similar to: count number of stop words in R

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

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

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

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

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

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:

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

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. Does anyone know a good link for that or have experience in this? Thanks for any help! Elahe

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

'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

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 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, 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

Thanks Jean, Does anyone know how to set these [hast1] and [hast2] as the colors of a plot? On Friday, April 22, 2016 7:39 AM, "Adams, Jean" <jvadams at usgs.gov> wrote: You can use the grepl() function to give you logicals for each criterion, then combine them as needed. For example: # example version of Command Command <- paste0("_localize_",

Hi all, I have tried a 5 fold cross validation using caret package with random forest method on iris dataset as example. Then I need ROC curve for each fold: > set.seed(1) > train_control <- trainControl(method="cv", number=5,savePredictions = TRUE,classProbs = TRUE) > output <- train(Species~., data=iris, trControl=train_control, method="rf") >

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,

In a data frame, I would like to assign two or more factors to one observation. Is it possible? If so, how? Example: In the variable $ dishes, there are several levels: "cup", "plate", "saucer". In my first observation, I see a saucer only. But in the second one, I see a cup and a saucer. In the third, however, there is only a cup, in the fourth a plate only and

Hi, I am trying to test some contrasts, using glht() in multcomp package on fixed effects in a linear mixed model fitted with lme() in nlme package. The command I used is: ## a simple randomized block design, ## type is fixed effect ## batch is random effect ## model with interaction dat.lme<-lme(info.index~type, random=~1|batch/type, data=dat) glht(dat.lme, linfct = mcp(type