我如何才能实现SQL的IN和NOT IN的等价?

我有一个所需值的列表。 场景如下:

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

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

我目前的做法如下:

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'])]

但这似乎是一个可怕的拼凑。有人能改进吗?


当前回答

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

实现:

df[df.countries.isin(countries)]

实施不像在其他国家:

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

其他回答

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

实现:

df[df.countries.isin(countries)]

实施不像在其他国家:

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

你也可以在.query()中使用.isin():

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

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

要否定你的查询,使用~:

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

更新:

另一种方法是使用比较操作符:

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

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

要对查询求反,使用!=:

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

使用.query()方法的替代解决方案:

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

我想过滤出dfbc行,有一个BUSINESS_ID,也是在dfProfilesBusIds的BUSINESS_ID

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

我通常对行进行泛型过滤,像这样:

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