On 05 Jan 2015, at 00:21 , Pete Brecknock 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() in the original q
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] <
> ml
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] <
ml-node+s789695n4701358...@n
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 variableswher
> 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, lappl
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 perfe
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 ?sapp
p
TIBCO Software
wdunlap tibco.com
> -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
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)
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 sa
Can anyone suggest a rationale for why sapply() returns different types
(list and matrix) in the two examples below? Is there any way to get
sapply() or any other apply() function to return a matrix in both cases?
simplify=TRUE doesn't change the outcome.
I understand why it is happening, I just
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
On 16-05-2013, at 17:31, "Patel, Shreena" 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 example to demonstrate:
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 example to demonstrate:
library(lme4)
# Function to fit model for a given lambda:
bc
On Wed, Aug 8, 2012 at 10:37 AM, alijk1989 [via R]
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,
> function(k){sum(sapply(1:
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,
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,r
On Wed, Aug 8, 2012 at 9:17 AM, alijk1989 [via R]
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
> (pmnorm(c(qnorm(Q[k]),qnorm(Q[j])),c(0,0),equicorr(2,rho[k,j]
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 <- function
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 <- sapply(LF[,1:4
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, 22.217782
Take a look at ?expand.grid
Michael
On Aug 4, 2012, at 5:03 PM, "alijk1989 [via R]"
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:
>
> e.g.
>
>> combn(1:3,2)
>
>
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
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
wrote:
> Wh
Why does this sapply code change df3 but not df1?
Thanks
df1 <- read.table(text="
cola colb colc cold cole
1NA59 NA 17
2NA6 NA 14 NA
3 3NA 11 15 19
4 48 12 NA 20
", header=TRUE)
df2 <-df1*2
df1
df2
df3 <-sapply(names(df1),function(x) {df
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 ex
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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__
potfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-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
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 al
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 infe
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 each element of my
vector.
On this example:
v = c(
n Fox ; "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() is a wrapper for lapply(), not apply() a
function to do it where * stands
> for any function.
>
> Thanks a lot!
>
>
> ____
> From: John Fox
> To: 'Alex Zhang'
> Cc: r-help@r-project.org
> Sent: Tuesday, December 27, 2011 4:06 PM
> Subject: RE: [R] sapply Call R
> "how" not just "what" ;). Is there a way using�a "*apply" function to
>> do it where * stands for any function.
>>
>> Thanks a lot!�
>>
>>
>>
>> From: John Fox
>> To: 'Alex Zhang
uot;*apply" function to
>do it where * stands for any function.
>
>Thanks a lot!�
>
>
>________
> From: John Fox
>To: 'Alex Zhang'
>Cc: r-help@r-project.org
>Sent: Tuesday, December 27, 2011 4:06 PM
>Subject: RE: [R] sapply C
a lot!
From: John Fox
To: 'Alex Zhang'
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
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
ole columns? Thanks.
>
>
>
> From: John Fox
> To: 'Alex Zhang'
> 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
>
>
s? That explains the behavior. Is there an
"*apply" function that will fee elements of the input data.frame into "FUN"
instead of whole columns? Thanks.
From: John Fox
To: 'Alex Zhang'
Cc: r-help@r-project.org
Sent: Tuesday, Dece
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 >
You are right that the problem is that "DummyFunc" isn't vectorized. R
looks for a single logical value in an "if" statement but "x>0" 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 exp
Dear all,
Happy new year!
I have a question re using sapply. Below is a dummy example that would
replicate the error I saw.
##Code Starts here
DummyFunc <- function(x) {
if (x > 0) {
return (x)
} else
{
return (-x)
}
}
Y = data.frame(val = c(-3:7))
sapply(Y, FUN = DummyFunc)
##Code ends her
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 s
Hello. I am wondering why I am getting NA for all
in cors=sapply(pred,cor,y=resp). I suppose that each column in pred has NAs in
them. Is there some way to fix this? Thanks
> str(pred)
'data.frame': 200 obs. of 13 variables:
$ mnO2: num 9.8 8 11.4 4.8 9 13.1 10.3 10.6 3.4 9.9 ...
$ Cl :
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 c
On Fri, Aug 12, 2011 at 5:08 PM, Katrina Bennett 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 columns (by
> ro
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.
Hi R-help,
I am working with US COOP network station data and the files are
concatenated in single rows for all years, but I need to pull these
apart into rows for each day. To do this, I need to extract part of
each row such as station id, year, mo, and repeat this against other
variables in the
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"
ava
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 wrote:
> Hello R-universe...
>
> I am having trouble writing a function which contains a l
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
Hello R-universe...
I am having trouble writing a function which contains a loop so I can sapply()
it to a list of data frames
Each data frame has 241 observations of 15 variables. My loop takes a random
sample of one row until the 40 consecutive rows after the sample have a
d2p(variable) sum
Hi
r-help-boun...@r-project.org napsal dne 04.05.2011 22:26:59:
> Erik Iverson
> Odeslal: r-help-boun...@r-project.org
>
> 04.05.2011 22:26
>
> Komu
>
> Dan Abner
>
>
> > Ultimately, I would like for this to be 1 conponent in a larger
function
> > that will produce PROC CONTENTS style ou
Perfect Erik! Thank you!
On Wed, May 4, 2011 at 4:22 PM, Erik Iverson 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 error me
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)))
data1.cont
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
Tha
Hello everyone,
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 ty
Sent from my iPhone
On Mar 9, 2011, at 6:13 PM, David Winsemius wrote:
>
>
> Sent from my iPhone
>
> On Mar 9, 2011, at 5:59 PM, Tomii wrote:
>
>> I try to calculate descriptive statistics for one of the variables in the
>> data frame, however command sapply calculates these statistics fo
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 5:59 PM, Tomii 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 (as well as
> o
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 (as well as
other statistics) for all column?
Here are commands and results:
> as1$tr
:45 +0100
>From: r-help-boun...@r-project.org (on behalf of nfdi...@gmail.com (Ernest
>Adrogué i Calveras))
>Subject: [R] sapply puzzlement
>To: r-help@r-project.org
>
>Hi,
>
>I have this data.frame with two variables in it,
>
>> z
> V1 V2
>1 10 8
>2 NA
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,] 1.333
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
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
--
View this message in context:
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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
V1V2
9.67 10.714286
My intention was substracting mean
y=sapply(1:nrow(x), function(i) sapply(1:ncol(x), function(j)
getrc(i, j)))
Arvin Zolfaghari, Ph.D.
Aon Benfield | Aon Benfield Analytics
55 Bishopsgate, London, EC2N 3BD, UK
t: +44 20 7522 8241 | f: +44 20 7522 3846| M: +44 78 8940 6637
e: arvin.zolfagh...@aonbenfield.com | w:
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 number
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 =
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
> for (i in 1:length(df)) {
+ r = r + length(n[[i]]$CO
That works great, and is ever so much simpler. Thanks much!
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R-help@r-projec
So try:
format(as.matrix(temp))
On Tue, Aug 10, 2010 at 4:13 PM, GL wrote:
>
> Both of those approaches seem to return (" v75") instead of ("v75 ").
> --
> View this message in context:
> http://r.789695.n4.nabble.com/sapply-lapply-instead-of-loop-tp2320265p2320305.html
> Sent from the R help
Both of those approaches seem to return (" v75") instead of ("v75 ").
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R-
Try this:
formatC(as.matrix(temp))
On Tue, Aug 10, 2010 at 3:55 PM, GL 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.
>
> #
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" "
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.
###
#INPUT
###
> temp
DX1 DX2 DX3
1 13761 8125 49178
2 6
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
nam
Hi r-users,
I have this code here :
dt <- winter_pos_sum
bt <- c(24.96874, 19.67861, 23.51001, 19.86868); round(bt,2)
alp <- c(2.724234, 3.914649, 3.229146, 3.120719); round(alp,2)
bt_min <- min(bt) ; bt_min
p <- alp_sum ; p
t <- 50
t1 <- t+1
#first get the sum ove
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 wrote:
> Hi r-users,
>
> I have this code here, but I just wonder how do I use 'sapply' to make it
> more efficient
>
> lamda_cor <- eigen(winter_
Hi r-users,
I have this code here, but I just wonder how do I use 'sapply' to make it more
efficient
lamda_cor <- eigen(winter_cor)$values
> lamda_cor
[1] 1.3459066 1.0368399 0.8958128 0.7214407
lamda_cxn <- function(dt)
{ n <- length(dt)
term <- vector(length=n, mode="numeric")
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?
You're right. It's necessary for xyplot though to prevent grouping.
On Mar 20, 2010 10:43 AM, "Dieter Menne"
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
library(lattice)
aa <- dat
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
Or perhaps more clearly,
histogram(~a1 + b1 + c1, data = aa, outer = TRUE)
--sundar
On Fri, Mar 19, 2010 at 3:50 PM, Gabor Grothendieck wrote:
> Try this:
>
> histogram(~ values | ind, stack(aa))
>
>
> On Fri, Mar 19, 2010 at 5:44 PM, Santosh wrote:
> > Dear R-gurus
> >
> > aa <- data.frame(a
Try this:
histogram(~ values | ind, stack(aa))
On Fri, Mar 19, 2010 at 5:44 PM, Santosh 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) histogram(x,breaks=NULL))
> pri
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 wrote:
> Try this:
>
> junk <- sapply(aa,function(x) print(histogram(x,breaks=NULL)))
>
> or, s
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 wrote:
> Dear R-gurus
>
> aa <- data.frame(a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5))
> sapply(aa,function(x) hist
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) histogram(x,breaks=NULL))
print(px,split=c(1,1,1,1),more=F)
The above code does not seem to work. am I missing something?
Thanks,
Santosh
On Sun, Aug 30, 2009 at 5:08 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 th
Fortune candidate?
cheers,
Rolf Turner
On 30/08/2009 6:08 PM, Noah Silverman wrote:
Normally, I would just write a huge "for each" loop, but have read
that
is hugely inefficient with R.
On 31/08/2009, at 10:47 AM, Duncan Murdoch wrote:
Don't believe
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 2,0
Try this:
data$score <- ave(data$score, data$group, FUN = prop.table)
On Sun, Aug 30, 2009 at 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 1
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 2,000 distinct "groups".
I want to normalize (s
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 wrote:
> On Jun 17, 2009, at 10:06 AM, Girish A.R. wrote:
>
>
>
> > Hi folks,
>
> > I'm trying to consolidate
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:
y
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:
y ~ x1 y ~ x2 y ~ x3 y ~
x4
Chi-Squa
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 samp
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 wrote:
> Newbie question so
Newbie question sorry (have tried the help pages I promise)
I have a dataframe (date,stockprice) say and looking how I might get the
return of: dataframe (difference in days, change in stock price) using
sapply - I require a very simple function and don't really want to go down
the zoo and quan
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
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 outer
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