similar to: fill out a PDF form in R

Hi Elahe, I have no clue, but maybe you can dump the data fields using pdftk, and work with those in R. HTH Ulrik On Wed, 26 Jul 2017 at 13:50 Elahe chalabi via R-help <r-help at r-project.org> wrote: > Hi all, > > I would like to get ideas about how to fill out a PDF form in R and to > know if it's possible or not. I could not find something helpful in > Internet.

Hi all, I have the following data and I have a histogram for mms like ggplot(hist,aes(x=hist$mms))+ geom_histogram(binwidth=1,fill="white",color="black")and then I want to fill the color of histogram by probable=1 and probable=0, could anyone help me in this? My data: structure(list(probable = c(1L, 0L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 1L,

define your string as whatever object you want: data <- "Mhm . Alright . There's um a young boy that's getting a cookie jar . And it he's uh in bad shape because uh the thing is falling over . And in the picture the mother is washing dishes and doesn't see it . And so is the the water is overflowing in the sink . And the dishes might get falled over if you don't fell

You can define stop words as below. data <- tm_map(data, removeWords, stopwords("english")) Patrick Casimir, PhD Health Analytics, Data Science, Big Data Expert & Independent Consultant C: 954.614.1178 ________________________________ From: R-help <r-help-bounces at r-project.org> on behalf of Bert Gunter <bgunter.4567 at gmail.com> Sent: Monday, June 12, 2017

Thanks for your reply. I know the command data <- tm_map(data, removeWords, stopwords("english")) removes English stop words, I don't know how should I count stop words of my string: str="Mhm . Alright . There's um a young boy that's getting a cookie jar . And it he's uh in bad shape because uh the thing is falling over . And in the picture the mother is

You can use regular expressions. ?regex and/or the stringr package are good places to start. Of course, you have to define "stop words." 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 Breathed in his "Bloom County" comic strip ) On Mon, Jun 12, 2017 at 5:40

Defining data as you mentioned in your respond causes the following error: Error in UseMethod("tm_map", x) : no applicable method for 'tm_map' applied to an object of class "character" I can solve this error by using Corpus(VectorSource(my string)) and the using your command but I cannot see the number of stop words in my string! On Monday, June 12, 2017 8:36

It's not what I want, the first data frame has 499 observations and the second data frame is a subset of the first one but with 375 observations. I want something that returns the ID for training data frame On Wednesday, November 1, 2017 10:18 AM, Eric Berger <ericjberger at gmail.com> wrote: matches <- merge(training,data,by=intersect(names(training),names(data))) HTH, Eric

Hi all, How can I overlay these two histograms? ggplot(gg, aes(gg$Alz, fill = gg$veg)) + geom_histogram(alpha = 0.2) ggplot(tt, aes(tt$Cont, fill = tt$veg)) + geom_histogram(alpha = 0.2) thanks for any help! Elahe

Hi all, Is there a way in R to count the number of stop words (English) of a string using tm package? str="Mhm . Alright . There's um a young boy that's getting a cookie jar . And it he's uh in bad shape because uh the thing is falling over . And in the picture the mother is washing dishes and doesn't see it . And so is the the water is overflowing in the sink . And the

'grepl' returns a logical vector; you have to use this to get your subset. You can use: df_tq <- subset(df, grepl("t1", Command)) df_t2 <- subset(df, grepl("t2", Command)) # if you want to also get a subset that has both, use df_both <- subset(df, grepl("t1", Command) & grepl("t2", Command)) Jim Holtman Data Munger Guru What is

Any responds?! On Wednesday, August 23, 2017 5:50 AM, Elahe chalabi via R-help <r-help at r-project.org> wrote: Hi all, I would like to do cross validation in random forest using rfcv function. As the documentation for this package says: rfcv(trainx, trainy, cv.fold=5, scale="log", step=0.5, mtry=function(p) max(1, floor(sqrt(p))), recursive=FALSE, ...) however I

But they row.names() cannot give me the IDs On Wednesday, November 1, 2017 9:45 AM, David Wolfskill <r at catwhisker.org> wrote: On Wed, Nov 01, 2017 at 04:13:42PM +0000, Elahe chalabi via R-help wrote: > Hi all, > I have two data frames that one of them does not have the column ID: > > > str(data) > 'data.frame': 499 obs. of 608 variables:

I think I would use pdftk to extract the form data. All subsequent manipulation in R. HTH Ulrik Eric Berger <ericjberger at gmail.com> schrieb am Mi., 24. Jan. 2018, 08:11: > Hi Scott, > I have never done this myself but I read something recently on the > r-help distribution that was related. > I just did a quick search and found a few hits that might work for you. > >

now after this: df_both <- subset(df, grepl("t1", Command) & grepl("t2", Command)) I use factor to apply the subset to df but then the Command level becomes 0 df_both$Command=factor(df_both$Command) str(df_both) $ Protocol : Factor w/ 0 levels: Do you know what is the reason? Thanks for replying On Sunday, April 24, 2016 12:18 PM, jim

Hi all, I would like to do cross validation in random forest using rfcv function. As the documentation for this package says: rfcv(trainx, trainy, cv.fold=5, scale="log", step=0.5, mtry=function(p) max(1, floor(sqrt(p))), recursive=FALSE, ...) however I don't know how to build trianx and trainy for my data set, and I could not understand the way trainx is built in the package

Using Ulrik?s example data (and assuming I understand what is wanted), here is what I would do: ex.dat <- c("FName: fname1", "Fval: Fval1.name1", "Fval: ", "FName: fname2", "Fval: Fval2.name2", "FName: fname3") tst <- data.frame(x = ex.dat, stringsAsFactors=FALSE) sp <- strsplit(tst$x, ':', fixed=TRUE) chk <-

http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example Overlaying aesthetics is possible. Overlaying graphs is not. Without sample data, concrete examples will be unlikely to appear, so read the above link and pay attention to the dput function. -- Sent from my phone. Please excuse my brevity. On April 20, 2016 3:01:43 PM PDT, "ch.elahe via R-help"

Hi all, I have two data frames that one of them does not have the column ID: > str(data) 'data.frame': 499 obs. of 608 variables: $ ID : int 1 2 3 4 5 6 7 8 9 10 ... $ alright : int 1 0 0 0 0 0 0 1 2 1 ... $ bad : int 1 0 0 0 0 0 0 0 0 0 ... $ boy : int 1 2 1 1 0 2 2 4 2 1 ... $ cooki : int 1 2 2 1 0 1 1 4 2 3

@Don your solution does not solve Vijayan's scenario 2. I used spread and gather for that. An alternative solution to insert mising Fval - picking up with Don's newtst - is newtst <- c("FName: fname1", "Fval: Fval1.name1", "FName: fname2", "Fval: Fval2.name2", "FName: fname3", "FName: fname4", "Fval: fval4.fname4")