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从pandas.DataFrame使用复杂条件选择

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问题：从pandas.DataFrame使用复杂条件选择
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回答 1
回答 2
回答 3

问题：从pandas.DataFrame使用复杂条件选择

例如，我有简单的DF：

import pandas as pd
from random import randint

df = pd.DataFrame({'A': [randint(1, 9) for x in xrange(10)],
                   'B': [randint(1, 9)*10 for x in xrange(10)],
                   'C': [randint(1, 9)*100 for x in xrange(10)]})

我可以使用熊猫的方法和惯用法从“ A”中选择与B对应的值大于50的值，对于C对应的值大于900的值吗？

点击查看英文原文

For example I have simple DF:

import pandas as pd
from random import randint

df = pd.DataFrame({'A': [randint(1, 9) for x in xrange(10)],
                   'B': [randint(1, 9)*10 for x in xrange(10)],
                   'C': [randint(1, 9)*100 for x in xrange(10)]})

Can I select values from ‘A’ for which corresponding values for ‘B’ will be greater than 50, and for ‘C’ – not equal 900, using methods and idioms of Pandas?

回答 0

当然！建立：

>>> import pandas as pd
>>> from random import randint
>>> df = pd.DataFrame({'A': [randint(1, 9) for x in range(10)],
                   'B': [randint(1, 9)*10 for x in range(10)],
                   'C': [randint(1, 9)*100 for x in range(10)]})
>>> df
   A   B    C
0  9  40  300
1  9  70  700
2  5  70  900
3  8  80  900
4  7  50  200
5  9  30  900
6  2  80  700
7  2  80  400
8  5  80  300
9  7  70  800

我们可以应用列操作并获取布尔系列对象：

>>> df["B"] > 50
0    False
1     True
2     True
3     True
4    False
5    False
6     True
7     True
8     True
9     True
Name: B
>>> (df["B"] > 50) & (df["C"] == 900)
0    False
1    False
2     True
3     True
4    False
5    False
6    False
7    False
8    False
9    False

[更新，切换到新样式.loc]：

然后我们可以使用它们来索引对象。对于读取访问，可以链接索引：

>>> df["A"][(df["B"] > 50) & (df["C"] == 900)]
2    5
3    8
Name: A, dtype: int64

但是由于视图和执行写操作的副本之间的差异，您可能会遇到麻烦。您可以.loc改用：

>>> df.loc[(df["B"] > 50) & (df["C"] == 900), "A"]
2    5
3    8
Name: A, dtype: int64
>>> df.loc[(df["B"] > 50) & (df["C"] == 900), "A"].values
array([5, 8], dtype=int64)
>>> df.loc[(df["B"] > 50) & (df["C"] == 900), "A"] *= 1000
>>> df
      A   B    C
0     9  40  300
1     9  70  700
2  5000  70  900
3  8000  80  900
4     7  50  200
5     9  30  900
6     2  80  700
7     2  80  400
8     5  80  300
9     7  70  800

请注意，我不小心输入了== 900not != 900和或~(df["C"] == 900)，但是我懒得修复它。为读者练习。：^）

点击查看英文原文

Sure! Setup:

>>> import pandas as pd
>>> from random import randint
>>> df = pd.DataFrame({'A': [randint(1, 9) for x in range(10)],
                   'B': [randint(1, 9)*10 for x in range(10)],
                   'C': [randint(1, 9)*100 for x in range(10)]})
>>> df
   A   B    C
0  9  40  300
1  9  70  700
2  5  70  900
3  8  80  900
4  7  50  200
5  9  30  900
6  2  80  700
7  2  80  400
8  5  80  300
9  7  70  800

We can apply column operations and get boolean Series objects:

>>> df["B"] > 50
0    False
1     True
2     True
3     True
4    False
5    False
6     True
7     True
8     True
9     True
Name: B
>>> (df["B"] > 50) & (df["C"] == 900)
0    False
1    False
2     True
3     True
4    False
5    False
6    False
7    False
8    False
9    False

[Update, to switch to new-style .loc]:

And then we can use these to index into the object. For read access, you can chain indices:

>>> df["A"][(df["B"] > 50) & (df["C"] == 900)]
2    5
3    8
Name: A, dtype: int64

but you can get yourself into trouble because of the difference between a view and a copy doing this for write access. You can use .loc instead:

>>> df.loc[(df["B"] > 50) & (df["C"] == 900), "A"]
2    5
3    8
Name: A, dtype: int64
>>> df.loc[(df["B"] > 50) & (df["C"] == 900), "A"].values
array([5, 8], dtype=int64)
>>> df.loc[(df["B"] > 50) & (df["C"] == 900), "A"] *= 1000
>>> df
      A   B    C
0     9  40  300
1     9  70  700
2  5000  70  900
3  8000  80  900
4     7  50  200
5     9  30  900
6     2  80  700
7     2  80  400
8     5  80  300
9     7  70  800

Note that I accidentally typed == 900 and not != 900, or ~(df["C"] == 900), but I’m too lazy to fix it. Exercise for the reader. :^)

回答 1

另一种解决方案是使用查询方法：

import pandas as pd

from random import randint
df = pd.DataFrame({'A': [randint(1, 9) for x in xrange(10)],
                   'B': [randint(1, 9) * 10 for x in xrange(10)],
                   'C': [randint(1, 9) * 100 for x in xrange(10)]})
print df

   A   B    C
0  7  20  300
1  7  80  700
2  4  90  100
3  4  30  900
4  7  80  200
5  7  60  800
6  3  80  900
7  9  40  100
8  6  40  100
9  3  10  600

print df.query('B > 50 and C != 900')

   A   B    C
1  7  80  700
2  4  90  100
4  7  80  200
5  7  60  800

现在，如果要更改A列中的返回值，可以保存其索引：

my_query_index = df.query('B > 50 & C != 900').index

…并用于.iloc更改它们，即：

df.iloc[my_query_index, 0] = 5000

print df

      A   B    C
0     7  20  300
1  5000  80  700
2  5000  90  100
3     4  30  900
4  5000  80  200
5  5000  60  800
6     3  80  900
7     9  40  100
8     6  40  100
9     3  10  600
点击查看英文原文

Another solution is to use the query method:

import pandas as pd

from random import randint
df = pd.DataFrame({'A': [randint(1, 9) for x in xrange(10)],
                   'B': [randint(1, 9) * 10 for x in xrange(10)],
                   'C': [randint(1, 9) * 100 for x in xrange(10)]})
print df

   A   B    C
0  7  20  300
1  7  80  700
2  4  90  100
3  4  30  900
4  7  80  200
5  7  60  800
6  3  80  900
7  9  40  100
8  6  40  100
9  3  10  600

print df.query('B > 50 and C != 900')

   A   B    C
1  7  80  700
2  4  90  100
4  7  80  200
5  7  60  800

Now if you want to change the returned values in column A you can save their index:

my_query_index = df.query('B > 50 & C != 900').index

….and use .iloc to change them i.e:

df.iloc[my_query_index, 0] = 5000

print df

      A   B    C
0     7  20  300
1  5000  80  700
2  5000  90  100
3     4  30  900
4  5000  80  200
5  5000  60  800
6     3  80  900
7     9  40  100
8     6  40  100
9     3  10  600

回答 2

并记住要使用括号！

请记住，&运算符的优先级高于诸如>或的运算符<。这就是为什么

4 < 5 & 6 > 4

评估为False。因此，如果使用pd.loc，则需要在逻辑语句两边加上方括号，否则会出现错误。这就是为什么：

df.loc[(df['A'] > 10) & (df['B'] < 15)]

代替 

df.loc[df['A'] > 10 & df['B'] < 15]

这将导致

TypeError：无法将dtyped [float64]数组与类型为[bool]的标量进行比较

点击查看英文原文

And remember to use parenthesis!

Keep in mind that & operator takes a precedence over operators such as > or < etc. That is why 

4 < 5 & 6 > 4

evaluates to False. Therefore if you’re using pd.loc, you need to put brackets around your logical statements, otherwise you get an error. That’s why do:

df.loc[(df['A'] > 10) & (df['B'] < 15)]

instead of 

df.loc[df['A'] > 10 & df['B'] < 15]

which would result in

TypeError: cannot compare a dtyped [float64] array with a scalar of type [bool]

回答 3

您可以使用熊猫它具有一些内置功能进行比较。因此，如果您想选择“ B”和“ C”条件满足的“ A”值（假设您要返回一个DataFrame pandas对象）

df[['A']][df.B.gt(50) & df.C.ne(900)]

df[['A']] 将以DataFrame格式返回A列。

pandas的“ gt”函数将返回大于50的列B的位置，而“ ne”函数将返回不等于900的位置。

点击查看英文原文

You can use pandas it has some built in functions for comparison. So if you want to select values of “A” that are met by the conditions of “B” and “C” (assuming you want back a DataFrame pandas object)

df[['A']][df.B.gt(50) & df.C.ne(900)]

df[['A']] will give you back column A in DataFrame format.

pandas ‘gt’ function will return the positions of column B that are greater than 50 and ‘ne’ will return the positions not equal to 900.