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How to filter Pandas dataframe using 'in' and 'not in' like in SQL

How can I achieve the equivalents of SQL's IN and NOT IN?

I have a list with the required values. Here's the scenario:

df = pd.DataFrame({'country': ['US', 'UK', 'Germany', 'China']})
countries_to_keep = ['UK', 'China']

# pseudo-code:
df[df['country'] not in countries_to_keep]

My current way of doing this is as follows:

df = pd.DataFrame({'country': ['US', 'UK', 'Germany', 'China']})
df2 = pd.DataFrame({'country': ['UK', 'China'], 'matched': True})

# IN
df.merge(df2, how='inner', on='country')

# NOT IN
not_in = df.merge(df2, how='left', on='country')
not_in = not_in[pd.isnull(not_in['matched'])]

But this seems like a horrible kludge. Can anyone improve on it?

Related (performance / pandas internals): Pandas pd.Series.isin performance with set versus array
Use a list of values to select rows from a pandas dataframe is similar, but the negation ~ was added as an edit in 2019.

g
gpr

You can use pd.Series.isin.

For "IN" use: something.isin(somewhere)

Or for "NOT IN": ~something.isin(somewhere)

As a worked example:

import pandas as pd

>>> df
  country
0        US
1        UK
2   Germany
3     China
>>> countries_to_keep
['UK', 'China']
>>> df.country.isin(countries_to_keep)
0    False
1     True
2    False
3     True
Name: country, dtype: bool
>>> df[df.country.isin(countries_to_keep)]
  country
1        UK
3     China
>>> df[~df.country.isin(countries_to_keep)]
  country
0        US
2   Germany

If you're actually dealing with 1-dimensional arrays (like in you're example) then on you're first line use a Series instead of a DataFrame, like @DSM used: df = pd.Series({'countries':['US','UK','Germany','China']})
@TomAugspurger: like usual, I'm probably missing something. df, both mine and his, is a DataFrame. countries is a list. df[~df.countries.isin(countries)] produces a DataFrame, not a Series, and seems to work even back in 0.11.0.dev-14a04dd.
This answer is confusing because you keep reusing the countries variable. Well, the OP does it, and that's inherited, but that something is done badly before does not justify doing it badly now.
@ifly6 : Agreed, I made the same mistake and realized it when I got a error : "'DataFrame' object has no attribute 'countries'
For people who are confused by the tilde (like me): stackoverflow.com/questions/8305199/…
M
MaxU - stop genocide of UA

Alternative solution that uses .query() method:

In [5]: df.query("countries in @countries_to_keep")
Out[5]:
  countries
1        UK
3     China

In [6]: df.query("countries not in @countries_to_keep")
Out[6]:
  countries
0        US
2   Germany

.query is so much more readable. Especially for the "not in" scenario, vs a distant tilde. Thanks!
What is @countries ? Another dataframe ? A list ?
@FlorianCastelain countries are the column you want to check on, OP called this column
@FlorianCastelain, somebody has renamed a variable in the original question: countries -> countries_to_keep, so my answer has become invalid. I've updated my answer correspondingly. countries_to_keep - is a list.
The most readable solution indeed. I wonder if syntax exists to avoid creating countries_to_keep. Is it possible to specify the list of values inside the query directly?
c
cs95

How to implement 'in' and 'not in' for a pandas DataFrame?

Pandas offers two methods: Series.isin and DataFrame.isin for Series and DataFrames, respectively.

Filter DataFrame Based on ONE Column (also applies to Series)

The most common scenario is applying an isin condition on a specific column to filter rows in a DataFrame.

df = pd.DataFrame({'countries': ['US', 'UK', 'Germany', np.nan, 'China']})
df
  countries
0        US
1        UK
2   Germany
3     China

c1 = ['UK', 'China']             # list
c2 = {'Germany'}                 # set
c3 = pd.Series(['China', 'US'])  # Series
c4 = np.array(['US', 'UK'])      # array

Series.isin accepts various types as inputs. The following are all valid ways of getting what you want:

df['countries'].isin(c1)

0    False
1     True
2    False
3    False
4     True
Name: countries, dtype: bool

# `in` operation
df[df['countries'].isin(c1)]

  countries
1        UK
4     China

# `not in` operation
df[~df['countries'].isin(c1)]

  countries
0        US
2   Germany
3       NaN

# Filter with `set` (tuples work too)
df[df['countries'].isin(c2)]

  countries
2   Germany

# Filter with another Series
df[df['countries'].isin(c3)]

  countries
0        US
4     China

# Filter with array
df[df['countries'].isin(c4)]

  countries
0        US
1        UK

Filter on MANY Columns

Sometimes, you will want to apply an 'in' membership check with some search terms over multiple columns,

df2 = pd.DataFrame({
    'A': ['x', 'y', 'z', 'q'], 'B': ['w', 'a', np.nan, 'x'], 'C': np.arange(4)})
df2

   A    B  C
0  x    w  0
1  y    a  1
2  z  NaN  2
3  q    x  3

c1 = ['x', 'w', 'p']

To apply the isin condition to both columns "A" and "B", use DataFrame.isin:

df2[['A', 'B']].isin(c1)

      A      B
0   True   True
1  False  False
2  False  False
3  False   True

From this, to retain rows where at least one column is True, we can use any along the first axis:

df2[['A', 'B']].isin(c1).any(axis=1)

0     True
1    False
2    False
3     True
dtype: bool

df2[df2[['A', 'B']].isin(c1).any(axis=1)]

   A  B  C
0  x  w  0
3  q  x  3

Note that if you want to search every column, you'd just omit the column selection step and do

df2.isin(c1).any(axis=1)

Similarly, to retain rows where ALL columns are True, use all in the same manner as before.

df2[df2[['A', 'B']].isin(c1).all(axis=1)]

   A  B  C
0  x  w  0

Notable Mentions: numpy.isin, query, list comprehensions (string data)

In addition to the methods described above, you can also use the numpy equivalent: numpy.isin.

# `in` operation
df[np.isin(df['countries'], c1)]

  countries
1        UK
4     China

# `not in` operation
df[np.isin(df['countries'], c1, invert=True)]

  countries
0        US
2   Germany
3       NaN

Why is it worth considering? NumPy functions are usually a bit faster than their pandas equivalents because of lower overhead. Since this is an elementwise operation that does not depend on index alignment, there are very few situations where this method is not an appropriate replacement for pandas' isin.

Pandas routines are usually iterative when working with strings, because string operations are hard to vectorise. There is a lot of evidence to suggest that list comprehensions will be faster here.. We resort to an in check now.

c1_set = set(c1) # Using `in` with `sets` is a constant time operation... 
                 # This doesn't matter for pandas because the implementation differs.
# `in` operation
df[[x in c1_set for x in df['countries']]]

  countries
1        UK
4     China

# `not in` operation
df[[x not in c1_set for x in df['countries']]]

  countries
0        US
2   Germany
3       NaN

It is a lot more unwieldy to specify, however, so don't use it unless you know what you're doing.

Lastly, there's also DataFrame.query which has been covered in this answer. numexpr FTW!


I like it, but what if I want to compare a column in df3 that isin df1 column? What would that look like?
K
Kos

I've been usually doing generic filtering over rows like this:

criterion = lambda row: row['countries'] not in countries
not_in = df[df.apply(criterion, axis=1)]

FYI, this is much slower than @DSM soln which is vectorized
@Jeff I'd expect that, but that's what I fall back to when I need to filter over something unavailable in pandas directly. (I was about to say "like .startwith or regex matching, but just found out about Series.str that has all of that!)
c
cs95

Collating possible solutions from the answers:

For IN: df[df['A'].isin([3, 6])]

For NOT IN:

df[-df["A"].isin([3, 6])] df[~df["A"].isin([3, 6])] df[df["A"].isin([3, 6]) == False] df[np.logical_not(df["A"].isin([3, 6]))]


This mostly repeats information from other answers. Using logical_not is a mouthful equivalent of the ~ operator.
c
cs95

I wanted to filter out dfbc rows that had a BUSINESS_ID that was also in the BUSINESS_ID of dfProfilesBusIds

dfbc = dfbc[~dfbc['BUSINESS_ID'].isin(dfProfilesBusIds['BUSINESS_ID'])]

You can negate the isin (as done in the accepted answer) rather than comparing to False
p
padu

Why is no one talking about the performance of various filtering methods? In fact, this topic often pops up here (see the example). I did my own performance test for a large data set. It is very interesting and instructive.

df = pd.DataFrame({'animals': np.random.choice(['cat', 'dog', 'mouse', 'birds'], size=10**7), 
                   'number': np.random.randint(0,100, size=(10**7,))})

df.info()

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 10000000 entries, 0 to 9999999
Data columns (total 2 columns):
 #   Column   Dtype 
---  ------   ----- 
 0   animals  object
 1   number   int64 
dtypes: int64(1), object(1)
memory usage: 152.6+ MB
%%timeit
# .isin() by one column
conditions = ['cat', 'dog']
df[df.animals.isin(conditions)]
367 ms ± 2.34 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
# .query() by one column
conditions = ['cat', 'dog']
df.query('animals in @conditions')
395 ms ± 3.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
# .loc[]
df.loc[(df.animals=='cat')|(df.animals=='dog')]
987 ms ± 5.17 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
df[df.apply(lambda x: x['animals'] in ['cat', 'dog'], axis=1)]
41.9 s ± 490 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
new_df = df.set_index('animals')
new_df.loc[['cat', 'dog'], :]
3.64 s ± 62.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
new_df = df.set_index('animals')
new_df[new_df.index.isin(['cat', 'dog'])]
469 ms ± 8.98 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
s = pd.Series(['cat', 'dog'], name='animals')
df.merge(s, on='animals', how='inner')
796 ms ± 30.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Thus, the isin method turned out to be the fastest and the method with apply() was the slowest, which is not surprising.


r
rachwa

You can also use .isin() inside .query():

df.query('country.isin(@countries_to_keep).values')

# Or alternatively:
df.query('country.isin(["UK", "China"]).values')

To negate your query, use ~:

df.query('~country.isin(@countries_to_keep).values')

Update:

Another way is to use comparison operators:

df.query('country == @countries_to_keep')

# Or alternatively:
df.query('country == ["UK", "China"]')

And to negate the query, use !=:

df.query('country != @countries_to_keep')

Good to know, although this is a bit less readable than this answer which uses in and not in inside query. Interesting that query supports both!
I
Ioannis Nasios
df = pd.DataFrame({'countries':['US','UK','Germany','China']})
countries = ['UK','China']

implement in:

df[df.countries.isin(countries)]

implement not in as in of rest countries:

df[df.countries.isin([x for x in np.unique(df.countries) if x not in countries])]

B
Billy Bonaros

A trick if you want to keep the order of the list:

df = pd.DataFrame({'country': ['US', 'UK', 'Germany', 'China']})
countries_to_keep = ['Germany', 'US']


ind=[df.index[df['country']==i].tolist() for i in countries_to_keep]
flat_ind=[item for sublist in ind for item in sublist]

df.reindex(flat_ind)

   country
2  Germany
0       US

L
LondonRob

My 2c worth: I needed a combination of in and ifelse statements for a dataframe, and this worked for me.

sale_method = pd.DataFrame(model_data["Sale Method"].str.upper())
sale_method["sale_classification"] = np.where(
    sale_method["Sale Method"].isin(["PRIVATE"]),
    "private",
    np.where(
        sale_method["Sale Method"].str.contains("AUCTION"), "auction", "other"
    ),
)