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

从答案中整理出可能的解决方案:

用于:df[df['A']。isin ([3], 6)]

对于不在:

df [-df[“A”]。isin ([3], 6)] df [~ df[“A”]。isin ([3], 6)] df [df[“A”]。isin([3,6]) == False] df [np。logical_not“A”(df[]。isin ([3], 6))]

其他回答

如果你想保持列表的顺序,有一个小技巧:

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

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)]
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])]

从答案中整理出可能的解决方案:

用于:df[df['A']。isin ([3], 6)]

对于不在:

df [-df[“A”]。isin ([3], 6)] df [~ df[“A”]。isin ([3], 6)] df [df[“A”]。isin([3,6]) == False] df [np。logical_not“A”(df[]。isin ([3], 6))]

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