Remove part of a string in dataframe column (R)
If your string is not too fancy/complex, it might be easiest to do something like:
gsub("C([0-9]+)_.*", "\\1", df$Col2)
# [1] "607989" "607989" "607989" "607989" "607989" "607989"
Start with a "C", followed by digits, followed by an underscore and then anything else. Digits are captured with ()
, and the replacement is set to that capture group (\\1
).
How to remove part of string value in column using stringr library?
You can use str_remove
which is shortcut for str_replace
with empty replacement (""
)
as.numeric(stringr::str_remove(df$column1, "value-"))
Or in base R -
as.numeric(sub("value-", "", df$column1))
How to remove part of characters in data frame column
There are multiple ways of doing this:
- Using
as.numeric
on a column of your choice.
raw$Zipcode <- as.numeric(raw$Zipcode)
- If you want it to be a
character
then you can usestringr
package.
library(stringr)
raw$Zipcode <- str_replace(raw$Zipcode, "^0+" ,"")
- There is another function called
str_remove
instringr
package.
raw$Zipcode <- str_remove(raw$Zipcode, "^0+")
- You can also use
sub
from base R.
raw$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)
.
How to remove part of a string in a column of dataframe in R?
You can do this with sub
and a regular expression.
df$S = sub("\\|.*", "", as.character(df$S))
df
S S1 S2 S3 S4
1 100130426 0.0000 0.0000 0.9066 0.0000
2 100133144 16.3644 9.2659 11.6228 12.0894
3 100134869 12.9316 17.3790 9.2294 11.0799
4 3457 1910.3000 2453.5000 2695.3700 1372.3624
5 9834 1660.1300 857.3000 1240.5300 1434.6463
6 ATP5L2 0.0000 0.0000 0.9066 0.0000
7 ATP5L 1510.2900 1270.7900 2965.5400 2397.1866
8 ATP5O 2176.1700 1868.9500 2004.5300 2360.3641
Details:
sub
substitutes the second argument for whatever matches the first argument. In this case, we want | and everything after it. You can't just write | because that has a special meaning in regular expressions so you "escape" it with by writing \\|. It is followed by .*. The . means "any character" and * means any number of times, so together \\|.* means | followed by any number of characters. We replace that with the empty string "". We apply this operation to as.character(df$S)
because your error message makes it look like your variable df$S
may be a factor, rather than a string.
How can I remove parts of string based on other column in R?
Replace empty patterns with ^$
dt$ToRemove[dt$ToRemove == ''] <- '^$'
and then use stringr::str_remove
which is vectorised.
dt$result <- stringr::str_remove(dt$SomeText, dt$ToRemove)
dt
# SomeText ToRemove result
#1 ABCDEF A BCDEF
#2 ABCDEF CDE ABF
#3 ABCDEF ^$ ABCDEF
r- how to remove a particular string from column values
As an example see this process:
# example data
x = c("Full Name A B", "Full Name F B")
y = c("Playing role G G", "Playing role G M")
dt = data.frame(x,y)
dt
# x y
# 1 Full Name A B Playing role G G
# 2 Full Name F B Playing role G M
library(dplyr)
dt %>% mutate_all(~gsub("Full Name |Playing role |Batting style |Bowling style ", "", .))
# x y
# 1 A B G G
# 2 F B G M
How to remove a specific character within a column in r?
You can use the gsub
function. Use a ^ to indicate the start of the string so that you don't remove 0s elsewhere.
x$VARIANT_ID <- gsub("^0", "", x$VARIANT_ID)
Removing some text string and characters from a column in dataframe in R
We can match the .
(\\.
- escaped as it is a metacharacter that matches any character) and one or more digits (\\d+
) till the end ($
) of the string and replace with blank (""
) and wrap with gsub
to match the backquote ("`") and remove it
df$Regression <- gsub("`", "", sub("\\.\\d+$", '', df$Regression))
df$Regression
[1] "TLC~7_A" "TLC~7_A" "TLC~7_A" "TLC~7_A" "TLC~7_A" "TLC~7_A"
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