From: John Fox j...@mcmaster.ca
To: 'Alex Zhang' alex.zh...@ymail.com
Cc: r-help@r-project.org
Sent: Tuesday, December 27, 2011 4:06 PM
Subject: RE: [R] sapply Call Returning the condition has length 1
Error
Dear Alex,
-Original Message-
From: Alex Zhang [mailto:alex.zh...@ymail.com
' alex.zh...@ymail.com
Cc: r-help@r-project.org
Sent: Tuesday, December 27, 2011 4:06 PM
Subject: RE: [R] sapply Call Returning the condition has length 1
Error
Dear Alex,
-Original Message-
From: Alex Zhang [mailto:alex.zh...@ymail.com]
Sent: December-27-11 3:34 PM
To: John Fox
' alex.zh...@ymail.com
Cc: r-help@r-project.org
Sent: Tuesday, December 27, 2011 4:06 PM
Subject: RE: [R] sapply Call Returning the condition has length 1 Error
Dear Alex,
-Original Message-
From: Alex Zhang [mailto:alex.zh...@ymail.com]
Sent: December-27-11 3:34 PM
To: John Fox
Cc
Zhang alex.zh...@ymail.com
Cc: John Fox j...@mcmaster.ca; r-help@r-project.org r-help@r-project.org
Sent: Tuesday, December 27, 2011 6:59 PM
Subject: Re: [R] sapply Call Returning the condition has length 1 Error
Your puzzle comes from a collision of two somewhat subtle facts that
i) sapply
Hi,
It is probably more confusing with several steps combined, but you are
correct that it is because there are NAs. It is fairly common for R
functions to return NA if there are any NA values unless you
explicitly set an argument on what to do with missing values. A quick
look at ?cor clearly
Hi Weidong Gu,
This works! For my clarity, and so I can repeat this process if need be:
The 'mat' generates a matrix using whatever is supplied to x (i.e.
coop.dat) using the columns from position 9:length(x) of 6 columns (by
row).
The 'rem.col' generates a matrix of the first 1:8 columns of 8
Katrina,
try this.
reorg-function(x){
mat-matrix(x[9:length(x)],ncol=6,byrow=T)
rem.col-matrix(rep(x[1:8],nrow(mat)),byrow=T,ncol=8)
return(data.frame(cbind(rem.col,mat)))
}
co-do.call('rbind',apply(coop.dat,1,function(x) reorg(x)))
You may need to tweak a bit to fit exactly what you want.
On Fri, Aug 12, 2011 at 5:08 PM, Katrina Bennett kebenn...@alaska.edu wrote:
Hi Weidong Gu,
This works! For my clarity, and so I can repeat this process if need be:
The 'mat' generates a matrix using whatever is supplied to x (i.e.
coop.dat) using the columns from position 9:length(x) of 6
I am not sure what purpose the while loop has. However, the main problem
seems to be that you need to put:
i-sample(1:(n-40),1) #This sample from 1 to n-40
rather than
i-sample(1:n-40,1) #this samples one 1:n and then subtracts 40
Otherwise, you may get negative index values
Best,
Daniel
Hi:
samp_func() doesn't return anything. Either (1) type test as the last
line of the function body or (2) don't assign the last sum to an
object.
HTH,
Dennis
On Thu, Aug 11, 2011 at 1:59 PM, Sean Bignami bignam...@gmail.com wrote:
Hello R-universe...
I am having trouble writing a function
The previous two posters basically covered everything, but since I'm on the
train with not too much to do, here's a more detailed response building on
what they said. The following code is shovel-ready and can be pasted
directly to your command line if you have your main data frame called d
Dan,
I am attempting to write a function to count the number of non-missing
values of each column in a data frame using the sapply function. I have the
following code which is receiving the error message below.
n.valid-sapply(data1,sum(!is.na))
Error in !is.na : invalid argument type
Ultimately, I would like for this to be 1 conponent in a larger function
that will produce PROC CONTENTS style output. Something like...
data1.contents-data.frame(Variable=names(data1),
Class=sapply(data1,class),
n.valid=sapply(data1,sum(!is.na)),
n.miss=sapply(data1,sum(is.na)))
Perfect Erik! Thank you!
On Wed, May 4, 2011 at 4:22 PM, Erik Iverson er...@ccbr.umn.edu wrote:
Dan,
I am attempting to write a function to count the number of non-missing
values of each column in a data frame using the sapply function. I have
the
following code which is receiving the
Hi
r-help-boun...@r-project.org napsal dne 04.05.2011 22:26:59:
Erik Iverson er...@ccbr.umn.edu
Odeslal: r-help-boun...@r-project.org
04.05.2011 22:26
Komu
Dan Abner dan.abne...@gmail.com
Ultimately, I would like for this to be 1 conponent in a larger
function
that will
Sent from my iPhone
On Mar 9, 2011, at 5:59 PM, Tomii dioge...@gmail.com wrote:
I try to calculate descriptive statistics for one of the variables in the
data frame, however command sapply calculates these statistics for every
value of the variable separately. How to make it calculate range
Hi:
Perhaps something like this?
m - matrix(rnorm(100, m = 10, s = 2), ncol = 5)
colnames(m) - paste('V', 1:5, sep = '')
# Summary function:
summs - function(x) c(mean = mean(x), sd = sd(x), range = diff(range(x)))
# Apply to columns of m and transpose the result:
t(apply(m, 2, summs))
For
Sent from my iPhone
On Mar 9, 2011, at 6:13 PM, David Winsemius dwinsem...@comcast.net wrote:
Sent from my iPhone
On Mar 9, 2011, at 5:59 PM, Tomii dioge...@gmail.com wrote:
I try to calculate descriptive statistics for one of the variables in the
data frame, however command sapply
sapply(z, function(row) ...) does not actually grab a row at a time out of
'z'. It grabs a column (because 'z' is a data.frame)
You may want:
t(apply(z, 1, function(row) row - means))
or:
t(t(z) - means)
Hope that helps,
-David Johnston
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View this message in context:
On Jan 27, 2011, at 7:16 PM, Ernest Adrogué i Calveras wrote:
Hi,
I have this data.frame with two variables in it,
z
V1 V2
1 10 8
2 NA 18
3 9 7
4 3 NA
5 NA 10
6 11 12
7 13 9
8 12 11
and a vector of means,
means - apply(z, 2, function (col) mean(na.omit(col)))
means
V1
In addition to what has already been suggested you could use ..
mapply(function(x,y) x-y, z,means)
which returns
V1 V2
[1,] 0.333 -2.7142857
[2,] NA 7.2857143
[3,] -0.667 -3.7142857
[4,] -6.667 NA
[5,] NA -0.7142857
[6,]
R works by going down the columns. If you make the rows into columns, it then
does what you want. You just have to make the columns back into rows to get the
original shape of your matrix.
So the code in one line is :
t(t(z) - means)
Original message
Date: Fri, 28 Jan 2011 01:16:45
On Oct 12, 2010, at 12:16 AM, rivercode wrote:
Hi,
I am trying to find the total number of rows for a list of
data.frames and
want to know if there is a better way than using a loop like:
df = { list of data.frame with varying number of rows...each one
has a
column called COL }
r = 0
On Oct 12, 2010, at 12:33 AM, David Winsemius wrote:
On Oct 12, 2010, at 12:16 AM, rivercode wrote:
Hi,
I am trying to find the total number of rows for a list of
data.frames and
want to know if there is a better way than using a loop like:
df = { list of data.frame with varying
will this do what you want:
newTemp[] - lapply(newTemp, function(.col){
+ # convert to character and pad to 5 space
+ sprintf(%5s, as.character(.col))
+ })
str(newTemp)
'data.frame': 5 obs. of 3 variables:
$ DX1: chr 13761 63371 51745 64081 ...
$ DX2: chr 8125 v75 77703 32826
Try this:
formatC(as.matrix(temp))
On Tue, Aug 10, 2010 at 3:55 PM, GL pfl...@shands.ufl.edu wrote:
Using the input below, can I do something more elegant (and more efficient)
than the loop also listed below to pad strings to a width of 5? The true
matrix is about 300K rows and 31 columns.
Both of those approaches seem to return ( v75) instead of (v75 ).
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__
So try:
format(as.matrix(temp))
On Tue, Aug 10, 2010 at 4:13 PM, GL pfl...@shands.ufl.edu wrote:
Both of those approaches seem to return ( v75) instead of (v75 ).
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Sent
That works great, and is ever so much simpler. Thanks much!
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__
Hello Roslina,
Maybe it is just me, but I have difficulty picking apart what you are
trying to do because, the data have the same names as the arguments in
your functions, and when you create the function term(), you have two
sets of arguments (for term() and for gam_sum() ) that have the same
Try this:
1 + (1 / log(length(lambda_cor))) * sum((l - lambda_cor /
length(lambda_cor)) * log(l))
On Sun, May 16, 2010 at 10:43 PM, Roslina Zakaria zrosl...@yahoo.comwrote:
Hi r-users,
I have this code here, but I just wonder how do I use 'sapply' to make it
more efficient
lamda_cor -
Dear R-gurus..
How do I implement the following:
a) Overlay frequency(instead of density) with line of density plot, vertical
lines of confidence intervals and reference levels?
b) Control the breaks (using nint?), order of the panel, and the layout,
place units for each conditioning variable?
Sundar Dorai-Raj-2 wrote:
Or perhaps more clearly,
histogram(~a1 + b1 + c1, data = aa, outer = TRUE)
Why outer=TRUE? Looks same for me without:
Dieter
library(lattice)
aa - data.frame(a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5))
histogram(~a1 + b1 + c1, data = aa)
--
View this
You're right. It's necessary for xyplot though to prevent grouping.
On Mar 20, 2010 10:43 AM, Dieter Menne dieter.me...@menne-biomed.de
wrote:
Sundar Dorai-Raj-2 wrote:
Or perhaps more clearly,
histogram(~a1 + b1 + c1, data = aa, o...
Why outer=TRUE? Looks same for me without:
Dieter
Try this:
junk - sapply(aa,function(x) print(histogram(x,breaks=NULL)))
or, shorter:
for(a in aa) print(histogram(a, breaks = NULL)
On Fri, Mar 19, 2010 at 5:44 PM, Santosh santosh2...@gmail.com wrote:
Dear R-gurus
aa - data.frame(a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5))
Thanks for your response.
how do I print them in an ordered manner, akin to using
print(px,split=c(2,2,1,1),more=T)) or par(mfrow=c(x,y))?
-Santosh
On Fri, Mar 19, 2010 at 2:58 PM, Gabor Grothendieck ggrothendi...@gmail.com
wrote:
Try this:
junk - sapply(aa,function(x)
Try this:
histogram(~ values | ind, stack(aa))
On Fri, Mar 19, 2010 at 5:44 PM, Santosh santosh2...@gmail.com wrote:
Dear R-gurus
aa - data.frame(a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5))
sapply(aa,function(x) histogram(x,breaks=NULL))
or px - sapply(aa,function(x)
Or perhaps more clearly,
histogram(~a1 + b1 + c1, data = aa, outer = TRUE)
--sundar
On Fri, Mar 19, 2010 at 3:50 PM, Gabor Grothendieck ggrothendi...@gmail.com
wrote:
Try this:
histogram(~ values | ind, stack(aa))
On Fri, Mar 19, 2010 at 5:44 PM, Santosh santosh2...@gmail.com wrote:
Try this:
data$score - ave(data$score, data$group, FUN = prop.table)
On Sun, Aug 30, 2009 at 6:08 PM, Noah Silvermann...@smartmediacorp.com wrote:
Hi,
I need a bit of guidance with the sapply function. I've read the help page,
but am still a bit unsure how to use it.
I have a large data
On 30/08/2009 6:08 PM, Noah Silverman wrote:
Hi,
I need a bit of guidance with the sapply function. I've read the help
page, but am still a bit unsure how to use it.
I have a large data frame with about 100 columns and 30,000 rows. One
of the columns is group of which there are about
On Sun, Aug 30, 2009 at 5:08 PM, Noah Silvermann...@smartmediacorp.com wrote:
Hi,
I need a bit of guidance with the sapply function. I've read the help page,
but am still a bit unsure how to use it.
I have a large data frame with about 100 columns and 30,000 rows. One of
the columns is
On Jun 17, 2009, at 10:06 AM, Girish A.R. wrote:
Hi folks,
I'm trying to consolidate the outputs (of anova() and lrm()) from
multiple runs of single-variable logistic regression. Here's how the
output looks:
Thanks, Marc! This is what I was looking for.
best,
-Girish
PS: Also appreciate your concern about this being a part of a variable
selection process.
On Jun 17, 9:01 pm, Marc Schwartz marc_schwa...@me.com wrote:
On Jun 17, 2009, at 10:06 AM, Girish A.R. wrote:
Hi folks,
I'm trying to
I'm not sure what you really want, so perhaps a simple example would
help (i.e. what a sample of the input looks like and what the output
you need looks like). My guess would be
sapply(df, diff)
but again, I'm not sure.
--sundar
On Sun, Feb 8, 2009 at 4:24 PM, glenn
Bullseye ! thanks a lot Sundar
-Original Message-
From: Sundar Dorai-Raj [mailto:sdorai...@gmail.com]
Sent: 09 February 2009 00:31
To: glenn
Cc: r-help@r-project.org
Subject: Re: [R] sapply
I'm not sure what you really want, so perhaps a simple example would
help (i.e. what a sample
Unfortunately, I have the same error message.
lapply(rowsplit, function(x)mean(x[,sapply(x, is.numeric)])) works but not
with median.
Strange, isn't it?
Any other idea?
Thanks in advance,
Ptit Bleu.
Henrique Dallazuanna wrote:
Try this:
lapply(l, function(x)median(x[,sapply(x,
You can provide a example of your data?
On Fri, Nov 7, 2008 at 9:14 AM, Ptit_Bleu [EMAIL PROTECTED] wrote:
Unfortunately, I have the same error message.
lapply(rowsplit, function(x)mean(x[,sapply(x, is.numeric)])) works but not
with median.
Strange, isn't it?
Any other idea?
Thanks in
I haven't looked at the detail, but I guess the answer is that mean works on
a data frame while median doesn't.
?mean
snip
For a data frame, a named vector with the appropriate method being applied
column by column.
-
I guess to use median you'll need nested '[l/s]apply's, the
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