I have a column ( samplename_date
) of a dataframe that looks as follows:
'008Q06-03 RGD17-48 3.8W Wm 1xtl' 03 July
I'm trying to split this into two columns, after the second single quote. I've been trying:
temp[['samplename','date']] = df['samplename_date'].str.split(''\\s', expand = True)
and variations thereof, but I can't seem to figure out how to handle the single quote in the regular expression pattern.
You need to escape single quote as well
temp[['samplename','date']] = df['samplename_date'].str.split('\'\s', expand = True)
You get
samplename date
0 '008Q06-03 RGD17-48 3.8W Wm 1xtl 03 July
I would personally use str.extract
temp[['samplename','date']] = df['samplename_date'].str.extract('\'(.*)\'\s(.*)', expand = True)
samplename date
0 008Q06-03 RGD17-48 3.8W Wm 1xtl 03 July
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.