I have this Pandas dataframe (df):
A B
0 1 green
1 2 red
2 s blue
3 3 yellow
4 b black
A type is object.
I'd select the record where A value are integer or numeric to have:
A B
0 1 green
1 2 red
3 3 yellow
Thanks
Call apply on the dataframe (note the double square brackets df[['A']] rather than df['A']) and call the string method isdigit(), we then set param axis=1 to apply the lambda function row-wise. What happens here is that the index is used to create a boolean mask.
In [66]:
df[df[['A']].apply(lambda x: x[0].isdigit(), axis=1)]
Out[66]:
A B
Index
0 1 green
1 2 red
3 3 yellow
Update
If you're using a version 0.16.0 or newer then the following will also work:
In [6]:
df[df['A'].astype(str).str.isdigit()]
Out[6]:
A B
0 1 green
1 2 red
3 3 yellow
Here we cast the Series to str using astype and then call the vectorised str.isdigit
Also note that convert_objects is deprecated and one should use to_numeric for the latest versions 0.17.0 or newer
You can use convert_objects, which when convert_numeric=True will forcefully set all non-numeric to nan. Dropping them and indexing gets your result.
This will be considerably faster that using apply on a larger frame as this is all implemented in cython.
In [30]: df[['A']].convert_objects(convert_numeric=True)
Out[30]:
A
0 1
1 2
2 NaN
3 3
4 NaN
In [31]: df[['A']].convert_objects(convert_numeric=True).dropna()
Out[31]:
A
0 1
1 2
3 3
In [32]: df[['A']].convert_objects(convert_numeric=True).dropna().index
Out[32]: Int64Index([0, 1, 3], dtype='int64')
In [33]: df.iloc[df[['A']].convert_objects(convert_numeric=True).dropna().index]
Out[33]:
A B
0 1 green
1 2 red
3 3 yellow
Note that convert_objects is deprecated
>>> df[['A']].convert_objects(convert_numeric=True)
__main__:1: FutureWarning: convert_objects is deprecated. Use the data-type specific converters pd.to_datetime, pd.to_timedelta and pd.to_numeric.
From 0.17.0: use pd.to_numeric, set errors='coerce' so that incorrect parsing returns NaN. Use notnull to return a boolean mask to use on the original dataframe:
>>> df[pd.to_numeric(df.A, errors='coerce').notnull()]
A B
0 1 green
1 2 red
3 3 yellow
Personally, I think it will be more succinct to just use the built-in map compared with .apply()
In [13]: df[map(pred, df['B'])]
