如何根据Pandas中某列的值从DataFrame中选择行?

在SQL中,我会使用:

SELECT *
FROM table
WHERE column_name = some_value

当前回答

我发现前面答案的语法是多余的,很难记住。Pandas在v0.13中引入了query()方法,我更喜欢它。对于您的问题,您可以使用df.query('col==val')。

转载自query()方法(实验):

In [167]: n = 10

In [168]: df = pd.DataFrame(np.random.rand(n, 3), columns=list('abc'))

In [169]: df
Out[169]:
          a         b         c
0  0.687704  0.582314  0.281645
1  0.250846  0.610021  0.420121
2  0.624328  0.401816  0.932146
3  0.011763  0.022921  0.244186
4  0.590198  0.325680  0.890392
5  0.598892  0.296424  0.007312
6  0.634625  0.803069  0.123872
7  0.924168  0.325076  0.303746
8  0.116822  0.364564  0.454607
9  0.986142  0.751953  0.561512

# pure python
In [170]: df[(df.a < df.b) & (df.b < df.c)]
Out[170]:
          a         b         c
3  0.011763  0.022921  0.244186
8  0.116822  0.364564  0.454607

# query
In [171]: df.query('(a < b) & (b < c)')
Out[171]:
          a         b         c
3  0.011763  0.022921  0.244186
8  0.116822  0.364564  0.454607

您还可以通过在环境中添加@来访问变量。

exclude = ('red', 'orange')
df.query('color not in @exclude')

其他回答

使用DuckDB选择行的DataFrames上的SQL语句

使用DuckDB,我们可以用SQL语句以高性能的方式查询panda DataFrames。

由于问题是如何根据列值从DataFrame中选择行?,问题中的示例是一个SQL查询,这个答案在本主题中看起来很合理。

例子:

In [1]: import duckdb

In [2]: import pandas as pd

In [3]: con = duckdb.connect()

In [4]: df = pd.DataFrame({"A": range(11), "B": range(11, 22)})

In [5]: df
Out[5]:
     A   B
0    0  11
1    1  12
2    2  13
3    3  14
4    4  15
5    5  16
6    6  17
7    7  18
8    8  19
9    9  20
10  10  21

In [6]: results = con.execute("SELECT * FROM df where A > 2").df()

In [7]: results
Out[7]:
    A   B
0   3  14
1   4  15
2   5  16
3   6  17
4   7  18
5   8  19
6   9  20
7  10  21

要添加:您还可以执行df.groupby('column_name').get_group('column_desired_value').reset_index()以生成具有特定值的指定列的新数据帧。例如。,

import pandas as pd
df = pd.DataFrame({'A': 'foo bar foo bar foo bar foo foo'.split(),
                   'B': 'one one two three two two one three'.split()})
print("Original dataframe:")
print(df)

b_is_two_dataframe = pd.DataFrame(df.groupby('B').get_group('two').reset_index()).drop('index', axis = 1) 
#NOTE: the final drop is to remove the extra index column returned by groupby object
print('Sub dataframe where B is two:')
print(b_is_two_dataframe)

运行此命令可以:

Original dataframe:
     A      B
0  foo    one
1  bar    one
2  foo    two
3  bar  three
4  foo    two
5  bar    two
6  foo    one
7  foo  three
Sub dataframe where B is two:
     A    B
0  foo  two
1  foo  two
2  bar  two

对于Pandas中给定值的多个列中仅选择特定列:

select col_name1, col_name2 from table where column_name = some_value.

选项位置:

df.loc[df['column_name'] == some_value, [col_name1, col_name2]]

或查询:

df.query('column_name == some_value')[[col_name1, col_name2]]

如果您想重复查询数据帧,并且速度对您很重要,最好的方法是将数据帧转换为字典,然后通过这样做,您可以将查询速度提高数千倍。

my_df = df.set_index(column_name)
my_dict = my_df.to_dict('index')

制作my_dict字典后,您可以浏览:

if some_value in my_dict.keys():
   my_result = my_dict[some_value]

如果column_name中有重复值,则无法创建字典。但您可以使用:

my_result = my_df.loc[some_value]

使用numpy.where可以获得更快的结果。

例如,使用unubtu的设置-

In [76]: df.iloc[np.where(df.A.values=='foo')]
Out[76]: 
     A      B  C   D
0  foo    one  0   0
2  foo    two  2   4
4  foo    two  4   8
6  foo    one  6  12
7  foo  three  7  14

时间比较:

In [68]: %timeit df.iloc[np.where(df.A.values=='foo')]  # fastest
1000 loops, best of 3: 380 µs per loop

In [69]: %timeit df.loc[df['A'] == 'foo']
1000 loops, best of 3: 745 µs per loop

In [71]: %timeit df.loc[df['A'].isin(['foo'])]
1000 loops, best of 3: 562 µs per loop

In [72]: %timeit df[df.A=='foo']
1000 loops, best of 3: 796 µs per loop

In [74]: %timeit df.query('(A=="foo")')  # slowest
1000 loops, best of 3: 1.71 ms per loop