How to check if a value is in the list in selection from pandas data frame?

Question

Looks ugly:

df_cut = df_new[
             (
             (df_new['l_ext']==31) |
             (df_new['l_ext']==22) |
             (df_new['l_ext']==30) |
             (df_new['l_ext']==25) |
             (df_new['l_ext']==64)
             )
            ]

Does not work:

df_cut = df_new[(df_new['l_ext'] in [31, 22, 30, 25, 64])]

Is there an elegant and working solution of the above "problem"?

score 241 · Accepted Answer · answered Aug 15 '13 at 10:05

241

Use isin

df_new[df_new['l_ext'].isin([31, 22, 30, 25, 64])]

answered Aug 15 '13 at 10:05

waitingkuo

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27

This is one of those less-intuitive `pandas` syntax features... You don't know unless you know. – openwonk Sep 07 '16 at 23:45
4

What's the negative of isin? I'm looking for a way to filter out the list elements. – Augmented Jacob Sep 21 '16 at 18:38
1

numpy.logical_not(foo.isin(x)) and another method suggested by Pandas author: http://stackoverflow.com/questions/14057007/remove-rows-not-isinx – Luke Barker Sep 28 '16 at 08:07
9

Negation using ~ should be enough for doing not isin. df_new[~df_new['l_ext'].isin([31, 22, 30, 25, 64])] – Arun Das Mar 27 '18 at 21:57
1

Be cautious, if the type of the column is ambiguous for example "object" which could be either numeric or string "isin" will automatically infer data type. – Semooze Oct 18 '21 at 14:40

jpp · Answer 2 · 2020-04-17T07:58:10.710

15

You can use pd.DataFrame.query:

select_values = [31, 22, 30, 25, 64]
df_cut = df_new.query('l_ext in @select_values')

In the background, this uses the top-level pd.eval function.

edited Apr 17 '20 at 07:58

answered Oct 08 '18 at 16:33

jpp

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Another alternative without having to create an extra variable: `df_new.query('l_ext.isin([31, 22, 30, 25, 64]).values')` – rachwa May 08 '22 at 20:07

How to check if a value is in the list in selection from pandas data frame?

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