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I have a column containing values of 3 strings separated by semicolons. I need to just extract the part of the string which comes before the first semicolon.
Type <- c("SNSR_RMIN_PSX150Y_CSH;SP_12;I0.00V50HX0HY3000")
What I want is: Get the first part of the string (till the first semicolon).
Desired output : SNSR_RMIN_PSX150Y_CSH
I tried gsub without success.
235
The substr() method extracts a part of a string. The substr() method begins at a specified position, and returns a specified number of characters. The substr() method does not change the original string. To extract characters from the end of the string, use a negative start position.
Extract text before or after space with formula in Excel You can quickly extract the text before space from the list only by using formula. Select a blank cell, and type this formula =LEFT(A1,(FIND(" ",A1,1)-1)) (A1 is the first cell of the list you want to extract text) , and press Enter button.
Use the substring() method to get the substring before a specific character, e.g. const before = str. substring(0, str. indexOf('_')); . The substring method will return a new string containing the part of the string before the specified character.
Use the split() method to cut string before the character in Python. The split() method splits a string into a list.
The stringi package works very fast here:
stri_extract_first_regex(Type, "^[^;]+")
## [1] "SNSR_RMIN_PSX150Y_CSH"
I benchmarked on the 3 main approaches here:
Unit: milliseconds
expr min lq mean median uq max neval
SAPPLY() 254.88442 267.79469 294.12715 277.4518 325.91576 419.6435 100
SUB() 182.64996 186.26583 192.99277 188.6128 197.17154 237.9886 100
STRINGI() 89.45826 91.05954 94.11195 91.9424 94.58421 124.4689 100
Here's the code for the Benchmarks:
library(stringi)
SAPPLY <- function() sapply(strsplit(Type, ";"), "[[", 1)
SUB <- function() sub(';.*$','', Type)
STRINGI <- function() stri_extract_first_regex(Type, "^[^;]+")
Type <- c("SNSR_RMIN_PSX150Y_CSH;SP_12;I0.00V50HX0HY3000")
Type <- rep(Type, 100000)
library(microbenchmark)
microbenchmark(
SAPPLY(),
SUB(),
STRINGI(),
times=100L)
When performance is important you can use substr in combination with regexpr from base.
substr(Type, 1, regexpr(";", Type, fixed=TRUE)-1)
#[1] "SNSR_RMIN_PSX150Y_CSH"
Timings: (Reusing the part from @tyler-rinker)
library(stringi)
SAPPLY <- function() sapply(strsplit(Type, ";"), "[[", 1)
SUB <- function() sub(';.*$','', Type)
SUB2 <- function() sub(';.*','', Type)
SUB3 <- function() sub('([^;]*).*','\\1', Type)
STRINGI <- function() stri_extract_first_regex(Type, "^[^;]+")
STRINGI2 <- function() stri_extract_first_regex(Type, "[^;]*")
SUBSTRREG <- function() substr(Type, 1, regexpr(";", Type)-1)
SUBSTRREG2 <- function() substr(Type, 1, regexpr(";", Type, fixed=TRUE)-1)
SUBSTRREG3 <- function() substr(Type, 1, regexpr(";", Type, fixed=TRUE, useBytes = TRUE)-1)
Type <- c("SNSR_RMIN_PSX150Y_CSH;SP_12;I0.00V50HX0HY3000")
Type <- rep(Type, 100000)
library(microbenchmark)
microbenchmark(SAPPLY(), SUB(), SUB2(), SUB3(), STRINGI()
, STRINGI2(), SUBSTRREG(), SUBSTRREG2(), SUBSTRREG3())
#Unit: milliseconds
# expr min lq mean median uq max neval
# SAPPLY() 382.23750 395.92841 412.82508 410.05236 427.58816 460.28508 100
# SUB() 111.92120 114.28939 116.41950 115.57371 118.15573 123.92400 100
# SUB2() 94.27831 96.50462 98.14741 97.38199 99.15260 119.51090 100
# SUB3() 167.77139 172.51271 175.07144 173.83121 176.27710 190.97815 100
# STRINGI() 38.27645 39.33428 39.94134 39.71842 40.50182 42.55838 100
# STRINGI2() 38.16736 39.19250 40.14904 39.63929 40.37686 56.03174 100
# SUBSTRREG() 45.04828 46.39867 47.13018 46.85465 47.71985 51.07955 100
# SUBSTRREG2() 10.67439 11.02963 11.29290 11.12222 11.43964 13.64643 100
# SUBSTRREG3() 10.74220 10.95139 11.39466 11.06632 11.46908 27.72654 100
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