So I want to take only the first three characters of a pandas column and match them. This is what I have come up with but the implementation is incorrect:
df.loc[df[0:2] == 'x, y] = 'x'
-
1Possible duplicate of how to do a left,right and mid of a string in a pandas dataframejpp– jpp2018-08-02 15:25:21 +00:00Commented Aug 2, 2018 at 15:25
Add a comment
|
1 Answer
You are close, need str and define column for replacement if df is DataFrame, also for x, y there is 4 characters with whitespace:
df.loc[df['col'].str[:4] == 'x, y', 'col'] = 'x'
#another solution
#df.loc[df['col'].str.startswith('x, y'), 'col'] = 'x'
If working with Series:
s[s.str[:4] == 'x, y'] = 'x'
Sample:
df = pd.DataFrame({'col':['x, y temp', 'sx, y', 'x, y', 's']})
print (df)
col
0 x, y temp
1 sx, y
2 x, y
3 s
#if want replace substring
df['col1'] = df['col'].str.replace('^x\, y', 'x')
#if want set new value if condition
df.loc[df['col'].str[:4] == 'x, y', 'col'] = 'x'
print (df)
col col1
0 x x temp <-col1 replace only substring
1 sx, y sx, y
2 x x
3 s s
