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'])]

你可以使用pd.Series.isin。

对于“IN”的用法:something.isin(某处)

或者对于"NOT IN": ~something.isin(某处)

举个例子:

>>> df
    country
0        US
1        UK
2   Germany
3     China
>>> countries_to_keep
['UK', 'China']
>>> df.country.isin(countries_to_keep)
0    False
1     True
2    False
3     True
Name: country, dtype: bool
>>> df[df.country.isin(countries_to_keep)]
    country
1        UK
3     China
>>> df[~df.country.isin(countries_to_keep)]
    country
0        US
2   Germany

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)]

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

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))]

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