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

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229

72

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

share|improve this question

edited Jul 17 '15 at 20:25

smci

15.5k678109

asked Nov 13 '13 at 17:11

LondonRobLondonRob

27.8k1676117

• 
  
  
  

  
  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 | 

229

72

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

share|improve this question

edited Jul 17 '15 at 20:25

smci

15.5k678109

asked Nov 13 '13 at 17:11

LondonRobLondonRob

27.8k1676117

• 
  
  
  

  
  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 | 

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

share|improve this question

edited Jul 17 '15 at 20:25

smci

15.5k678109

asked Nov 13 '13 at 17:11

LondonRobLondonRob

27.8k1676117

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

# pseudo-code:
df[df['countries'] not in countries]

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'])]

share|improve this question

edited Jul 17 '15 at 20:25

smci

15.5k678109

asked Nov 13 '13 at 17:11

LondonRobLondonRob

27.8k1676117

share|improve this question

edited Jul 17 '15 at 20:25

smci

15.5k678109

asked Nov 13 '13 at 17:11

LondonRobLondonRob

27.8k1676117

edited Jul 17 '15 at 20:25

smci

15.5k678109

edited Jul 17 '15 at 20:25

smci

15.5k678109

asked Nov 13 '13 at 17:11

LondonRobLondonRob

27.8k1676117

asked Nov 13 '13 at 17:11

LondonRobLondonRob

27.8k1676117

5 Answers
5

active

oldest

votes

494

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

share|improve this answer

edited Apr 15 '18 at 17:52

jpp

102k2165116

answered Nov 13 '13 at 17:13

DSMDSM

214k35411379

• 
  
  
  

  
  32
  

  
  
  
  
  
  

  
  
  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
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  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

24

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

share|improve this answer

answered Jul 19 '17 at 12:19

MaxUMaxU

124k12126182

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Note that this is currently marked as "experimental" in the docs...
  
  – LondonRob
  Jul 19 '17 at 14:49
  

  
  
  
  

add a comment | 

11

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)]

share|improve this answer

answered Nov 13 '13 at 17:14

KosKos

50.5k19123201

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  
  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 | 

3

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'])]

share|improve this answer

edited Mar 11 at 6:25

jezrael

351k26315391

answered Jul 13 '17 at 3:12

Sam HendersonSam Henderson

30135

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  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 | 

2

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])]

share|improve this answer

answered Apr 4 '18 at 11:51

Ioannis NasiosIoannis Nasios

3,75831036

add a comment | 

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494

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

share|improve this answer

edited Apr 15 '18 at 17:52

jpp

102k2165116

answered Nov 13 '13 at 17:13

DSMDSM

214k35411379

• 
  
  
  

  
  32
  

  
  
  
  
  
  

  
  
  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
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  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

494

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

share|improve this answer

edited Apr 15 '18 at 17:52

jpp

102k2165116

answered Nov 13 '13 at 17:13

DSMDSM

214k35411379

• 
  
  
  

  
  32
  

  
  
  
  
  
  

  
  
  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
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  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

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

share|improve this answer

edited Apr 15 '18 at 17:52

jpp

102k2165116

answered Nov 13 '13 at 17:13

DSMDSM

214k35411379

>>> 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

share|improve this answer

edited Apr 15 '18 at 17:52

jpp

102k2165116

answered Nov 13 '13 at 17:13

DSMDSM

214k35411379

edited Apr 15 '18 at 17:52

jpp

102k2165116

edited Apr 15 '18 at 17:52

jpp

102k2165116

answered Nov 13 '13 at 17:13

DSMDSM

214k35411379

answered Nov 13 '13 at 17:13

DSMDSM

214k35411379

24

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

share|improve this answer

answered Jul 19 '17 at 12:19

MaxUMaxU

124k12126182

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Note that this is currently marked as "experimental" in the docs...
  
  – LondonRob
  Jul 19 '17 at 14:49
  

  
  
  
  

add a comment | 

24

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

share|improve this answer

answered Jul 19 '17 at 12:19

MaxUMaxU

124k12126182

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  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

share|improve this answer

answered Jul 19 '17 at 12:19

MaxUMaxU

124k12126182

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

share|improve this answer

answered Jul 19 '17 at 12:19

MaxUMaxU

124k12126182

answered Jul 19 '17 at 12:19

MaxUMaxU

124k12126182

answered Jul 19 '17 at 12:19

MaxUMaxU

124k12126182

11

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)]

share|improve this answer

answered Nov 13 '13 at 17:14

KosKos

50.5k19123201

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  
  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 | 

11

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)]

share|improve this answer

answered Nov 13 '13 at 17:14

KosKos

50.5k19123201

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  
  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)]

share|improve this answer

answered Nov 13 '13 at 17:14

KosKos

50.5k19123201

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

share|improve this answer

answered Nov 13 '13 at 17:14

KosKos

50.5k19123201

answered Nov 13 '13 at 17:14

KosKos

50.5k19123201

answered Nov 13 '13 at 17:14

KosKos

50.5k19123201

3

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'])]

share|improve this answer

edited Mar 11 at 6:25

jezrael

351k26315391

answered Jul 13 '17 at 3:12

Sam HendersonSam Henderson

30135

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  4
  

  
  
  
  
  
  

  
  
  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

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'])]

share|improve this answer

edited Mar 11 at 6:25

jezrael

351k26315391

answered Jul 13 '17 at 3:12

Sam HendersonSam Henderson

30135

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  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'])]

share|improve this answer

edited Mar 11 at 6:25

jezrael

351k26315391

answered Jul 13 '17 at 3:12

Sam HendersonSam Henderson

30135

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

share|improve this answer

edited Mar 11 at 6:25

jezrael

351k26315391

answered Jul 13 '17 at 3:12

Sam HendersonSam Henderson

30135

edited Mar 11 at 6:25

jezrael

351k26315391

edited Mar 11 at 6:25

jezrael

351k26315391

answered Jul 13 '17 at 3:12

Sam HendersonSam Henderson

30135

answered Jul 13 '17 at 3:12

Sam HendersonSam Henderson

30135

2

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])]

share|improve this answer

answered Apr 4 '18 at 11:51

Ioannis NasiosIoannis Nasios

3,75831036

add a comment | 

2

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])]

share|improve this answer

answered Apr 4 '18 at 11:51

Ioannis NasiosIoannis Nasios

3,75831036

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])]

share|improve this answer

answered Apr 4 '18 at 11:51

Ioannis NasiosIoannis Nasios

3,75831036

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

df[df.countries.isin(countries)]

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

share|improve this answer

answered Apr 4 '18 at 11:51

Ioannis NasiosIoannis Nasios

3,75831036

answered Apr 4 '18 at 11:51

Ioannis NasiosIoannis Nasios

3,75831036

answered Apr 4 '18 at 11:51

Ioannis NasiosIoannis Nasios

3,75831036