为什么没有人谈论各种过滤方法的性能?事实上，这个主题经常出现在这里(参见示例)。我为一个大型数据集做了自己的性能测试。这是非常有趣和有教育意义的。

df = pd.DataFrame({'animals': np.random.choice(['cat', 'dog', 'mouse', 'birds'], size=10**7), 
                   'number': np.random.randint(0,100, size=(10**7,))})

df.info()

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 10000000 entries, 0 to 9999999
Data columns (total 2 columns):
 #   Column   Dtype 
---  ------   ----- 
 0   animals  object
 1   number   int64 
dtypes: int64(1), object(1)
memory usage: 152.6+ MB

%%timeit
# .isin() by one column
conditions = ['cat', 'dog']
df[df.animals.isin(conditions)]

367 ms ± 2.34 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
# .query() by one column
conditions = ['cat', 'dog']
df.query('animals in @conditions')

395 ms ± 3.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
# .loc[]
df.loc[(df.animals=='cat')|(df.animals=='dog')]

987 ms ± 5.17 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
df[df.apply(lambda x: x['animals'] in ['cat', 'dog'], axis=1)]

41.9 s ± 490 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
new_df = df.set_index('animals')
new_df.loc[['cat', 'dog'], :]

3.64 s ± 62.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
new_df = df.set_index('animals')
new_df[new_df.index.isin(['cat', 'dog'])]

469 ms ± 8.98 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
s = pd.Series(['cat', 'dog'], name='animals')
df.merge(s, on='animals', how='inner')

796 ms ± 30.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

因此，isin方法是最快的，而带有apply()的方法是最慢的，这并不奇怪。

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)]

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