Hello,
It also works with class "factor":
df <- data.frame(variable = c("12.6%", "30.9%", "61.4%"))
class(df$variable)
#[1] "factor"
as.numeric(gsub(pattern = "%", "", df$variable))
#[1] 12.6 30.9 61.4
This is because sub() and gsub() return a character vector and the
instruction becomes an equivalent of what the help page ?factor
documents in section Warning:
To transform a factor f to approximately its original numeric values,
as.numeric(levels(f))[f] is recommended and slightly more efficient than
as.numeric(as.character(f)).
Also, I would still prefer
as.numeric(sub(pattern = "%$","",df$variable))
#[1] 12.6 30.9 61.4
The pattern is more strict and there is no need to search&replace
multiple occurrences of '%'.
Hope this helps,
Rui Barradas
On 18/08/2018 23:08, GALIB KHAN wrote:
So running the code in my head....as long as that column's data type is a
vector of characters then it should work.
Did you try it out?
On Sat, Aug 18, 2018, 5:02 PM Jeff Reichman <reichm...@sbcglobal.net> wrote:
Given it’s a variable would I just change the 12.6 in
as.numeric(gsub(pattern = "%","","12.6%"))
To the variable name say … as.numeric(gsub(pattern = "%","",df$variable))
*From:* GALIB KHAN <gh...@scarletmail.rutgers.edu>
*Sent:* Saturday, August 18, 2018 4:23 PM
*To:* reichm...@sbcglobal.net
*Cc:* r-help@r-project.org
*Subject:* Re: [R] Converting chr to num
Hey there,
as.numeric(gsub(pattern = "%","","12.6%"))
On Sat, Aug 18, 2018 at 4:20 PM, Jeff Reichman <reichm...@sbcglobal.net>
wrote:
R-Help Forum
How do I convert a chr variable that contains percentages to an integer
Example 12.6% (chr) to 12.6 (int)
Jeff
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