similar to: grep

With today's R 2.10.0(devel) I get: > anyDuplicated(c(1,NA,3,NA,5), incomp=NA) # expect 0 Warning: stack imbalance in 'anyDuplicated', 20 then 21 Warning: stack imbalance in '.Internal', 19 then 20 Warning: stack imbalance in '{', 17 then 18 [1] 0 > anyDuplicated(c(1,NA,3,NA,3), incomp=NA) # expect 5 Warning: stack imbalance in 'anyDuplicated', 20 then 21

Good Morning, I am trying to create a new column of character strings based on the first two letters in a string in another column. I believe that I need to use some combination of ifelse and grep but I am not totally sure how to combine them. I am not totally sure why the command below isn't working. Obviously it isn't finding anything that matches my criteria but I am not sure why. Any

Hi, In man page for duplicated: Value: ?duplicated()?: For a vector input, a logical vector of the same length as ?x?. For a data frame, a logical vector with one element for each row. For a matrix or array, a logical array with the same dimensions and dimnames. When 'x' is a matrix or array, the returned value is NOT a logical array: > m <-

Does anybody know how to negate a string in a grep command, i.e., what I need is to return only strings that do NOT contain a second string anywhere in the entire string. for txt <- c("boo","goo","doob","foo","boofoo") I need a grep command that returns strings with "oo" except when "b" is present anywhere. I know that

Could not get "which" to work, but my grep worked. Thanks. > which(grep("very|somewhat",names(goprobit.p$est))) Error in which(grep("very|somewhat", names(goprobit.p$est))) : argument to 'which' is not logical > grep("very|somewhat",names(goprobit.p$est)) [1] 6 7 8 9 10 11 12 13 28 29 30 31 32 33 34 35 50 51 52 53 54 55 56 57 On 7/12/2024

Hi all, Could you please help me? I am looking for books/pointers/resources/tutorials on visualizing complex/big data and on understanding multivariate relations in complicated data. More specifically, we have categorical variables and are interested in how to visualize the categorical data and visualize data conditioned upon categorical values. Could anybody please give me some pointers?

I have a fairly large data set with six variables set up like the following dummy: # Create fake data df <- data.frame(code = c(rep(1001, 8), rep(1002, 8)), year = rep(c(rep(2011, 4), rep(2012, 4)), 2), fund = rep(c("10E", "10E", "10E", "27E"), 4), func = rep(c("110000",

R-Devel, I store and retrieve a large amount of financial data (millions of rows) in a PostgreSQL database keyed by date (and represented in R by class Date). Unfortunately, I frequently find that a great deal of processing time is spent converting dates from character representations to Date class representations in R, presumably because strptime is not fast for large vectors (>10,000

When 'getOption("checkPackageLicense")' is 'TRUE' and the user calls 'library(grid)' for the first time, R asks the user to either accept or decline the package license. This should not be necessary as the package license is "Part of R ..." with "..." denoting the R version number, and R is free and open source. The unnecessary license

Suppose I have a vector ["A_cont_1", "A_cont_12", "B_treat_8", "AB_cont_22", "cont_21_Aa"], I hope I can extract the strings which include 3 short strings, say "A", "cont" and "2", that is to say, "A_cont_12", "AB_cont_22" and "cont_21_Aa" will be extract, using a relatively short code

I am using grep to locate colnames to automate a report build and have run into a problem when a colname is not found. The use of integer(0) in a conditional statement seems to be a no no as it has length 0. Below is a self-contained trivial example. I would like to get something like "NA" or -1 for the position when it is not found OR learn a way to use integer(0) or some

'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

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

Dear r-help members, I have a number in the form of a string, say: a<-"-01020.909200" I'd like to extract "1020." as well as ".9092" Front<-grep(pattern="[1-9]+[0-9]*\\.", value=TRUE, x=a, fixed=FALSE) End<-grep(pattern="\\.[0-9]*[1-9]+", value=TRUE, x=a, fixed=FALSE) However, both strings give "-01020.909200", exactly

Hi all, I have one factor variable in my df and I want to extract the names from it which contain both "t2" and "pd": 'data.frame': 36919 obs. of 162 variables $TE :int 38,41,11,52,48,75,..... $TR :int 100,210,548,546,..... $Command :factor W/2229 levels

Dear All, I am experiencing some problems with a script of mine. It crashes with this message Error in grepl(fut_string, past_string) : invalid regular expression

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Hi I am relatively new to R and am trying to figure out to plot 3d scatter plot using defined colors based on x-axis and y-axis values. Right now in the code below, I assign colors based on certain values in the names of the x-axis. Now if I want to extend the condition to assign a color based on the names of both x-axis and y-axis values, what should I be doing? Any help or ideas would be

I do not follow you Gabriel. Package name must not use digit numbers. Tarbal will use them, taken from the DESCRIPTION file, version field. That's why I consider the weird case name you presented as irrelevant, and not to be considered. Le ven. 9 ao?t 2019 ? 20:41, Gabriel Becker <gabembecker at gmail.com> a ?crit : > > > On Fri, Aug 9, 2019 at 11:05 AM neonira Arinoem

Inspired by this old Stack Overflow question https://stackoverflow.com/questions/19765610/when-does-locale-affect-rs-regular-expressions I was wondering why this is TRUE: Sys.setlocale("LC_ALL", "et_EE") grepl("[A-Z]", "T") TRE's documentation at <https://laurikari.net/tre/documentation/regex-syntax/> says that a range "is shorthand for