Re: [R] lapply and runif issue?

Your lapply is making the call runif(n=3, min=i) for i in 1:3. That runif's 3 argument is 'max', with default value 1 so that is equivalent to calling runif(n=3, min=i, max=1) When i>max, outside the domain of the family of uniform distributions, runif returns NaN's. Bill Dunlap TIBCO Soft

Re: [R] lapply and runif issue?

Thanks, Ista. That explains it. What I missed is the following "note" in ?lapply: "This means that the recorded call is always of the form FUN(X[[i]], ...), with i replaced by the current (integer or double) index. " That being the case, X[[i]] gets passed to the first available argument, which

Re: [R] lapply and runif issue?

Hi Bert, On Tue, Nov 14, 2017 at 8:11 PM, Bert Gunter wrote: > Could someone please explain the following? I did check bug reports, but > did not recognize the issue there. I am reluctant to call it a bug, as it > is much more likely my misunderstanding. Ergo my request for clarification: > > ##

[R] lapply and runif issue?

Could someone please explain the following? I did check bug reports, but did not recognize the issue there. I am reluctant to call it a bug, as it is much more likely my misunderstanding. Ergo my request for clarification: ## As expected: > lapply(1:3, rnorm, n = 3) [[1]] [1] 2.481575 1.998182 1.

Re: [R] lapply

> On Jul 30, 2016, at 7:53 PM, roslinazairimah zakaria > wrote: > > Dear r-users, > > I would like to use lapply for the following task: > > ## Kolmogorov-Smirnov > ks.test(stn_all[,1][stn_all[,1] > 0],stn_all_gen[,1][stn_all_gen[,1] > 0]) > ks.test(stn_all[,2][stn_all[,2] > 0],stn_all_gen[,2

[R] lapply

Dear r-users, I would like to use lapply for the following task: ## Kolmogorov-Smirnov ks.test(stn_all[,1][stn_all[,1] > 0],stn_all_gen[,1][stn_all_gen[,1] > 0]) ks.test(stn_all[,2][stn_all[,2] > 0],stn_all_gen[,2][stn_all_gen[,2] > 0]) ks.test(stn_all[,3][stn_all[,3] > 0],stn_all_gen[,3][stn_all

[R] lapply on list of lists

Dear R list users, I have three lists of data frames, respectively temp_list, wind_list and snow_list. The elements of these three lists are temp_list$station1, temp_list$station2 and temp_list$station3 with columns date and temp; wind_list$station1, wind_list$station2 and wind_list$station3 wit

Re: [R] lapply function

You need the vectorizes ifelse() instead of if(). Also watch out for order of operations in the last line, and there is already a base R function named scale(). And spelling of arguments, of course. On Saturday, May 30, 2015, Sohail Khan wrote: > Hi R Gurus, > > I am writing a simple function t

[R] lapply function

Hi R Gurus, I am writing a simple function that take a numeric vector column from a data frame and scales the vector column with certain criteria. I would then pass this function to a list of dataframes by lappy. Question is how do I write a function that works on a numeric vector. My function

Re: [R] lapply returns NULL ?

; , "ce" , r-help@r-project.org Subject: Re: [R] lapply returns NULL ? I think that removing them is something the OP doesn't understand how to do. The lapply function ALWAYS produces an output element for every input element. If this is not what you want then you need to choose

Re: [R] lapply returns NULL ?

Thanks Jeff et. all, This is exactly what I needed. -Original Message- From: "Jeff Newmiller" [jdnew...@dcn.davis.ca.us] Date: 07/12/2014 10:38 AM To: "Uwe Ligges" , "ce" , r-help@r-project.org Subject: Re: [R] lapply returns NULL ? I think that rem

Re: [R] lapply returns NULL ?

I think that removing them is something the OP doesn't understand how to do. The lapply function ALWAYS produces an output element for every input element. If this is not what you want then you need to choose a looping structure that is not so tightly linked to the input, such as a for loop (unt

Re: [R] lapply returns NULL ?

Hello, Try the following. res <- lapply(foo, function(x) if(x[1] == 1 ) x ) res[!sapply(res, is.null)] Hope this helps, Rui Barradas Em 12-07-2014 14:25, ce escreveu: Dear all, I have a list of arrays : foo<-list(A = c(1,3), B =c(1, 2), C = c(3, 1)) foo $A [1] 1 3 $B [1] 1 2 $C [1

Re: [R] lapply returns NULL ?

On 12.07.2014 15:25, ce wrote: Dear all, I have a list of arrays : foo<-list(A = c(1,3), B =c(1, 2), C = c(3, 1)) foo $A [1] 1 3 $B [1] 1 2 $C [1] 3 1 if( foo$C[1] == 1 ) foo$C[1] lapply(foo, function(x) if(x[1] == 1 ) x ) $A [1] 1 3 $B [1] 1 2 $C NULL I don't want to lis

[R] lapply returns NULL ?

Dear all, I have a list of arrays : foo<-list(A = c(1,3), B =c(1, 2), C = c(3, 1)) > foo $A [1] 1 3 $B [1] 1 2 $C [1] 3 1 > if( foo$C[1] == 1 ) foo$C[1] > lapply(foo, function(x) if(x[1] == 1 ) x ) $A [1] 1 3 $B [1] 1 2 $C NULL I don't want to list $C NULL in the output. How I can

Re: [R] Lapply to create sub categories based on categorical data

Hi, Try: x <- c(rep("A",0.1*1),rep("B",0.2*1),rep("C",0.65*1),rep("D",0.05*1)) set.seed(24) categorical_data <- sample(x,1) set.seed(49) p_val <- runif(1,0,1) combi <- data.frame(V1=categorical_data,V2=p_val) variables <- unique(combi$V1)  res <- lapply(levels(variables)

Re: [R] lapply?

Hi, the output of lapply() is a list; see ?lapply and ?sapply. # if you know the length of your list in advance, # this definition is better: uu <- vector("list", 2) # list elements uu[[1]] <- c(1,2,3) uu[[2]] <- c(3,4,5) # some options to achieve what you want: matrix(unlist(uu), 2, 3, T) do

Re: [R] lapply?

Thanks all! So many ways On Thu, Nov 14, 2013 at 10:35 AM, Rui Barradas wrote: > Hello, > > You are applying rbind to each element of the list, not rbinding it with > the others. Try instead > > do.call(rbind, uu) > > Hope this helps, > > Rui Barradas > > Em 14-11-2013 15:20, Brian Smith es

Re: [R] lapply?

Hello, You are applying rbind to each element of the list, not rbinding it with the others. Try instead do.call(rbind, uu) Hope this helps, Rui Barradas Em 14-11-2013 15:20, Brian Smith escreveu: Hi, I was trying to use lapply to create a matrix from a list: uu <- list() uu[[1]] <- c(1,2

Re: [R] lapply?

On 14-11-2013, at 16:20, Brian Smith wrote: > Hi, > > I was trying to use lapply to create a matrix from a list: > > uu <- list() > uu[[1]] <- c(1,2,3) > uu[[2]] <- c(3,4,5) > > The output I desire is a matrix with 2 rows and 3 columns, so I try: > > xx <- lapply(uu,rbind) > > Obviously, I'

[R] lapply?

Hi, I was trying to use lapply to create a matrix from a list: uu <- list() uu[[1]] <- c(1,2,3) uu[[2]] <- c(3,4,5) The output I desire is a matrix with 2 rows and 3 columns, so I try: xx <- lapply(uu,rbind) Obviously, I'm not doing something right, but what!? [[alternative HTML versi

Re: [R] lapply to multivariate function?

Hello, I have no experience with packages foreach and doMC. But I believe that paralel computing only pays if the datasets are really large, due to the setup time. Maybe "thousands of observations" is not that large. Rui Barradas Em 01-09-2013 22:21, Ignacio Martinez escreveu: Thanks a lot

Re: [R] lapply to multivariate function?

Thanks a lot Rui. Loops make sense to me. I made one modification to your code. I have thousands of observation, so I would like to run it in parallel. This is my reproducible example: # Make Data Frame for video actions between given times for user X DataVideoActionT <- function (userX, Time1, Ti

Re: [R] lapply to multivariate function?

Hello, Your example doesn't really run, but for what I've seen, if your second data frame is named dat2, something along the lines of n <- nrow(dat2) res <- list("vector", n) for(i in 1:n){ res[[i]] <- with(dat2, DataVideoActionT(anon_ID[i], Time1[i], TimeM[i], TimeL[i])) } do.call(rbind,

Re: [R] lapply to multivariate function?

I hope this reproduceble example helps understand what I'm trying to do. This is the function: # Make Data Frame for video actions between given times for user X DataVideoActionT <- function (userX, Time1, Time2, Time3){ #Get data for user X videoActionsX<-subset(videoLectureActions, username

Re: [R] lapply to multivariate function?

Oh, another possibility is ?mapply, which I should have pointed out in my previous reply. Sorry. -- Bert On Sun, Sep 1, 2013 at 8:30 AM, Bert Gunter wrote: > Rui et.al.: > > But apply will not work if the data frame has columns of different > classes/types, as appears to be the case here. Viz,

Re: [R] lapply to multivariate function?

Rui et.al.: But apply will not work if the data frame has columns of different classes/types, as appears to be the case here. Viz, from ?apply: "If X is not an array but an object of a class with a non-null dim value (such as a data frame),apply

Re: [R] lapply to multivariate function?

Hello, Maybe you need apply, not lapply. It seems you want to apply() a function to the first dimension of your data.frame, something like apply(dat, 1, fun) #apply by rows Hope this helps, Rui Barradas Em 01-09-2013 15:00, Ignacio Martinez escreveu: I have a Data Frame that contains, be

[R] lapply to multivariate function?

I have a Data Frame that contains, between other things, the following fields: userX, Time1, Time2, Time3. The number of observations is 2000. I have a function that has as inputs userX, Time1, Time2, Time3 and return a data frame with 1 observation and 19 variables. I want to apply that function

Re: [R] lapply and SpatialGridDataFrame error

Irucka Embry mail2world.com> writes: > > Hi all, I have a set of 54 files that I need to convert from ASCII grid > format to .shp files to .bnd files for BayesX. > > I have the following R code to operate on those files: > > library(maptools) > library(Grid2Polygons) > library(BayesX) > librar

[R] lapply and SpatialGridDataFrame error

Hi all, I have a set of 54 files that I need to convert from ASCII grid format to .shp files to .bnd files for BayesX. I have the following R code to operate on those files: library(maptools) library(Grid2Polygons) library(BayesX) library(BayesXsrc) library(R2BayesX) readfunct <- function(x) { u

Re: [R] lapply (and friends) with data.frames are slow

Hi David, Yes, it is - although the SO question was more directed at figuring out why sapply seemed slower, the question to R-help is more nuanced in "is this coercion really necessary for data.frames?", and I figured it might take some more knowledge of R internals / the difference between lists

Re: [R] lapply (and friends) with data.frames are slow

On Jan 5, 2013, at 11:38 AM, Kevin Ushey wrote: Hey guys, I noticed something curious in the lapply call. I'll copy+paste the function call here because it's short enough: lapply <- function (X, FUN, ...) { FUN <- match.fun(FUN) if (!is.vector(X) || is.object(X)) X <- as.list(X)

Re: [R] lapply (and friends) with data.frames are slow

On Sat, Jan 5, 2013 at 7:38 PM, Kevin Ushey wrote: > Hey guys, > > I noticed something curious in the lapply call. I'll copy+paste the > function call here because it's short enough: > > lapply <- function (X, FUN, ...) > { > FUN <- match.fun(FUN) > if (!is.vector(X) || is.object(X)) >

[R] lapply (and friends) with data.frames are slow

Hey guys, I noticed something curious in the lapply call. I'll copy+paste the function call here because it's short enough: lapply <- function (X, FUN, ...) { FUN <- match.fun(FUN) if (!is.vector(X) || is.object(X)) X <- as.list(X) .Internal(lapply(X, FUN)) } Notice that lapp

[R] lapply and kernelUD (adehabitatHR package): Home Range kernel estimation for a list of individuals

Dear R experts, I'm using the adehabitatHR package in order to perform a kernel analysis and estimate the home range of my input data (GPS relocations of 42 individuals). I've done the analysis for one of the individuals and it worked perfectly (see code below). But now I'm trying to use a list a

Re: [R] lapply with different size lists?

Hello, When you don't know what's going on, break long instructions into simpler ones. R is good at doing a lot in one line, to debug don't do that. b = function (m, mat) { n=nrow (mat) p=ceiling (n/m) lapply (1:p, function (l,n,m) { inf = ((l-1)*m)+1 if (lIf you wa

[R] lapply with different size lists?

Hello, I have 2 functions (a and b) a = function(n) { matrix (runif(n*2,0.0,1), n) } > > > b = function (m, matrix) { > n=nrow (matrix) > p=ceiling (n/m) > lapply (1:p, function (l,n,m) { > inf = ((l-1)*m)+1 > if (l

Re: [R] lapply and paste

On Thu, Mar 29, 2012 at 7:44 AM, Ed Siefker wrote: > Thank you, I was confused about that.  What exactly is lapply for then, > if R handles this kind of thing automatically?  Are there functions that are > not "vectorized"? > There are, especially ones you write yourself that don't need to be vec

Re: [R] lapply and paste

Yes, there are non-vectorized functions e.g, integrate(), or you can use lapply() to apply a vectorized function to each element of a list (which is not what suff was) individually: x <- list(1:3, 1:4, 1:5, 2:7) mean(x) # bad lapply(x, mean) #good Michael On Mar 28, 2012, at 2:44 PM, Ed Siefke

Re: [R] lapply and paste

suff isn't a list, so lapply() isn't the right choice. How about instead: > suff<- c("C1", "C2", "C3") > sapply(suff, function(x)paste("filename_", x, ".ext", sep="")) C1C2C3 "filename_C1.ext" "filename_C2.ext" "filename_C3.ext" On Wed, Mar 28, 2012

Re: [R] lapply and paste

Thank you, I was confused about that. What exactly is lapply for then, if R handles this kind of thing automatically? Are there functions that are not "vectorized"? On Wed, Mar 28, 2012 at 1:37 PM, R. Michael Weylandt wrote: > I think you're confused about the need for lapply -- paste is > vec

Re: [R] lapply and paste

I think you're confused about the need for lapply -- paste is vectorized so this paste("filename_", suff, ".ext", sep = "") will work. But if you want to use lapply (for whatever reason) try this: lapply(suff, function(x) paste("filename_", x, ".ext", sep = "") Michael On Wed, Mar 28, 2012 at

[R] lapply and paste

I have a list of suffixes I want to turn into file names with extensions. suff<- c("C1", "C2", "C3") paste("filename_", suff[[1]], ".ext", sep="") [1] "filename_C1.ext" How do I use lapply() on that call to paste()? What's the right way to do this: filenames <- lapply(suff, paste, ...) ? Can

Re: [R] lapply to change variable names and variable values

Thanks both! That solves ! You've made a very happy newbie! Simon On 2012-03-12, at 2:52 PM, Sarah Goslee wrote: > Hi Simon, > > On Mon, Mar 12, 2012 at 2:37 PM, Simon Kiss wrote: >> Hi: I'm sure this is a very easy problem. I've consulted Data Manipulation >> With R and the R Book and can't fi

Re: [R] lapply to change variable names and variable values

Your function doesn't return the new data frame but rather the new names. Note, e.g. x <- 1:2 names(x) <- letters[1:2] .Last.value # Not x! Try this: .xx<- lapply(.xx, function(x) {colnames(x)<-c('State', 'Year'); x}) or more explicitly .xx<- lapply(.xx, function(x) {colnames(x)<-c('State', '

Re: [R] lapply to change variable names and variable values

Hi Simon, On Mon, Mar 12, 2012 at 2:37 PM, Simon Kiss wrote: > Hi: I'm sure this is a very easy problem. I've consulted Data Manipulation > With R and the R Book and can't find an answer. > > Sample list of data frames looks as follows: > > .xx<-list(df<-data.frame(Var1=rep('Alabama', 400), Var2

Re: [R] lapply to change variable names and variable values

Hi, On Mon, Mar 12, 2012 at 2:37 PM, Simon Kiss wrote: > Hi: I'm sure this is a very easy problem. I've consulted Data Manipulation > With R and the R Book and can't find an answer. > > Sample list of data frames looks as follows: > > .xx<-list(df<-data.frame(Var1=rep('Alabama', 400), Var2=rep(c

[R] lapply to change variable names and variable values

Hi: I'm sure this is a very easy problem. I've consulted Data Manipulation With R and the R Book and can't find an answer. Sample list of data frames looks as follows: .xx<-list(df<-data.frame(Var1=rep('Alabama', 400), Var2=rep(c(2004, 2005, 2006, 2007), 400)), df2<-data.frame(Var1=rep('Tennes

Re: [R] lapply to list of variables

Hi hi, It is much easier to deal with lists than a large number of separate objects. So the first answer to your question > How can I apply a function to a list of variables. .. might be to convert your "list of variables" to a regular list. Instead of ... monday <- 1:3 tuesday <- 4:7 wednesday

Re: [R] lapply to list of variables

Hi: Here's another way of doing this on the simplified version of your example: L <- vector('list', 3) # initialize a list of three components ## populate it for(i in seq_along(L)) L[[i]] <- rnorm(20, 10, 3) ## name the components names(L) <- c('Monday', 'Tuesday', 'Wednesday') ## replace va

Re: [R] lapply to list of variables

On 08.11.2011 17:59, Ana wrote: Hi Can someone help me with this? How can I apply a function to a list of variables. something like this listvar=list("Monday","Tuesday","Wednesday") This is a list of length one character vectors rather than a "list of variables". func=function(x){x

[R] lapply to list of variables

Hi Can someone help me with this? How can I apply a function to a list of variables. something like this listvar=list("Monday","Tuesday","Wednesday") func=function(x){x[which(x<=10)]=NA} lapply(listvar, func) were Monday=[213,56,345,33,34,678,444] Tuesday=[213,56,345,33,34,678,444] ... in m

Re: [R] lapply and Two TimeStamps as input

alaios wrote: > > Dear all, > > I have a function that recognizes the following format for timestamps > "%Y-%m-%d %H:%M:%S" > > my function takes two input arguments the TimeStart and TimeEnd > I would like to help me create the right list with pairs of TimeStart and > TimeEnd which I can feed

Re: [R] lapply and Two TimeStamps as input

Provide some sample data. For instance, a data frame with two columns and the function. On Oct 31, 2011, at 6:37 PM, Alaios wrote: > Dear all, > > I have a function that recognizes the following format for timestamps > "%Y-%m-%d %H:%M:%S" > > my function takes two input arguments the TimeSt

[R] lapply and Two TimeStamps as input

Dear all, I have a function that recognizes the following format for timestamps "%Y-%m-%d %H:%M:%S" my function takes two input arguments the TimeStart and TimeEnd I would like to help me create the right list with pairs of TimeStart and TimeEnd which I can feed to lapply (I am using mclapply ac

Re: [R] lapply to return vector

do.call(rbind, lapply(...)) HTH, D. On Sat, Oct 22, 2011 at 1:44 AM, Alaios wrote: > Dear all I have wrote the following line > > > return(as.vector(lapply(as.data.frame(data),min,simplify=TRUE))); > > > I want the lapply to return a vector as it returns a list with elements as > shown below >

[R] lapply to return vector

Dear all I have wrote the following line return(as.vector(lapply(as.data.frame(data),min,simplify=TRUE))); I want the lapply to return a vector as it returns a list with elements as shown below List of 30001 $ V1: num -131 $ V2: num -131 $ V3: num -137 $ V4: num -129 $ V5:

Re: [R] lapply, if statement and concatenating to a list

Hi Lorenzo, On Thu, May 5, 2011 at 8:38 AM, Lorenzo Cattarino wrote: > Hi R users > > I was wondering on how to use lapply & co when the applied function has a > conditional statement and the output is a 'growing' object. > See example below: > > list1 <- list('A','B','C') > list2 <- c() > > myf

[R] lapply, if statement and concatenating to a list

Hi R users I was wondering on how to use lapply & co when the applied function has a conditional statement and the output is a 'growing' object. See example below: list1 <- list('A','B','C') list2 <- c() myfun <- function(x,list2) { one_elem <- x cat('one_elem= ', one_elem, '\n') random <

Re: [R] lapply sequence

On 20/04/2011 7:26 AM, Dean Marks wrote: Good day, My question is: Does the lapply function guarantee a particular sequence in which elements are mapped? And, are we guaranteed that lapply will always be sequential (i.e. never map elements in parallel) ? No. The reason I ask is if I use lapp

[R] lapply sequence

Good day, My question is: Does the lapply function guarantee a particular sequence in which elements are mapped? And, are we guaranteed that lapply will always be sequential (i.e. never map elements in parallel) ? The reason I ask is if I use lapply with the mapping function set to something that

[R] lapply over list and the respective name of the list item

Hi all, I find myself sometimes in the situation where I lapply over a list and in the particular function I'd like to use the name and or position of the respective list item. What I usually do is to use mapply on the list and the names of the list / a position list: o <- list(A=1:3, B=1:2, C=1)

Re: [R] lapply, strsplit, and list elements

tical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 > -Original Message- > From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com] > Sent: Friday, February 04, 2011 12:22 PM > To: Greg Snow > Cc: Dick Harray; r-help@r-project.org > Subject: Re: [R

Re: [R] lapply, strsplit, and list elements

On Fri, Feb 4, 2011 at 1:27 PM, Greg Snow wrote: > Try this: > >> x <- c("349/077,349/074,349/100,349/117", > +          "340/384.2,340/513,367/139,455/128,D13/168", > +          "600/437,128/903,128/904") >> >> library(gsubfn) >> out <- strapply(x, '([0-9]+)(?=/)') >> out > [[1]] > [1] "349" "349

Re: [R] lapply, strsplit, and list elements

Try this: strsplit(x, "/\\d+\\.\\d+,|/\\d+,|/\\d+") On Fri, Feb 4, 2011 at 1:37 PM, Dick Harray wrote: > Hi there, > > I have a problem about lapply, strsplit, and accessing list elements, > which I don't understand or cannot solve: > > I have e.g. a character vector with three elements: > > x

Re: [R] lapply, strsplit, and list elements

nter Intermountain Healthcare greg.s...@imail.org 801.408.8111 > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > project.org] On Behalf Of Dick Harray > Sent: Friday, February 04, 2011 8:37 AM > To: r-help@r-project.org > Subject: [R]

Re: [R] lapply, strsplit, and list elements

> -Original Message- > From: r-help-boun...@r-project.org > [mailto:r-help-boun...@r-project.org] On Behalf Of Dick Harray > Sent: Friday, February 04, 2011 7:37 AM > To: r-help@r-project.org > Subject: [R] lapply, strsplit, and list elements > > Hi there, >

[R] lapply, strsplit, and list elements

Hi there, I have a problem about lapply, strsplit, and accessing list elements, which I don't understand or cannot solve: I have e.g. a character vector with three elements: x = c("349/077,349/074,349/100,349/117", "340/384.2,340/513,367/139,455/128,D13/168", "600/437,128/903,1

Re: [R] lapply getting names of the list

On Dec 9, 2010, at 2:21 PM, David Winsemius wrote: On Dec 9, 2010, at 12:44 PM, Sashi Challa wrote: Hello All, I have a toy dataframe like this. It has 8 columns separated by tab. NameSampleIDAl1 Al2 X Y R Th rs191191A1 A B 0.9

[R] lapply getting names of the list

rsday, December 09, 2010 10:07 AM To: Sashi Challa Cc: r-help@R-project.org Subject: Re: [R] lapply getting names of the list Hi Sashi, On Thu, Dec 9, 2010 at 9:44 AM, Sashi Challa wrote: > Hello All, > > I have a toy dataframe like this. It has 8 columns separated by tab. > > Name  

Re: [R] lapply getting names of the list

On Dec 9, 2010, at 12:44 PM, Sashi Challa wrote: Hello All, I have a toy dataframe like this. It has 8 columns separated by tab. NameSampleIDAl1 Al2 X Y R Th rs191191A1 A B 0.999 0.090.780.090 abc928291 A1 B

Re: [R] lapply getting names of the list

Hi Sashi, On Thu, Dec 9, 2010 at 9:44 AM, Sashi Challa wrote: > Hello All, > > I have a toy dataframe like this. It has 8 columns separated by tab. > > Name    SampleID        Al1     Al2     X       Y       R       Th > rs191191        A1      A       B       0.999   0.09    0.78    0.090 > abc9

[R] lapply getting names of the list

Hello All, I have a toy dataframe like this. It has 8 columns separated by tab. NameSampleIDAl1 Al2 X Y R Th rs191191A1 A B 0.999 0.090.780.090 abc928291 A1 B J 0.3838 0.3839 0.028 0.888 abcnab A1

Re: [R] lapply to subsets

Yes, that is what I what... Thanks. Feng On Wed, Oct 13, 2010 at 6:38 AM, Michael Bedward wrote: > Hello Feng, > > I think you just want this... > > lapply(A, function(x) apply(x[,,-c(1,2)], c(1,2), mean)) > > Michael > > > On 13 October 2010 04:00, Feng Li wrote: > > Dear R, > > > > I have a s

Re: [R] lapply to subsets

Hello Feng, I think you just want this... lapply(A, function(x) apply(x[,,-c(1,2)], c(1,2), mean)) Michael On 13 October 2010 04:00, Feng Li wrote: > Dear R, > > I have a silly question concerns with *apply. Say I have a list called A, > > A <- list(a  =  array(1:20, c(2, 2, 5)), b  = array(1

[R] lapply to subsets

Dear R, I have a silly question concerns with *apply. Say I have a list called A, A <- list(a = array(1:20, c(2, 2, 5)), b = array(1:30, c(2, 3, 5))) I wish to calculate the mean of A$a, and A$b w.r.t. their third dimension so I did lapply(A,apply,c(1,2),mean) Now if I still wish to do the

Re: [R] lapply and boxplots with variable names

Try this: library(lattice) bwplot(values ~ TimePeriod | ind, cbind(stack(my.data), TimePeriod = my.data$TimePeriod)) On Tue, Jun 22, 2010 at 1:45 PM, Shawn Morrison < shawn.morri...@dryasresearch.com> wrote: > Hi all, > > I have a dataset with several variables, each of which is a separate > col

Re: [R] lapply and boxplots with variable names

Thanks Josh, I do want to see each plot. I took your code and modified it (below) and it appears to do what I wanted: my.data <- data.frame(var1=rnorm(1000), var2=rnorm(1000), TimePeriod=factor(rep((LETTERS[1:4]), 250))) str(my.data) lapply(names(my.data[ , 1:2]), function(y) { quartz() bo

Re: [R] lapply and boxplots with variable names

Hello Shawn, Does this do what you want? I'm assuming you want to look at each plot, so I added a call to par(). ### my.data <- data.frame(var1=rnorm(1000), var2=rnorm(1000), TimePeriod=factor(rep((LETTERS[1:4]), 250))) str(my.data) lapply(names(my.data[ , 1:2]), function(y) { old.par <-

Re: [R] lapply and boxplots with variable names

Many thanks Phil, that does help. If I could ask a follow-up, how do I put each plot in its own device window? Right now, all I get is the boxplot for var2 (var1 gets overwritten?). I tried putting quartz() before the boxplot command but got an error message. Cheers, Shawn On 22/06/10 11:13

Re: [R] lapply and boxplots with variable names

Shawn - Does this example help? (Please don't use cbind when creating a data frame, since it first creates a matrix, which means everything must be of the same mode.) var1 = rnorm(1000) var2 = rnorm(1000) TimePeriod = rep((LETTERS[1:4]), 250) my.data = data.frame(var1,var2,TimePeriod) lapply

[R] lapply and boxplots with variable names

Hi all, I have a dataset with several variables, each of which is a separate column. For each variable, I want to produce a boxplot and include the name of the variable (ie, column name) on each plot. I have included a sample dataset below. Can someone tell me where I am going wrong? Thank

Re: [R] lapply or data.table to find a unit's previous transaction

William, Try a rolling join in data.table, something like this (untested) : setkey(Data, UnitID, TranDt)# sort by unit then date previous = transform(Data, TranDt=TranDt-1) Data[previous,roll=TRUE]# lookup the prevailing date before, if any, for each row within that row's UnitID Tha

Re: [R] lapply or data.table to find a unit's previous transaction

On Wed, Jun 2, 2010 at 10:29 PM, William Rogers wrote: > I have a dataset of property transactions that includes the > transaction ID (TranID), property ID (UnitID), and transaction date > (TranDt). I need to create a data frame (or data table) that includes > the previous transaction date, if one

[R] lapply or data.table to find a unit's previous transaction

I have a dataset of property transactions that includes the transaction ID (TranID), property ID (UnitID), and transaction date (TranDt). I need to create a data frame (or data table) that includes the previous transaction date, if one exists. This is an easy problem in SQL, where I just run a sub-

Re: [R] lapply with functions with changing parameters

Henrique, thx, your suggestion worked perfectly fine for me. On 01.06.2010, at 23:01, Henrique Dallazuanna wrote: > lapply(mydf[-6], ccf, y = mydf[6]) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read

Re: [R] lapply with functions with changing parameters

Bunny, lautloscrew.com wrote: Dear all, I am trying to avoid a for loop here and wonder if the following is possible: I have a data.frame with 6 columns and i want to get a cross-correlogram (by using ccf) . Obivously ccf only accepts two columns at once and then returms a list. In fact, with

Re: [R] lapply with functions with changing parameters

Try this: lapply(mydf[-6], ccf, y = mydf[6]) On Tue, Jun 1, 2010 at 5:50 PM, Bunny, lautloscrew.com < bu...@lautloscrew.com> wrote: > Dear all, > > I am trying to avoid a for loop here and wonder if the following is > possible: > > I have a data.frame with 6 columns and i want to get a cross-co

[R] lapply with functions with changing parameters

Dear all, I am trying to avoid a for loop here and wonder if the following is possible: I have a data.frame with 6 columns and i want to get a cross-correlogram (by using ccf) . Obivously ccf only accepts two columns at once and then returms a list. In fact, with a for loop i´d do the follow

Re: [R] lapply - function with arguments

another thought possibly fn = function(n, a=1, b=3) return(n*(a+b)) sapply(1:3, fn) -- View this message in context: http://n4.nabble.com/lapply-function-with-arguments-tp1838373p1838506.html Sent from the R help mailing list archive at Nabble.com.

Re: [R] lapply - function with arguments

Thank you Jim 2010/4/13 jim holtman > lapply(yourList, f, a=1, b=2) > > On Tue, Apr 13, 2010 at 9:11 AM, Randall Wrong > wrote: > >> Dear R users, >> >> I have created a function f of n, a and b : f(n,a,b) >> >> I would like to apply this function several times to some values of n. a >> and

Re: [R] lapply - function with arguments

lapply(yourList, f, a=1, b=2) On Tue, Apr 13, 2010 at 9:11 AM, Randall Wrong wrote: > Dear R users, > > I have created a function f of n, a and b : f(n,a,b) > > I would like to apply this function several times to some values of n. a > and > b are held constant. I was thinking of using lapply. Ho

[R] lapply - function with arguments

Dear R users, I have created a function f of n, a and b : f(n,a,b) I would like to apply this function several times to some values of n. a and b are held constant. I was thinking of using lapply. How can I do this ? Thank you very much Randall [[alternative HTML version deleted]]

[R] lapply and list indexing basics (after realizing I wasn't previously subscribed...sorry)

library(MASS) dat <- data.frame( col1=as.factor(sample(1:4, 100, T)), col2=as.factor(sample(1:4, 100, T)), col3=as.factor(sample(1:4, 100, T)), isi=rnorm(100) ) dat <- split(dat, as.factor(sample(1:3, 100, T))) lapply(dat, function(x, densfun) fitdistr(x$isi, densfun), 'normal')

Re: [R] lapply and list indexing basics

It would have been nice if you had at least posted what the structure of myList is. Assuming that this is the list of your data frames, then the following might work: lapply(myList, function(x) fitdistr(x$isi, densfun='gamma',start=list(scale=1, shape=2))) On Sun, Mar 7, 2010 at 7:30 PM, Dgnn

[R] lapply and list indexing basics

I have split my original dataframe to generate a list of dataframes each of which has 3 columns of factors and a 4th column of numeric data. I would like to use lapply to apply the fitdistr() function to only the 4th column (x$isi) of the dataframes in the list. Is there a way to do this or am I

[R] lapply and list indexing basics (after realizing I wasn't previously subscribed...sorry)

I have split my original dataframe to generate a list of dataframes each of which has 3 columns of factors and a 4th column of numeric data. I would like to use lapply to apply the fitdistr() function to only the 4th column (x$isi) of the dataframes in the list. Is there a way to do this or am

Re: [R] lapply with data frame

below! From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of bill.venab...@csiro.au [bill.venab...@csiro.au] Sent: 01 March 2010 17:18 To: n...@smartmediacorp.com; r-help@r-project.org Subject: [ExternalEmail] Re: [R] lapply with data frame Data frames are

Re: [R] lapply with data frame

r-help-boun...@r-project.org] On Behalf Of Noah Silverman [n...@smartmediacorp.com] Sent: 28 February 2010 12:37 To: r-help@r-project.org Subject: [R] lapply with data frame I'm a bit confused on how to use lapply with a data.frame. For example. lapply(data, function(x) print(x)) WHAT exa

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