Pandas convert float in scientific notation to string

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

I just stumbled upon this problem after reading a dataframe from a json file using the read_json method and unfortunately it does not have a keep_default_na parameter.

The solution was to convert the long floats to np.int64 before converting them to str.

In [53]: tweet_id_sample = tweets.iloc[0]['id']
         tweet_id_sample
Out[53]: 8.924206435553362e+17

In [54]: tweet_id_sample.astype(str)
Out[54]: '8.924206435553362e+17'

In [55]: tweet_id_sample.astype(np.int64).astype(str)
Out[55]: '892420643555336192'

In [56]: # This overflows
         tweet_id_sample.astype(int)
Out[56]: -2147483648