CODEWITHSUNDEEP
rreplacecharacter
Gabriel Voelcker
Gabriel Voelcker

Reputation: 1

How to delete the 0 before a number?

I'm having the following difficulty in R: a dataframe has a column with some IDs, which are in the character format. What's the most concise way I can delete the 0s in front?

Example:

This is what I have:

ID <- as.character(c("001001","0001002","01003","001004","1005"))
order <- c("a","b","c","d","e")
df <- as.data.frame(cbind(ID, order))

This is what I want:

ID2 <- as.character(c("1001","1002","1003","1004","1005"))
order2 <- c("a","b","c","d","e")
df2 <- as.data.frame(cbind(ID, order))

I've tried replacing strings but it deletes the 0s I don't want (ex: the ID2[1] = 11).

Thanks in advance!

Upvotes: 0

Views: 233

Answers (2)

Frank Zhang
Frank Zhang

Reputation: 1688

use trimws from base R

trimws(c("001001","0001002","01003","001004","1005"),which = "left",whitespace = "0")
#> [1] "1001" "1002" "1003" "1004" "1005"

Created on 2020-06-30 by the reprex package (v0.3.0)

Upvotes: 7

akrun
akrun

Reputation: 886938

It is easier to do this if we convert to integer or numeric class as numeric values cannot have 0 prefix. After the conversion, just wrap with as.character if we need the class to remain as character

df$ID <- as.character(as.integer(df$ID))
df$ID
#[1] "1001" "1002" "1003" "1004" "1005"

It could also be done in a regex way (unnecessary though)

df$ID <- sub("^0+", "", df$ID)

In the above code, we match one or more 0s (0+) at the start (^) of the string and replace with blank ("")

if the IDs have characters other than digits, an option is also to capture the digits after the prefix 0's and replace with the backreference (\\1) of the captured groups. This would make sure that strings "0xyz" remains as such

df$ID <- sub("^0+(\\d+)$", "\\1", df$ID)

Upvotes: 2

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