如何基于列值从DataFrame中选择行?

问题描述 投票:1598回答:10

如何基于Python Pandas中某些列的值从DataFrame中选择行?

在SQL中,我将使用:

SELECT *
FROM table
WHERE colume_name = some_value

我试图查看熊猫文档,但没有立即找到答案。

python pandas dataframe
10个回答
3065
投票

要选择列值等于标量some_value的行,请使用==

df.loc[df['column_name'] == some_value]

要选择其列值为可迭代的行some_values,请使用isin

df.loc[df['column_name'].isin(some_values)]

&组合多个条件:

df.loc[(df['column_name'] >= A) & (df['column_name'] <= B)]

注意括号。由于Python的operator precedence rules&的绑定比<=>=更紧密。因此,最后一个示例中的括号是必需的。没有括号

df['column_name'] >= A & df['column_name'] <= B

解析为

df['column_name'] >= (A & df['column_name']) <= B

这导致Truth value of a Series is ambiguous error


要选择列值不等于的行some_value,请使用!=

df.loc[df['column_name'] != some_value]

isin返回一个布尔序列,因此要选择some_values中值为not的行,请使用~取反布尔序列:

df.loc[~df['column_name'].isin(some_values)]

例如,

import pandas as pd
import numpy as np
df = pd.DataFrame({'A': 'foo bar foo bar foo bar foo foo'.split(),
                   'B': 'one one two three two two one three'.split(),
                   'C': np.arange(8), 'D': np.arange(8) * 2})
print(df)
#      A      B  C   D
# 0  foo    one  0   0
# 1  bar    one  1   2
# 2  foo    two  2   4
# 3  bar  three  3   6
# 4  foo    two  4   8
# 5  bar    two  5  10
# 6  foo    one  6  12
# 7  foo  three  7  14

print(df.loc[df['A'] == 'foo'])

产量

     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

如果要包含多个值,请将它们放在列出(或更普遍地说,是任何可迭代的)并使用isin

print(df.loc[df['B'].isin(['one','three'])])

产量

     A      B  C   D
0  foo    one  0   0
1  bar    one  1   2
3  bar  three  3   6
6  foo    one  6  12
7  foo  three  7  14

但是请注意,如果您希望多次执行此操作,首先创建一个索引,然后使用df.loc

df = df.set_index(['B'])
print(df.loc['one'])

产量

       A  C   D
B              
one  foo  0   0
one  bar  1   2
one  foo  6  12

或者,要包含索引中的多个值,请使用df.index.isin

df.loc[df.index.isin(['one','two'])]

产量

       A  C   D
B              
one  foo  0   0
one  bar  1   2
two  foo  2   4
two  foo  4   8
two  bar  5  10
one  foo  6  12

2
投票

您也可以使用.apply:

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

248
投票

tl; dr

相当于]的熊猫>

select * from table where column_name = some_value

table[table.column_name == some_value]

多个条件:

table[(table.column_name == some_value) | (table.column_name2 == some_value2)]

table.query('column_name == some_value | column_name2 == some_value2')

代码示例

import pandas as pd

# Create data set
d = {'foo':[100, 111, 222], 
     'bar':[333, 444, 555]}
df = pd.DataFrame(d)

# Full dataframe:
df

# Shows:
#    bar   foo 
# 0  333   100
# 1  444   111
# 2  555   222

# Output only the row(s) in df where foo is 222:
df[df.foo == 222]

# Shows:
#    bar  foo
# 2  555  222

在上面的代码中,df[df.foo == 222]行基于列值给出行,在这种情况下为222

也可能有多个条件:

df[(df.foo == 222) | (df.bar == 444)]
#    bar  foo
# 1  444  111
# 2  555  222

但是在那一点上,我建议使用query函数,因为它不那么冗长,并且会产生相同的结果:

df.query('foo == 222 | bar == 444')

211
投票

有几种方法可以从熊猫数据框中选择行:


52
投票

我发现先前答案的语法是多余的,很难记住。熊猫在v0.13中引入了for j in spec.columns: d = pd.concat([df] * j, ignore_index=True) for i in spec.index: stmt = '{}(d)'.format(i) setp = 'from __main__ import d, {}'.format(i) spec.at[i, j] = timeit(stmt, setp, number=50) 方法,我更喜欢它。对于您的问题,您可以执行query()


19
投票

使用exclude = ('red', 'orange') df.query('color not in @exclude') 可以实现更快的结果。

例如,使用numpy.where-


16
投票

这是一个简单的示例

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

13
投票

对于在熊猫中给定值,仅从多个列中选择特定列:

from pandas import DataFrame

# Create data set
d = {'Revenue':[100,111,222], 
     'Cost':[333,444,555]}
df = DataFrame(d)


# mask = Return True when the value in column "Revenue" is equal to 111
mask = df['Revenue'] == 111

print mask

# Result:
# 0    False
# 1     True
# 2    False
# Name: Revenue, dtype: bool


# Select * FROM df WHERE Revenue = 111
df[mask]

# Result:
#    Cost    Revenue
# 1  444     111

10
投票

df.query['column_name' == 'some_value'][[col_name1, col_name2]] .query结合使用可带来更大的灵活性:

[2019年8月更新的答案


9
投票

附加到这个著名的问题(虽然为时已晚):您也可以执行 Sender email 1 [email protected] 2 [email protected] 来制作一个新的数据帧,其中指定列具有特定值。例如:

df.groupby('column_name').get_group('column_desired_value').reset_index()
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