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?
This question is related to
python
pandas
dataframe
sql-function
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]))]
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.
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
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
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 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'])]
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
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
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)]
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])]
Source: Stackoverflow.com