You can use map
or apply
, as mentioned in this comment:
print (df.userid.map(lambda x: '{:.0f}'.format(x)))
0 nan
1 109117800000
2 113785600000
Name: userid, dtype: object
df.userid = df.userid.map(lambda x: '{:.0f}'.format(x))
print (df)
userid
0 nan
1 109117800000
2 113785600000
I wondered whether map
would be faster, but it is the same:
#[300000 rows x 1 columns]
df = pd.concat([df]*100000).reset_index(drop=True)
#print (df)
In [40]: %timeit (df.userid.map(lambda x: '{:.0f}'.format(x)))
1 loop, best of 3: 211 ms per loop
In [41]: %timeit (df.userid.apply(lambda x: '{:.0f}'.format(x)))
1 loop, best of 3: 210 ms per loop
Another solution is to_string
, but it is slow:
print(df.userid.to_string(float_format='{:.0f}'.format))
0 nan
1 109117800000
2 113785600000
In [41]: (df.userid.to_string(float_format='{:.0f}'.format))
1 loop, best of 3: 2.52 s per loop
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