CODEWITHSUNDEEP
rdataframeextract
velvetrock
velvetrock

Reputation: 585

Extract words from a column of data.frame

In my data, there is a column like :

df <- data.frame(status = c("GET/sfuksd1567","GET/sjsh787","POST/hsfhuks","GET/sfukfiezd17","POST/fshks"), stringsAsFactors = FALSE)

I want to create another column automatically which is the indicator of the variable status and it only extracts the "GET" or "POST", like df$ind=c("GET","GET","POST","GET","POST").

I've tried the function substr, but I didn't success.

Original data:

> df
           status
1  GET/sfuksd1567
2     GET/sjsh787
3    POST/hsfhuks
4 GET/sfukfiezd17
5      POST/fshks

Expected result:

> df
           status  ind
1  GET/sfuksd1567  GET
2     GET/sjsh787  GET
3    POST/hsfhuks POST
4 GET/sfukfiezd17  GET
5      POST/fshks POST

Upvotes: 2

Views: 4242

Answers (3)

akrun
akrun

Reputation: 886938

We can use stri_extract_first_words from library(stringi)

library(stringi)
stri_extract_first_words(df$status)
#[1] "GET"  "GET"  "POST" "GET"  "POST"

Another option from tidyr is extract

extract(df, status, into='ind', '([^/]+)/.*', remove=FALSE)

Benchmarks

Using the stri_extract_first_words, the benchmarks are:

david <- function() sub('/.*', '', df$status)
etienne <- function() sapply(strsplit(df$status,'/'),`[[`,1)
akrun <- function()stri_extract_first_words(df$status)
df <-  df[sample(1:nrow(df), 1e6, replace=TRUE),, drop=FALSE]
library(microbenchmark)
microbenchmark(david(), etienne(), akrun(), unit='relative', times=20L)
#Unit: relative
#      expr      min       lq     mean   median       uq      max neval
#   david() 1.826192 1.824263 1.781562 1.814156 1.788085 1.699008    20
# etienne() 4.935629 5.159218 5.136180 5.198875 5.137107 5.930806    20
#   akrun() 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000    20

NOTE: There are other options in @David Arenburg's post. I am guessing the sub version to be faster. I could be wrong.

Upvotes: 2

David Arenburg
David Arenburg

Reputation: 92282

You could simply remove everything after the backslash using regex

df$ind <- sub("/.*", "", df$status)
df
#            status  ind
# 1  GET/sfuksd1567  GET
# 2     GET/sjsh787  GET
# 3    POST/hsfhuks POST
# 4 GET/sfukfiezd17  GET
# 5      POST/fshks POST

Or if you don't like regex, you could try

library(tidyr)
separate(df, "status", c("ind", "status"))

Or

library(data.table) ## V1.9.6+
setDT(df)[, tstrsplit(status, "/")]

Or

read.table(text = df$status, sep = "/")

The last three options will just split the status columns into two separate ones.

Upvotes: 11

etienne
etienne

Reputation: 3678

We have :

df<-data.frame(status=c("GET/sfuksd1567","GET/sjsh787","POST/hsfhuks","GET/sfukfiezd17","POST/fshks"),stringsAsFactors=F)

You can do:

df$ind<-sapply(1:nrow(df),function(x){strsplit(df$status,'/')[[x]][1]})

or 

df$ind<-sapply(strsplit(df$status,'/'),`[[`,1)

Both return

df
           status  ind
1  GET/sfuksd1567  GET
2     GET/sjsh787  GET
3    POST/hsfhuks POST
4 GET/sfukfiezd17  GET
5      POST/fshks POST

Benchmark :

microbenchmark(david=sub("/.*", "", df$status),etienne=sapply(strsplit(df$status,'/'),`[[`,1))

Unit: microseconds
    expr    min      lq     mean  median     uq     max neval cld
   david 25.198 25.8985 27.64456 26.5980 27.298 116.189   100  a 
 etienne 62.294 63.3440 65.13979 63.8695 65.094 128.088   100   b

Upvotes: 3

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