Conditionally Remove Leading or Trailing '.' Character in R

conditionally remove leading or trailing `.` character in R

Use regular expressions:

test <- c('.name.1.','name.2','.name.3.')
gsub('^\\.|\\.$', '', test)
# [1] "name.1" "name.2" "name.3"

The two backslashes, \\, in the regular expression escape the dot, ., which would actually mean any character. The caret, ^, marks the beginning of the string, the dollar, $, the end of the string. The pipe, |, is a logical "or". So in essence the regular expression matches a dot at the beginning of the string or a dot at the end of the string and replaces it with an empty string.

More information on regular expressions can be found here and information on gsub and related functions here.

conditionally remove first character

We can use str_remove to match the letter 'D' at the start (^) of the string

library(dplyr)
library(stringr)
DF %>%
   mutate(code2 = str_remove(code, '^D'))

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Or in base R with sub

DF$code2 <- sub('^D', '', DF$code)

Remove first character of string with condition in R

We can specify the ^0+ as pattern i.e. one or more 0s at the start (^) of the string instead of . (. in regex matches any character)

data_individual$Zipcode <- sub("^0+", "", data_individual$Zipcode)

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Or with tidyverse

library(stringr)
data_individual$Zipcode <- str_remove(data_individual$Zipcode, "^0+")

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Another option without regex would be to convert to numeric as numeric values doesn't support prefix 0 (assuming all zipcodes include only digits)

data_individual$Zipcode <- as.numeric(data_individual$Zipcode)

Remove trailing /'

How about this?

my.url <- "http://test.test/test.test.html/"
gsub("/$", "", my.url)

it returns

[1] "http://test.test/test.test.html"

you can use $ to look for elements at the end of the line

Remove character based on condition and position in string - R

While akrun's approach is the one that should be used. Here is stringr composition:

1. with word(df, 1) we take the left part of the string
2. with word(df, -1) we take the right part (here we use
   2a. str_remove_all with regex ^0+ to remove leading zeros.
3. Finally we use str_c to combine both parts:

library(stringr)
str_c(word(df,1), str_remove_all(word(df, -1), '^0+'))

[1] "2015808"    "201341"     "20155"      "2015301585" "2015311585" "2014380096" "2013100041"

R remove first character from string

If we need to remove the first character, use sub, match one character (. represents a single character), replace it with ''.

sub('.', '', listfruit)
#[1] "applea"  "bananab" "ranggeo"

Or for the first and last character, match the character at the start of the string (^.) or the end of the string (.$) and replace it with ''.

gsub('^.|.$', '', listfruit)
#[1] "apple"  "banana" "rangge"

We can also capture it as a group and replace with the backreference.

sub('^.(.*).$', '\\1', listfruit)
#[1] "apple"  "banana" "rangge"

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Another option is with substr

substr(listfruit, 2, nchar(listfruit)-1)
#[1] "apple"  "banana" "rangge"

Conditionally removing from array if last character is []

You can remove the unwanted tickers using regex.

For example, with grep :

stocktickers = c("PETR4", "VALE3", "MDNE3","BDLL4F ","QCOM34F", "SANB11","USIM5")
grep("(F|\\s)$",stocktickers, value = TRUE, invert = TRUE)
#[1] "PETR4"  "VALE3"  "MDNE3"  "SANB11" "USIM5" 

This removes the values from stocktickers where it ends either with "F" or whitespace.

remove last two characters only if it equals a condition

Try this

str_replace_all(strings, 'sl$','')

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