Re: [R] na.rm = T treatment by ggplot2's geom_bar

Important to notice: this seems to be an issue only with an unordered factor on the X axis. When the variable is numeric or an ordered factor, then it works as described in Help. On Thu, Jul 27, 2017 at 11:58 AM, Dimitri Liakhovitski < dimitri.liakhovit...@gmail.com> wrote: > Тhank

Re: [R] na.rm = T treatment by ggplot2's geom_bar

y brevity. > > On July 27, 2017 8:18:03 AM PDT, Dimitri Liakhovitski < > dimitri.liakhovit...@gmail.com> wrote: > >Thank you, Bert! > > > >I do NOT have an object named "T" in scope (I checked - and besides, it > >would never occur to me to use th

Re: [R] na.rm = T treatment by ggplot2's geom_bar

Jeff Newmiller wrote: > I suspect this is by design. Questions about "why" should probably cc the > contributed package maintainer(s). > -- > Sent from my phone. Please excuse my brevity. > > On July 27, 2017 7:49:47 AM PDT, Dimitri Liakhovitski < > dimitri.liakhovit

Re: [R] na.rm = T treatment by ggplot2's geom_bar

it." > -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) > > > On Thu, Jul 27, 2017 at 7:49 AM, Dimitri Liakhovitski > wrote: > > To clarify: my question is not about "who could I exclude NAs from being > > counted" - I know how

Re: [R] na.rm = T treatment by ggplot2's geom_bar

To clarify: my question is not about "who could I exclude NAs from being counted" - I know how to do that. My question is: Why na.rm = T is not working for geom_bar in this case? On Thu, Jul 27, 2017 at 8:24 AM, Dimitri Liakhovitski < dimitri.liakhovit...@gmail.com> wrote: &

[R] na.rm = T treatment by ggplot2's geom_bar

lot(data = md, mapping = aes(x = a)) + geom_bar(na.rm = T) But I still have NAs in the picture. Why? What am I missing? Thank you! -- Dimitri Liakhovitski [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE

Re: [R] Error with installed.packages with R 3.4.0 on Windows

://www.R-project.org/ >>> >> posting-guide.html >> >>> and provide commented, minimal, self-contained, reproducible code. >>> >> >> -- >> Peter Dalgaard, Professor, >> Center for Statistics, Copenhagen Business School >> Solbje

Re: [R] update.packages() error R 3.4.0

_ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-conta

Re: [R] Why is merge sorting even when sort = F?

51 PM, Jeff Newmiller wrote: > If you are still wondering, try re-reading my answer. FALSE is more > efficient, TRUE is sorted. Lack of sorting has nothing to do with preserving > order. > -- > Sent from my phone. Please excuse my brevity. > > On March 8, 2017 8:55:06 AM

Re: [R] Why is merge sorting even when sort = F?

ordering afterward, or you can roll your own > possibly-less-efficient merge using match and indexing: > > info[ match( grades2$grade, info$grade ), ] > -- > Sent from my phone. Please excuse my brevity. > > On March 8, 2017 8:07:27 AM PST, Dimitri Liakhovitski > wrote: >

[R] Why is merge sorting even when sort = F?

I have in info: This solution resorts everything in the order of column 'grade': merge(grades2, info, by = "grade", all.x = T, all.y = F) Could you please explain why this solution also resorts - despite sort = FALSE? merge(grades2, info, by = "gr

Re: [R] weird behavior of DoubleMetaphone

e(temp[word]) } return(paste(temp, collapse = "")) } The next 4 lines return the same thing: metaphone("netflix") metaphone("net flex") mymetaphone("netflix") mymetaphone("net flex") Dimitri On Wed, Dec 14, 2016 at 2:35 PM, Dimitri Liakhovitski

[R] weird behavior of DoubleMetaphone

;NTFL" mymetaphone("net flex") # returns "NTFLKS" Why such a difference between DoubleMetaphone("net flex")$primary and mymetaphone("net flex")? Why isn't DoubleMetaphone("net flex")$primary returning "NTFLKS"? Thank you! -- Dimi

Re: [R] creating possible cominations of a vector's elements

c("Var2", "Var1")] > mapply(function(x, y) paste(vec[x:y], collapse=" "), > x=idx[, 1], y=idx[, 2]) > } > > David C > > -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of David L &

Re: [R] creating possible cominations of a vector's elements

vector" > [10] "vector" > > This makes a vector of strings but if you want a list use as.list(mapply()) > > David L. Carlson > Department of Anthropology > Texas A&M University > > > > -Original Message- > From: R-help [mailto:r-help-bo

[R] creating possible cominations of a vector's elements

Hello! I have a vector of strings 'x' that was based on a longer string 'mystring' (the actual length of x is unknown). mystring <- "this is my vector" x <- strsplit(mystr, " ")[[1]] I am looking for an elegant way of creating an object (e.g., a list) that contains the following strings: "this"

Re: [R] are R packages safe?

s reply, even the official R distribution comes with no > warranty, and that will be the case with most OSS. > > Regards, > > Marc > > >> On Dec 8, 2016, at 12:08 PM, Dimitri Liakhovitski >> wrote: >> >> Thank you, Marc. >> That's helpf

Re: [R] are R packages safe?

hu, Dec 8, 2016 at 10:05 AM, Dimitri Liakhovitski > wrote: >> I just thought maybe there is something - about the process of >> submitting packages or anything like that - that shows that at least >> some diligence is being done to ensure that a given package is not >>

Re: [R] are R packages safe?

sured > that all "official" R packages have been "vetted" and are safe? > > Thanks a lot! > > -- > Dimitri Liakhovitski > > > > Dimitri, > > You are going to need to define "safe". > > Also, note that the notion of "official

Re: [R] are R packages safe?

with ABSOLUTELY NO WARRANTY." > > Don't think it could be clearer than that! > > Cheers, > Bert > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka Berkeley

[R] are R packages safe?

Guys, suddenly, I am being asked for a proof that R packages that are not '"base" are safe. I've never been asked this question before. Is there some documentation on CRAN that discusses how it's ensured that all "official" R packages have been "vetted&qu

[R] double metaphone for non-English languages

and discovered that DoubleMetaphone can't process non-ASCII characters (like Umlauts in German, accents in French). What would you recommend I use instead? Use 'phonetic' from stringdist? Thank you! -- Dimitri Liakhovitski __ R-help@r-project.org

[R] stringi behaves differently in 2 similar situations

t in the second case? Thanks a lot! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html a

Re: [R] using pmax in presence of NAs

Thank you, Peter! On Wed, Nov 16, 2016 at 4:21 PM, peter dalgaard wrote: > >> On 16 Nov 2016, at 21:58 , Dimitri Liakhovitski >> wrote: >> >> Hello! >> >> I need to calculate the maximum of each row of a data frame. >> This works: >> >

Re: [R] using pmax in presence of NAs

gt; do.call(pmax, c(x, na.rm=TRUE)) > [1] 111 112 113 114 15 > > On Wed, Nov 16, 2016 at 3:58 PM, Dimitri Liakhovitski > wrote: >> Hello! >> >> I need to calculate the maximum of each row of a data frame. >> This works: >> >> x <- data.frame(a =

Re: [R] using pmax in presence of NAs

To clarify: I know I could do: apply(x, 1, max, na.rm = T) But I was wondering if one can modify the pmax one... On Wed, Nov 16, 2016 at 3:58 PM, Dimitri Liakhovitski wrote: > Hello! > > I need to calculate the maximum of each row of a data frame. > This works: > > x <-

[R] using pmax in presence of NAs

non-NAs in each row: x <- data.frame(a = c(1:5), b=11:15, c=c(111:114,NA)) x I'd like it to return: [1] 111 112 113 114 15 Thanks a lot! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see http

Re: [R] multiplying a matrix by a vector

operation is performed column-wise: > > t(y * t(x)) > > Hope this helps, > > Rui Barradas > > > Em 03-11-2016 21:05, Dimitri Liakhovitski escreveu: >> >> Hello! >> >> I have a matrix x and a vector y: >> >> x <- matrix(1:6, ncol = 2)

Re: [R] multiplying a matrix by a vector

>> >>> sweep(x, 2, y, "*") >> [,1] [,2] >> [1,]2 12 >> [2,]4 15 >> [3,]6 18 >>> >> >> >> On Thu, Nov 3, 2016 at 5:05 PM, Dimitri Liakhovitski >> wrote: >>> Hello! >>> >>> I h

[R] multiplying a matrix by a vector

Or I could repeat each element of y and multiply two matrices - that's better: rep.row<-function(x,n){ matrix(rep(x,each=n),nrow=n) } y <- rep.row(y, nrow(x)) x * y However, maybe there is a more elegant r-like way of doing it? Thank yo

Re: [R] removing all non-numeric characters from a string, but not "."

Thank you very much, gentlemen! On Tue, Jul 26, 2016 at 5:48 PM, peter dalgaard wrote: > >> On 26 Jul 2016, at 23:28 , Dimitri Liakhovitski >> wrote: >> >> Hello! >> >> I have a string x: >> x <- c("x - 84", "y - 293.04",

[R] removing all non-numeric characters from a string, but not "."

Hello! I have a string x: x <- c("x - 84", "y - 293.04", "z = 12.5") I want to remove all the non-numeric stuff from it. The following works: gsub("[^0-9]", "", x) However, it strips my numbers of "." Help - how could I do the same

Re: [R] Shift all values above certain value by 1

a frame? Thanks! On Fri, Jan 22, 2016 at 11:34 AM, Dimitri Liakhovitski wrote: > Hello! > > # I have a data frame x: > x <- data.frame(a = 1:10, b = 2:11, c = 3:12, other = rnorm(10)) > > # First, I need to change every value 7 in columns a:c to 4 > # Then, I need t

[R] Shift all values above certain value by 1

Hello! # I have a data frame x: x <- data.frame(a = 1:10, b = 2:11, c = 3:12, other = rnorm(10)) # First, I need to change every value 7 in columns a:c to 4 # Then, I need to decrease by 1 all values in columns a:c that are >7 What would be the fastest way of doing it? Thank you! -- D

Re: [R] (no subject)

har", names(x))]) > newx <- x > newx$char <- row(y)[y==1] > > # merge and define winner > res <- merge(newx, info2) > res$winner <- with(res, ifelse(char==val, 1, 0)) > res > > > On Wed, Dec 23, 2015 at 3:35 PM, Dimitri Liakhovitski > wrote: >>

[R] (no subject)

mber of games and the number of sets per game. # However, we can assume that the number of sets per game is always the same, # and the number of players per set is always the same. -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNS

Re: [R] Expanding matrix into dummies

and sticking things into it." > -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) > > > On Tue, Dec 22, 2015 at 3:03 PM, Marc Schwartz wrote: >> >>> On Dec 22, 2015, at 4:42 PM, Dimitri Liakhovitski >>> wrote

[R] Expanding matrix into dummies

# I have a matrix x: k <- 20 N <- 5 set.seed(123) x <- matrix(c(sample(1:k, N, replace = F), sample(1:k, N, replace = F), sample(1:k, N, replace = F), sample(1:k, N, replace = F), sample(1:k, N, replace = F), sample(1:k, N, repl

Re: [R] Trying to avoid the loop while merging two data frames

Taking it back - no need for all.x = T, all.y = F On Tue, Dec 22, 2015 at 3:56 PM, Dimitri Liakhovitski wrote: > Actually, the correct merge line should be: > my.merge <- merge(myinfo, mydata, by="version", all.x = T, all.y = F) > > On Tue, Dec 22, 2015 at 3:50 PM, Dimi

Re: [R] Trying to avoid the loop while merging two data frames

Actually, the correct merge line should be: my.merge <- merge(myinfo, mydata, by="version", all.x = T, all.y = F) On Tue, Dec 22, 2015 at 3:50 PM, Dimitri Liakhovitski wrote: > You are right, guys, merge is working. Somehow I was under the > erroneous impression that beca

Re: [R] Trying to avoid the loop while merging two data frames

erge my.merge <- merge(myinfo, mydata, by="version") names(my.merge) result2 <- my.merge[,c("myid", "version", "a", "b", "c", "d")] names(result2) result2.order <- arrange(result2, myid, version, a, b, c, d) dim(result2.or

Re: [R] Trying to avoid the loop while merging two data frames

> What is the problem that you are trying to solve? > Tell me what you want to do, not how you want to do it. > > On Tue, Dec 22, 2015 at 12:27 PM, Dimitri Liakhovitski > wrote: >> >> Hello! >> I have a solution for my task that is based on a loop. However, it

[R] Trying to avoid the loop while merging two data frames

] <- myinfo[myinfo$version == result[[i]]$version, ] result[[i]]$myid <- id result[[i]] <- result[[i]][c(5, 1:4)] } result <- do.call(rbind, result) head(result) # This is the desired result -- Dimitri Liakhovitski __ R-help@r-pr

Re: [R] Why mean is not working in by?

t; aggregate() calls its FUN with each column of a data.frame as the argument. > > Bill Dunlap > TIBCO Software > wdunlap tibco.com > > On Tue, Dec 8, 2015 at 3:08 PM, Dimitri Liakhovitski > wrote: >> >> Sorry, I omitted the first line: >> >> myvars <

Re: [R] Why mean is not working in by?

by=iris[,"Species", drop=FALSE], FUN=mean) > > provide the answers I would expect. If you want clearer advice, you > need to provide an actually reproducible example, and tell us more > about what you expect to get. > > Sarah > > > On Tue, Dec 8, 2015 at 5:30 PM, Dimi

[R] Why mean is not working in by?

;Species"], FUN = var) by(data = iris[myvars], INDICES = iris["Species"], FUN = max) by(data = iris[myvars], INDICES = iris["Species"], FUN = min) by(data = iris[myvars], INDICES = iris["Species"], FUN = sd) by(data = iris[myvars],

Re: [R] "haven" - read_spss: How to avoid extracting value labels instead of long labels?

the development version of rio > (https://github.com/leeper/rio) has an (non-exported) function for > cleaning up meta data from haven imports. See > https://github.com/leeper/rio/blob/master/R/utils.R#L86 > > Best, > Ista > > On Thu, Nov 12, 2015 at 8:37 PM, Dimitri Liakh

Re: [R] "haven" - read_spss: How to avoid extracting value labels instead of long labels?

1 2 And I actually need to be able to identify if label is empty. Thank you for looking into it! Dimitri On Thu, Nov 12, 2015 at 5:55 PM, Dimitri Liakhovitski wrote: > Looks like a little bug in 'haven': > > When I actually look at the att

[R] extraction of parameters for intercept from "lavaan" object

057? Or should there also be an intercept present in this formula? I did not specify any intercepts anywhere in my model? If there should be an intercept, then where can I find it in the object sem1? Thanks a lot! -- Dimitri Liakhovitski __ R-help@r

Re: [R] "haven" - read_spss: How to avoid extracting value labels instead of long labels?

ny suggestions? Thanks! On Thu, Nov 12, 2015 at 11:56 AM, Dimitri Liakhovitski wrote: > Hello! > > I don't have an example file, but I think my question should be clear > without it. > I have an SPSS file. I read it in using 'haven': > > library(haven) >

[R] "haven" - read_spss: How to avoid extracting value labels instead of long labels?

VERY/SOMEWHAT FAMILIAR" "NOT AT ALL FAMILIAR" Question: How could I avoid the extraction of the Value Labels for the columns that have no long labels? Thank you very much! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list --

Re: [R] A bunch of packages keeps wanting to get updated

Thank you, Duncan. Indeed those packages were sitting in a temp directory. I had to install them manually from a 'zip file. On Thu, Nov 5, 2015 at 2:40 PM, Duncan Murdoch wrote: > On 05/11/2015 1:00 PM, Dimitri Liakhovitski wrote: >> >> I am using a windows laptop, R 3.2.2 &g

Re: [R] A bunch of packages keeps wanting to get updated

v 5, 2015 at 2:00 PM, Dimitri Liakhovitski wrote: > I am using a windows laptop, R 3.2.2 > > I am using R-gui. > > When I go to Packages -> Update packages and then select a Cran mirror > (in the US) - it tells me to update the following packages: > > GLMMGibbs > RDCO

[R] A bunch of packages keeps wanting to get updated

I didn't just update them. What could be the reason? I tried to totally delete the folders of those packages, install them from scratch (successfully), still - they keep appearing in Update Pacakges widnow. Any advice? Thank you! -- Dimitri Liakhovitski ___

[R] F-test of equality of variances - but weighted?

ar1/var2 With df's of weighted n1-1 for group 1 and weighted n2-1 for group 2? Would it be kosher from statistical perspective? Thanks a lot! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://s

Re: [R] na.action in xtabs - how to include NAs?

#x27;s output for a call to table. > > xtabs(formula = ~a + b, data = x, na.action = na.pass, exclude = NULL) > Bill Dunlap > TIBCO Software > wdunlap tibco.com > > > On Wed, Sep 30, 2015 at 7:56 AM, Dimitri Liakhovitski > wrote: >> Please help: >> >> #

[R] na.action in xtabs - how to include NAs?

x27;t figure out how to force xtabs to include NAs. # All my attempts below fail to include NAs: xtabs(~ a + b, x, na.action(na.pass)) xtabs(~ a + b, x, na.action = "na.pass") xtabs(~ a + b, x, na.action(na.pass(x))) xtabs(~ a + b, x, exclude = NULL) Th

Re: [R] dplyr complete.cases(.) works one way but not another

SE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://s

[R] dplyr complete.cases(.) works one way but not another

ter(complete.cases(.)) But this command doesn't work: filter(mydata, complete.cases(.)) Error: object '.' not found Why doesn't it work? Thank you! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and

Re: [R] Hep with regex! - combination of ^, |, \\, _ and $

ttp://harvest.nps.edu > > -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Berend > Hasselman > Sent: Friday, September 18, 2015 1:30 AM > To: Dimitri Liakhovitski > Cc: r-help > Subject: Re: [R] Hep with regex! - combination of ^,

Re: [R] Hep with regex! - combination of ^, |, \\, _ and $

For the last one, looks like this one works: x[grep("^(q10|q12).*\\_1$", x)] On Thu, Sep 17, 2015 at 5:46 PM, Dimitri Liakhovitski wrote: > Duncan, > Of course my verbal descriptions and my code don't match my regexp - > otherwise I wouldn't be asking the question

Re: [R] Hep with regex! - combination of ^, |, \\, _ and $

/2015 5:11 PM, Dimitri Liakhovitski wrote: >> (x <- c("q10_1", "q10_2", "q10_11", "q12_1", "q12_2", "q13_1", "q13_11")) >> >> # Which strings start with "q10" or "q12? - WORKS >> x[grep

[R] Hep with regex! - combination of ^, |, \\, _ and $

[grep("1$", x)] # Which strings end with "_1"? - WORKS x[grep("\\_1$", x)] # Which strings start with "q10" AND contain a "1"? - WORKS x[grep("^q10.+1", x)] # Which strings start with "q10" AND end with a "_1"

[R] Converting a .wav file into an mp3 file in R

Hello, I know how to read in mp3 files, e.g., using tuneR. But is it possible to read in a .wav file - as below and then compress it to mp3 format? library(tuneR) mywav <- readWave("myfile.wav") Thanks a lot for any hints! -- Dimitri

Re: [R] "unfurling" rankings into a matrix of preferences

t;Data is not information. Information is not knowledge. And knowledge > is certainly not wisdom." > -- Clifford Stoll > > > On Mon, Jul 27, 2015 at 2:32 PM, Dimitri Liakhovitski > wrote: >> With NAs it'd be: >> >> rk <- c(2,NA,4,3,1, NA) >>

Re: [R] "unfurling" rankings into a matrix of preferences

With NAs it'd be: rk <- c(2,NA,4,3,1, NA) outer(rk, rk, "<") + 0 Wow, I still can't believe it - just one line! On Mon, Jul 27, 2015 at 5:31 PM, Dimitri Liakhovitski wrote: > Wow! > > On Mon, Jul 27, 2015 at 5:28 PM, Bert Gunter wrote: >> ## I leav

Re: [R] "unfurling" rankings into a matrix of preferences

[3,]0100 > [4,]1110 > > > > Cheers, > Bert > > Bert Gunter > > "Data is not information. Information is not knowledge. And knowledge > is certainly not wisdom." >-- Clifford Stoll > > > On Mon, Jul 27, 2015 at 12:38

[R] "unfurling" rankings into a matrix of preferences

elegant way? Thank you very much! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htm

Re: [R] For Hadley Wickham: Need for a small fix in haven::read_spss

s1, attr_default, "label", NA_character_) > > (code untested, but you get the idea) > > Hadley > > On Mon, Jul 20, 2015 at 8:56 AM, Dimitri Liakhovitski > wrote: >> Hadley, >> >> you've added function labelled to haven, which is great. However,

[R] For Hadley Wickham: Need for a small fix in haven::read_spss

a labels. When I run lapply(spss1, attr, "label") I see under those 2 variables "NULL" - which is true and valid. However, would it be possible to have instead of NULL an NA? This way the length of varnames and mylables would the same and one could put them side by side (e.g., in

Re: [R] Weighted skewness and curtosis

knowledge. And knowledge > is certainly not wisdom." >-- Clifford Stoll > > > On Thu, Jul 16, 2015 at 8:37 AM, Dimitri Liakhovitski > wrote: >> Unfortunately not - more like 0.7654, 1.2345. >> I understand that I could multiply each number by 100, round it t

Re: [R] Weighted skewness and curtosis

: > > On Jul 16, 2015, at 8:10 AM, Dimitri Liakhovitski wrote: > >> Is there an R package that allows one to calculate skewness and >> curtosis - but weighted with individual level weights (one weight per >> observation)? >> > > Integer weights? > &

[R] Weighted skewness and curtosis

Is there an R package that allows one to calculate skewness and curtosis - but weighted with individual level weights (one weight per observation)? Thank you! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more

Re: [R] dplyr, summarize_each, mean - dealing with NAs

I found the answer: md %>% group_by(device1, device2) %>% summarise_each(funs(mean(., na.rm = TRUE))) On Thu, Jun 25, 2015 at 4:35 PM, Dimitri Liakhovitski wrote: > Just want to clarify - I know how to do it using base R. I just want > to figure out how to do it in dplyr. This i wh

Re: [R] dplyr, summarize_each, mean - dealing with NAs

(dplyr) > md %>% group_by(device1, device2) %>% summarise_each(funs(mean)) > > However, I am getting some NAs. I want the NAs to be ignored (na.rm = > TRUE) - I tried, but the function doesn't want to accept this > argument. > Both these lines result in error: > >

[R] dplyr, summarize_each, mean - dealing with NAs

o be ignored (na.rm = TRUE) - I tried, but the function doesn't want to accept this argument. Both these lines result in error: md %>% group_by(device1, device2) %>% summarise_each(funs(mean), na.rm = TRUE) md %>% group_by(device1, device2) %>% summarise_each(funs(mean, na.rm

Re: [R] dplyr - counting a number of specific values in each column - for all columns at once

Thank you guys - it's a great learning: 'summarise_each' and 'funs' On Tue, Jun 16, 2015 at 3:47 PM, Hadley Wickham wrote: > On Tue, Jun 16, 2015 at 12:24 PM, Dimitri Liakhovitski > wrote: >> Hello! >> >> I have a data frame: >> >> m

Re: [R] dplyr - counting a number of specific values in each column - for all columns at once

.edu > Department of Ecology VOICE: (360) 407-6815 > PO Box 47600FAX:(360) 407-7534 > Olympia, WA 98504-7600 > > USPS: PO Box 47600, Olympia, WA 98504-7600 > Parcels:300 Desmond Drive, Lacey, WA 98503-1274 &g

Re: [R] dplyr - counting a number of specific values in each column - for all columns at once

NA md[4,1] <- NA md vapply(md[myvars], function(x) sum(x==5,na.rm=TRUE),1L) But the result should be by device. On Tue, Jun 16, 2015 at 1:56 PM, Dimitri Liakhovitski wrote: > Thank you, Bert. > I'll be honest - I am just learning dplyr and was wondering if one > could do it in dplyr. &

Re: [R] dplyr - counting a number of specific values in each column - for all columns at once

t; > Cheers, > Bert > > Bert Gunter > > "Data is not information. Information is not knowledge. And knowledge is > certainly not wisdom." >-- Clifford Stoll > > On Tue, Jun 16, 2015 at 10:24 AM, Dimitri Liakhovitski > wrote: >> >> Hello! >>

Re: [R] dplyr - counting a number of specific values in each column - for all columns at once

300 Desmond Drive, Lacey, WA 98503-1274 > > > On Tue, 16 Jun 2015, Dimitri Liakhovitski wrote: > >> Hello! >> >> I have a data frame: >> >> md <- data.frame(a = c(3,5,4,5,3,5), b = c(5,5,5,4,4,1), c = >> c(1,3,4,3,5,5), >> device =

[R] dplyr - counting a number of specific values in each column - for all columns at once

) - so that I can't specify those counts.a, counts.b, etc. manually - dozens of times. Does dplyr allow to run the count of 5s on all 'myvars' columns at once? -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCR

Re: [R] reshape data frame when one column has unequal number of entries

"))) > > Duncan > > Duncan Mackay > Department of Agronomy and Soil Science > University of New England > Armidale NSW 2351 > Email: home: mac...@northnet.com.au > > -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Di

[R] reshape data frame when one column has unequal number of entries

= as.character(x$b) x str(x) I need to produce this data frame: 1 a 1 b 1 c 2 d 2 e 3 f Is it possible without looping? Thank you! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch

Re: [R] regexpr - ignore all special characters and punctuation in a string

f happiness: Part 2." >> b = "What a nice day today: Story of happiness (Part 2)" >> sa = gsub("[^A-Za-z0-9]", "", a) >> sb = gsub("[^A-Za-z0-9]", "", b) >> a==b >> # [1] FALSE >> sa==sb >> # [1] TR

Re: [R] regexpr - ignore all special characters and punctuation in a string

I think I found a partial answer: str_replace_all(x, "[[:punct:]]", " ") On Mon, Apr 20, 2015 at 9:59 AM, Dimitri Liakhovitski wrote: > Hello! > > Please point me in the right direction. > I need to match 2 strings, but focusing ONLY on characters, ignor

[R] regexpr - ignore all special characters and punctuation in a string

ppiness: Part 2." == "What a nice day today: Story of happiness (Part 2)" -- Thank you! Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read

Re: [R] Reading .LIST files into R

IMDB. I guess I'll just have to open it as any text file. On Tue, Apr 14, 2015 at 4:48 PM, Sarah Goslee wrote: > What produced this file? > > On Tue, Apr 14, 2015 at 4:40 PM, Dimitri Liakhovitski > wrote: >> Is it possible to read a LIST file into R? Any package? >&g

[R] Reading .LIST files into R

Is it possible to read a LIST file into R? Any package? I've done some googling, but there are just too many hits for a regular 'list'. Appreciate any pointers! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUB

Re: [R] Color US counties on US map using a numeric variable for color intensity

es=c("yellow","red"),na.color="white") > > Jim > > > On Fri, Apr 3, 2015 at 8:08 AM, Dimitri Liakhovitski > wrote: >> >> Jean, I think I fixed it: >> >> newpal <- colorRamp(c("yellow", "red")) >> missin

Re: [R] Color US counties on US map using a numeric variable for color intensity

nal$Mean.Wait, na.rm=T)), maxColorValue=255)) map('county', fill=TRUE, col=newcol, resolution=0, lty=0, bg="transparent") map('state', lwd=1, add=TRUE) One understanding question: what exactly does this rgb line do and why do we have to say "maxColorValue=255

Re: [R] Color US counties on US map using a numeric variable for color intensity

ol, > resolution=0, lty=0, bg="transparent") > map('state', lwd=1, add=TRUE) > > Jean > > > On Thu, Apr 2, 2015 at 12:03 PM, Dimitri Liakhovitski > wrote: >> >> I have a data frame 'mydata.final' (see below) that contains U

[R] Color US counties on US map using a numeric variable for color intensity

in 'wait': map('county', fill = TRUE, col = allcolors[mydata.final$wait], resolution = 0, lty = 0, bg = "transparent") map('state', lwd=1, add=TRUE) My goal is: instead of splitting 'Mean.Wait' into 5 ordered categories

Re: [R] Why does R replace all row values with NAs

frame(a=1:10,b=2:11,c=c(1,NA,3,NA,5,NA,7,NA,NA,10)) > > x[is.true(x$c >= 6), ] > a b c > 7 7 8 7 > 10 10 11 10 > > > Bill Dunlap > TIBCO Software > wdunlap tibco.com > > On Fri, Feb 27, 2015 at 7:27 AM, Dimitri Liakhovitski > wrote: >&g

Re: [R] Why does R replace all row values with NAs

Thank you very much, Duncan. All this being said: What would you say is the most elegant and most safe way to solve such a seemingly simple task? Thank you! On Fri, Feb 27, 2015 at 10:02 AM, Duncan Murdoch wrote: > On 27/02/2015 9:49 AM, Dimitri Liakhovitski wrote: >> So, Dun

Re: [R] Why does R replace all row values with NAs

, but I'll replace all the values in this row with NAs? On Fri, Feb 27, 2015 at 9:13 AM, Duncan Murdoch wrote: > On 27/02/2015 9:04 AM, Dimitri Liakhovitski wrote: >> I know how to get the output I need, but I would benefit from an >> explanation why R behaves the way it doe

[R] Why does R replace all row values with NAs

ith NAs? x[x$c<6,] # Leaves rows with c=NA, but makes the whole row an NA. Why??? x[(x$c<6) | is.na(x$c),] # output I need - I have to be super-explicit Thank you very much! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UN

[R] get_map in ggplot doesn't allow the exact specification of my box's corners?

unding box to center/zoom specification. (experimental) Does it mean that there is no way for me to create a map with these exact corners? Thank you! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https:/

Re: [R] More elegant way of stacking the data

.c2 2 b c2 3 2 > 3.c2 3 c c2 4 3 > 4.c2 4 d c2 5 4 > 5.c2 5 e c2 6 5 > 1.c3 1 a c3 3 1 > 2.c3 2 b c3 4 2 > 3.c3 3 c c3 5 3 > 4.c3 4 d c3 6 4 > 5.c3 5 e c3 7 5 > 1.c4 1 a c4 4 1 > 2.c4 2 b c4 5 2 > 3.c4 3 c c4 6 3 >

[R] More elegant way of stacking the data

mylist[[i]]<-df[c("a","b",myname)] names(mylist[[i]])<-c("a","b","c") } out<-do.call(rbind,mylist) out Thank you! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list https://stat.ethz.c

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