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How to remove part of characters in data frame column

Time:01-28

I have a data frame as follows (part of a larger set):

data frame

for the column raw$Zipcode I need to remove the two 00 before the Zipcode number for Swedish zipcodes (preferably through a function). I am very new to R and have found gsub and strsplit but can't seem to make it work:

raw2 <- unlist(strsplit(raw$ZipCode, split="00", fixed=TRUE))[2]

The zeroes are characters as other countries in the data set have letters. How can I remove the first two zeroes in all cases of the two first character letters being zeroes in a column?

CodePudding user response:

There are multiple ways of doing this:

  1. Using as.numeric on a column of your choice.
raw$Zipcode <- as.numeric(raw$Zipcode)
  1. If you want it to be a character then you can use stringr package.
library(stringr)
raw$Zipcode <- str_replace(raw$Zipcode, "^0 " ,"")
  1. There is another function called str_remove in stringr package.
raw$Zipcode <- str_remove(raw$Zipcode, "^0 ")
  1. You can also use sub from base R.
raw$Zipcode <- sub("^0 ", "", raw$Zipcode)

But if you want to remove n number of leading zeroes, replace with {n} to remove them.

For instance to remove two 0's use sub("^0{2}", "", raw$Zipcode).

CodePudding user response:

v <- c("00345", "00045", "12345", "12005")

sub("^0{2}", "", v)

# [1] "345"   "045"   "12345" "12005"
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