Custom boolean filtering in Pandas?

Question

I have a dataframe

                  0         1         2         3       Marketcap
0  1.707280  0.666952  0.638515 -0.061126  2.291747     1.71B
1 -1.017134  1.353627  0.618433  0.008279  0.148128     1.82B
2 -0.774057 -0.165566 -0.083345  0.741598 -0.139851      1.1M
3 -0.630724  0.250737  1.308556 -1.040799  1.064456    30.92M
4  2.029370  0.899612  0.261146  1.474148 -1.663970   476.74k
5  2.029370  0.899612  0.261146  1.474148 -1.663970        -1

Is there some sort of custom filter method, that would let Python know B > M > K?

Say I want to filter, df[df.Marketcap > 35.00M], is there a clever or clean way to do this? Having the M or B makes the value very readable and easy to differentiate.

Thank you.

EDIT: Reopened the thread as Max U's answer while excellent seems to produce a pandas bug, which we opened an issue on Github.

Answer 1

Source DF:

In [176]: df
Out[176]:
                    0         1         2         3 Market Cap
0  1.707280  0.666952  0.638515 -0.061126  2.291747      1.71B
1 -1.017134  1.353627  0.618433  0.008279  0.148128      1.82B
2 -0.774057 -0.165566 -0.083345  0.741598 -0.139851       1.1M
3 -0.630724  0.250737  1.308556 -1.040799  1.064456     30.92M
4  2.029370  0.899612  0.261146  1.474148 -1.663970    476.74k
5  2.029370  0.899612  0.261146  1.474148 -1.663970         -1

Solution:

to_replace = ['\d+\s*[Kk]','\d+\s*[Mm]','\d+\s*[Bb]', '-1', 'N/A']
value = [1000,1000000,1000000000, 1, 1]

mask = df.assign(
    f=df['Market Cap'].replace(to_replace, value, regex=True),
    Marketcap=pd.to_numeric(df['Market Cap'].str.replace(r'[^\d\.]', ''), errors='coerce')
).eval("Marketcap * f < 35000000")

df[mask]

Result:

In [178]: df[mask]
Out[178]:
                    0         1         2         3 Market Cap
2 -0.774057 -0.165566 -0.083345  0.741598 -0.139851       1.1M
3 -0.630724  0.250737  1.308556 -1.040799  1.064456     30.92M
4  2.029370  0.899612  0.261146  1.474148 -1.663970    476.74k
5  2.029370  0.899612  0.261146  1.474148 -1.663970         -1

PS if you want to leave non-numeric values (like N/A) in the resulting data set change:

pd.to_numeric(df['Market Cap'].str.replace(r'[^\d\.]', ''), errors='coerce')

to

pd.to_numeric(df['Market Cap'].str.replace(r'[^\d\.]', ''), errors='coerce').fillna('0')

Answer 2

UPDATE:

In [44]: df
Out[44]:
          0         1         2         3         4 Marketcap
0  1.707280  0.666952  0.638515 -0.061126  2.291747     1.71B
1 -1.017134  1.353627  0.618433  0.008279  0.148128     1.82B
2 -0.774057 -0.165566 -0.083345  0.741598 -0.139851      1.1M
3 -0.630724  0.250737  1.308556 -1.040799  1.064456    30.92M
4  2.029370  0.899612  0.261146  1.474148 -1.663970   476.74k
5  2.029370  0.899612  0.261146  1.474148 -1.663970        -1

In [45]: df[pd.eval(df.Marketcap.replace(['[Kk]','[Mm]','[Bb]'],
                                         ['*10**3','*10**6','*10**9'], regex=True) \
                      .add(' < 35*10**6'))]
Out[45]:
          0         1         2         3         4 Marketcap
2 -0.774057 -0.165566 -0.083345  0.741598 -0.139851      1.1M
3 -0.630724  0.250737  1.308556 -1.040799  1.064456    30.92M
4  2.029370  0.899612  0.261146  1.474148 -1.663970   476.74k
5  2.029370  0.899612  0.261146  1.474148 -1.663970        -1

---

I'd do it this way:

In [13]: df[pd.eval(df.Marketcap.replace(['M','B'],['','*1000'], regex=True).add(' > 35'))]
Out[13]:
          0         1         2         3         4 Marketcap
0  1.707280  0.666952  0.638515 -0.061126  2.291747     1.71B
1 -1.017134  1.353627  0.618433  0.008279  0.148128     1.82B
4  2.029370  0.899612  0.261146  1.474148 -1.663970    100.9M

Explanation:

In [14]: df.Marketcap.replace(['M','B'],['','*1000'], regex=True)
Out[14]:
0    1.71*1000
1    1.82*1000
2          1.1
3        30.92
4        100.9
Name: Marketcap, dtype: object

In [15]: df.Marketcap.replace(['M','B'],['','*1000'], regex=True).add(' > 35')
Out[15]:
0    1.71*1000 > 35
1    1.82*1000 > 35
2          1.1 > 35
3        30.92 > 35
4        100.9 > 35
Name: Marketcap, dtype: object

In [16]: pd.eval(df.Marketcap.replace(['M','B'],['','*1000'], regex=True).add(' > 35'))
Out[16]: array([True, True, False, False, True], dtype=object)

Answer 3

This isn't super clean, but it does the trick and doesn't use any python iteration:

Code:

# Create a separate column (which you can omit later) that converts 'Marketcap' strings to numbers
df['cap'] = df.loc[df['Marketcap'].str.contains('B'), 'Marketcap'].str.replace('B','').astype(float) * 1000
df['cap'].fillna(df.loc[df['Marketcap'].str.contains('M'), 'Marketcap'].str.replace('M',''), inplace = True)

# For pandas pre-0.20.0 (<May 2017)
print df.ix[df['cap'].astype(float) > 35, :-1]

# For pandas 0.20.0+ (.ix[] deprecated)
print df.iloc[df[df['cap'].astype(float) > 35].index, :-1]

# Or, alternate pandas 0.20.0+ option (thanks @Psidom)
print df[df['cap'].astype(float) > 35].iloc[:,:-1]

Output:

          0         1         2         3         4 Marketcap
0  1.707280  0.666952  0.638515 -0.061126  2.291747     1.71B
1 -1.017134  1.353627  0.618433  0.008279  0.148128     1.82B
4  2.029370  0.899612  0.261146  1.474148 -1.663970    100.9M