How to implement 'in' and 'not in' for Pandas dataframe

192

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({'countries':['US','UK','Germany','China']})

countries = ['UK','China']



# pseudo-code:

df[df['countries'] not in countries]

My current way of doing this is as follows:

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

countries = pd.DataFrame({'countries':['UK','China'], 'matched':True})



# IN

df.merge(countries,how='inner',on='countries')



# NOT IN

not_in = df.merge(countries,how='left',on='countries')

not_in = not_in[pd.isnull(not_in['matched'])]

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

edited Jul 17 '15 at 20:25

smci

14.8k672104

asked Nov 13 '13 at 17:11

LondonRob

26.4k1471112

1

I think your solution is the best solution. Yours can cover IN, NOT_IN of multiple columns.

– Bruce Jung
Mar 17 '15 at 1:55

Do you want to test on single column or multiple columns?

– smci
Jul 17 '15 at 20:26

Related (performance / pandas internals): Pandas pd.Series.isin performance with set versus array

– jpp
Jun 28 '18 at 0:06

add a comment |

192

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({'countries':['US','UK','Germany','China']})

countries = ['UK','China']



# pseudo-code:

df[df['countries'] not in countries]

My current way of doing this is as follows:

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

countries = pd.DataFrame({'countries':['UK','China'], 'matched':True})



# IN

df.merge(countries,how='inner',on='countries')



# NOT IN

not_in = df.merge(countries,how='left',on='countries')

not_in = not_in[pd.isnull(not_in['matched'])]

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

edited Jul 17 '15 at 20:25

smci

14.8k672104

asked Nov 13 '13 at 17:11

LondonRob

26.4k1471112

1

I think your solution is the best solution. Yours can cover IN, NOT_IN of multiple columns.

– Bruce Jung
Mar 17 '15 at 1:55

Do you want to test on single column or multiple columns?

– smci
Jul 17 '15 at 20:26

Related (performance / pandas internals): Pandas pd.Series.isin performance with set versus array

– jpp
Jun 28 '18 at 0:06

add a comment |

192

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({'countries':['US','UK','Germany','China']})

countries = ['UK','China']



# pseudo-code:

df[df['countries'] not in countries]

My current way of doing this is as follows:

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

countries = pd.DataFrame({'countries':['UK','China'], 'matched':True})



# IN

df.merge(countries,how='inner',on='countries')



# NOT IN

not_in = df.merge(countries,how='left',on='countries')

not_in = not_in[pd.isnull(not_in['matched'])]

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

edited Jul 17 '15 at 20:25

smci

14.8k672104

asked Nov 13 '13 at 17:11

LondonRob

26.4k1471112

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({'countries':['US','UK','Germany','China']})

countries = ['UK','China']



# pseudo-code:

df[df['countries'] not in countries]

My current way of doing this is as follows:

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

countries = pd.DataFrame({'countries':['UK','China'], 'matched':True})



# IN

df.merge(countries,how='inner',on='countries')



# NOT IN

not_in = df.merge(countries,how='left',on='countries')

not_in = not_in[pd.isnull(not_in['matched'])]

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

python pandas dataframe sql-function

edited Jul 17 '15 at 20:25

smci

14.8k672104

asked Nov 13 '13 at 17:11

LondonRob

26.4k1471112

edited Jul 17 '15 at 20:25

smci

14.8k672104

asked Nov 13 '13 at 17:11

LondonRob

26.4k1471112

edited Jul 17 '15 at 20:25

smci

14.8k672104

edited Jul 17 '15 at 20:25

smci

14.8k672104

edited Jul 17 '15 at 20:25

smci

14.8k672104

asked Nov 13 '13 at 17:11

LondonRob

26.4k1471112

asked Nov 13 '13 at 17:11

LondonRob

26.4k1471112

asked Nov 13 '13 at 17:11

LondonRob

26.4k1471112

1

I think your solution is the best solution. Yours can cover IN, NOT_IN of multiple columns.

– Bruce Jung
Mar 17 '15 at 1:55

Do you want to test on single column or multiple columns?

– smci
Jul 17 '15 at 20:26

Related (performance / pandas internals): Pandas pd.Series.isin performance with set versus array

– jpp
Jun 28 '18 at 0:06

add a comment |

1

I think your solution is the best solution. Yours can cover IN, NOT_IN of multiple columns.

– Bruce Jung
Mar 17 '15 at 1:55

Do you want to test on single column or multiple columns?

– smci
Jul 17 '15 at 20:26

Related (performance / pandas internals): Pandas pd.Series.isin performance with set versus array

– jpp
Jun 28 '18 at 0:06

I think your solution is the best solution. Yours can cover IN, NOT_IN of multiple columns.

– Bruce Jung
Mar 17 '15 at 1:55

Do you want to test on single column or multiple columns?

– smci
Jul 17 '15 at 20:26

Related (performance / pandas internals): Pandas pd.Series.isin performance with set versus array

– jpp
Jun 28 '18 at 0:06

add a comment |

5 Answers
5

active

oldest

votes

429

You can use pd.Series.isin.

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

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

As a worked example:

>>> df

  countries

0        US

1        UK

2   Germany

3     China

>>> countries

['UK', 'China']

>>> df.countries.isin(countries)

0    False

1     True

2    False

3     True

Name: countries, dtype: bool

>>> df[df.countries.isin(countries)]

  countries

1        UK

3     China

>>> df[~df.countries.isin(countries)]

  countries

0        US

2   Germany

edited Apr 15 '18 at 17:52

jpp

95.7k2157109

answered Nov 13 '13 at 17:13

DSM

206k35397372

29

isin is not inverse sin()? :D

– Kos
Nov 13 '13 at 17:15

1

Just an FYI, the @LondonRob had his as a DataFrame and yours is a Series. DataFrame's isin was added in .13.

– TomAugspurger
Nov 13 '13 at 18:07

Any suggestions for how to do this with pandas 0.12.0? It's the current released version. (Maybe I should just wait for 0.13?!)

– LondonRob
Nov 13 '13 at 18:41

2

@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.

– DSM
Nov 14 '13 at 16:10

2

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
May 18 '18 at 22:20

|
show 4 more comments

Alternative solution that uses .query() method:

In [5]: df.query("countries in @countries")

Out[5]:

  countries

1        UK

3     China



In [6]: df.query("countries not in @countries")

Out[6]:

  countries

0        US

2   Germany

answered Jul 19 '17 at 12:19

MaxU

120k12112167

3

Note that this is currently marked as "experimental" in the docs...

– LondonRob
Jul 19 '17 at 14:49

add a comment |

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)]

answered Nov 13 '13 at 17:14

Kos

49.8k19120196

6

FYI, this is much slower than @DSM soln which is vectorized

– Jeff
Nov 13 '13 at 17:47

@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!)

– Kos
Nov 14 '13 at 7:42

add a comment |

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

Finally got it working:

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

answered Jul 13 '17 at 3:12

Sam Henderson

17115

3

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

– cricket_007
Jul 19 '17 at 12:17

add a comment |

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])]

answered Apr 4 '18 at 11:51

Ioannis Nasios

3,6713832

add a comment |

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

429

You can use pd.Series.isin.

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

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

As a worked example:

>>> df

  countries

0        US

1        UK

2   Germany

3     China

>>> countries

['UK', 'China']

>>> df.countries.isin(countries)

0    False

1     True

2    False

3     True

Name: countries, dtype: bool

>>> df[df.countries.isin(countries)]

  countries

1        UK

3     China

>>> df[~df.countries.isin(countries)]

  countries

0        US

2   Germany

edited Apr 15 '18 at 17:52

jpp

95.7k2157109

answered Nov 13 '13 at 17:13

DSM

206k35397372

29

isin is not inverse sin()? :D

– Kos
Nov 13 '13 at 17:15

1

Just an FYI, the @LondonRob had his as a DataFrame and yours is a Series. DataFrame's isin was added in .13.

– TomAugspurger
Nov 13 '13 at 18:07

Any suggestions for how to do this with pandas 0.12.0? It's the current released version. (Maybe I should just wait for 0.13?!)

– LondonRob
Nov 13 '13 at 18:41

2

@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.

– DSM
Nov 14 '13 at 16:10

2

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
May 18 '18 at 22:20

|
show 4 more comments

429

You can use pd.Series.isin.

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

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

As a worked example:

>>> df

  countries

0        US

1        UK

2   Germany

3     China

>>> countries

['UK', 'China']

>>> df.countries.isin(countries)

0    False

1     True

2    False

3     True

Name: countries, dtype: bool

>>> df[df.countries.isin(countries)]

  countries

1        UK

3     China

>>> df[~df.countries.isin(countries)]

  countries

0        US

2   Germany

edited Apr 15 '18 at 17:52

jpp

95.7k2157109

answered Nov 13 '13 at 17:13

DSM

206k35397372

29

isin is not inverse sin()? :D

– Kos
Nov 13 '13 at 17:15

1

Just an FYI, the @LondonRob had his as a DataFrame and yours is a Series. DataFrame's isin was added in .13.

– TomAugspurger
Nov 13 '13 at 18:07

Any suggestions for how to do this with pandas 0.12.0? It's the current released version. (Maybe I should just wait for 0.13?!)

– LondonRob
Nov 13 '13 at 18:41

2

@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.

– DSM
Nov 14 '13 at 16:10

2

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
May 18 '18 at 22:20

|
show 4 more comments

429

You can use pd.Series.isin.

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

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

As a worked example:

>>> df

  countries

0        US

1        UK

2   Germany

3     China

>>> countries

['UK', 'China']

>>> df.countries.isin(countries)

0    False

1     True

2    False

3     True

Name: countries, dtype: bool

>>> df[df.countries.isin(countries)]

  countries

1        UK

3     China

>>> df[~df.countries.isin(countries)]

  countries

0        US

2   Germany

edited Apr 15 '18 at 17:52

jpp

95.7k2157109

answered Nov 13 '13 at 17:13

DSM

206k35397372

You can use pd.Series.isin.

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

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

As a worked example:

>>> df

  countries

0        US

1        UK

2   Germany

3     China

>>> countries

['UK', 'China']

>>> df.countries.isin(countries)

0    False

1     True

2    False

3     True

Name: countries, dtype: bool

>>> df[df.countries.isin(countries)]

  countries

1        UK

3     China

>>> df[~df.countries.isin(countries)]

  countries

0        US

2   Germany

edited Apr 15 '18 at 17:52

jpp

95.7k2157109

answered Nov 13 '13 at 17:13

DSM

206k35397372

edited Apr 15 '18 at 17:52

jpp

95.7k2157109

edited Apr 15 '18 at 17:52

jpp

95.7k2157109

edited Apr 15 '18 at 17:52

jpp

95.7k2157109

answered Nov 13 '13 at 17:13

DSM

206k35397372

answered Nov 13 '13 at 17:13

DSM

206k35397372

answered Nov 13 '13 at 17:13

DSM

206k35397372

29

isin is not inverse sin()? :D

– Kos
Nov 13 '13 at 17:15

1

Just an FYI, the @LondonRob had his as a DataFrame and yours is a Series. DataFrame's isin was added in .13.

– TomAugspurger
Nov 13 '13 at 18:07

Any suggestions for how to do this with pandas 0.12.0? It's the current released version. (Maybe I should just wait for 0.13?!)

– LondonRob
Nov 13 '13 at 18:41

2

@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.

– DSM
Nov 14 '13 at 16:10

2

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
May 18 '18 at 22:20

|
show 4 more comments

29

isin is not inverse sin()? :D

– Kos
Nov 13 '13 at 17:15

1

Just an FYI, the @LondonRob had his as a DataFrame and yours is a Series. DataFrame's isin was added in .13.

– TomAugspurger
Nov 13 '13 at 18:07

Any suggestions for how to do this with pandas 0.12.0? It's the current released version. (Maybe I should just wait for 0.13?!)

– LondonRob
Nov 13 '13 at 18:41

2

@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.

– DSM
Nov 14 '13 at 16:10

2

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
May 18 '18 at 22:20

isin is not inverse sin()? :D

– Kos
Nov 13 '13 at 17:15

Just an FYI, the @LondonRob had his as a DataFrame and yours is a Series. DataFrame's isin was added in .13.

– TomAugspurger
Nov 13 '13 at 18:07

Any suggestions for how to do this with pandas 0.12.0? It's the current released version. (Maybe I should just wait for 0.13?!)

– LondonRob
Nov 13 '13 at 18:41

@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.

– DSM
Nov 14 '13 at 16:10

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
May 18 '18 at 22:20

|
show 4 more comments

Alternative solution that uses .query() method:

In [5]: df.query("countries in @countries")

Out[5]:

  countries

1        UK

3     China



In [6]: df.query("countries not in @countries")

Out[6]:

  countries

0        US

2   Germany

answered Jul 19 '17 at 12:19

MaxU

120k12112167

3

Note that this is currently marked as "experimental" in the docs...

– LondonRob
Jul 19 '17 at 14:49

add a comment |

Alternative solution that uses .query() method:

In [5]: df.query("countries in @countries")

Out[5]:

  countries

1        UK

3     China



In [6]: df.query("countries not in @countries")

Out[6]:

  countries

0        US

2   Germany

answered Jul 19 '17 at 12:19

MaxU

120k12112167

3

Note that this is currently marked as "experimental" in the docs...

– LondonRob
Jul 19 '17 at 14:49

add a comment |

Alternative solution that uses .query() method:

In [5]: df.query("countries in @countries")

Out[5]:

  countries

1        UK

3     China



In [6]: df.query("countries not in @countries")

Out[6]:

  countries

0        US

2   Germany

answered Jul 19 '17 at 12:19

MaxU

120k12112167

Alternative solution that uses .query() method:

In [5]: df.query("countries in @countries")

Out[5]:

  countries

1        UK

3     China



In [6]: df.query("countries not in @countries")

Out[6]:

  countries

0        US

2   Germany

answered Jul 19 '17 at 12:19

MaxU

120k12112167

answered Jul 19 '17 at 12:19

MaxU

120k12112167

answered Jul 19 '17 at 12:19

MaxU

120k12112167

answered Jul 19 '17 at 12:19

MaxU

120k12112167

3

Note that this is currently marked as "experimental" in the docs...

– LondonRob
Jul 19 '17 at 14:49

add a comment |

3

Note that this is currently marked as "experimental" in the docs...

– LondonRob
Jul 19 '17 at 14:49

Note that this is currently marked as "experimental" in the docs...

– LondonRob
Jul 19 '17 at 14:49

add a comment |

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)]

answered Nov 13 '13 at 17:14

Kos

49.8k19120196

6

FYI, this is much slower than @DSM soln which is vectorized

– Jeff
Nov 13 '13 at 17:47

@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!)

– Kos
Nov 14 '13 at 7:42

add a comment |

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)]

answered Nov 13 '13 at 17:14

Kos

49.8k19120196

6

FYI, this is much slower than @DSM soln which is vectorized

– Jeff
Nov 13 '13 at 17:47

@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!)

– Kos
Nov 14 '13 at 7:42

add a comment |

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)]

answered Nov 13 '13 at 17:14

Kos

49.8k19120196

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)]

answered Nov 13 '13 at 17:14

Kos

49.8k19120196

answered Nov 13 '13 at 17:14

Kos

49.8k19120196

answered Nov 13 '13 at 17:14

Kos

49.8k19120196

answered Nov 13 '13 at 17:14

Kos

49.8k19120196

6

FYI, this is much slower than @DSM soln which is vectorized

– Jeff
Nov 13 '13 at 17:47

@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!)

– Kos
Nov 14 '13 at 7:42

add a comment |

6

FYI, this is much slower than @DSM soln which is vectorized

– Jeff
Nov 13 '13 at 17:47

@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!)

– Kos
Nov 14 '13 at 7:42

FYI, this is much slower than @DSM soln which is vectorized

– Jeff
Nov 13 '13 at 17:47

@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!)

– Kos
Nov 14 '13 at 7:42

add a comment |

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

Finally got it working:

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

answered Jul 13 '17 at 3:12

Sam Henderson

17115

3

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

– cricket_007
Jul 19 '17 at 12:17

add a comment |

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

Finally got it working:

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

answered Jul 13 '17 at 3:12

Sam Henderson

17115

3

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

– cricket_007
Jul 19 '17 at 12:17

add a comment |

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

Finally got it working:

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

answered Jul 13 '17 at 3:12

Sam Henderson

17115

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

Finally got it working:

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

answered Jul 13 '17 at 3:12

Sam Henderson

17115

answered Jul 13 '17 at 3:12

Sam Henderson

17115

answered Jul 13 '17 at 3:12

Sam Henderson

17115

answered Jul 13 '17 at 3:12

Sam Henderson

17115

3

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

– cricket_007
Jul 19 '17 at 12:17

add a comment |

3

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

– cricket_007
Jul 19 '17 at 12:17

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

– cricket_007
Jul 19 '17 at 12:17

add a comment |

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])]

answered Apr 4 '18 at 11:51

Ioannis Nasios

3,6713832

add a comment |

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])]

answered Apr 4 '18 at 11:51

Ioannis Nasios

3,6713832

add a comment |

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])]

answered Apr 4 '18 at 11:51

Ioannis Nasios

3,6713832

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])]

answered Apr 4 '18 at 11:51

Ioannis Nasios

3,6713832

answered Apr 4 '18 at 11:51

Ioannis Nasios

3,6713832

answered Apr 4 '18 at 11:51

Ioannis Nasios

3,6713832

answered Apr 4 '18 at 11:51

Ioannis Nasios

3,6713832

add a comment |

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