python – 带有pd.Series布尔值的索引numpy数组

2019年8月3日 281次阅读

我发现了一段我不太懂的代码.它基本上是这样的：

array = np.ones((5, 4))*np.nan
s1 = pd.Series([1,4,0,4,5], index=[0,1,2,3,4])
I = s1 == 4
print(I)

0    False
1     True
2    False
3     True
4    False
dtype: bool

我真的理解这部分,它在4的索引处返回一个带有True的bo.Series布尔值.现在,作者使用I来索引数组：

array[I,0] = 3
array[I,1] = 7
array[I,2] = 2
array[I,3] = 5
print(array)

[[  3.   7.   2.   5.]
 [  3.   7.   2.   5.]
 [ nan  nan  nan  nan]
 [ nan  nan  nan  nan]
 [ nan  nan  nan  nan]]

新阵列对我来说毫无意义,我想返回：

[[ nan  nan  nan  nan]
 [  3.   7.   2.   5.]
 [ nan  nan  nan  nan]
 [  3.   7.   2.   5.]
 [ nan  nan  nan  nan]]

有人可以解释这里发生了什么,以及如何更改上面的代码以返回我需要的东西？

最佳答案 解释在于numpy数组和pandas系列以不同方式处理逻辑索引.前者将True视为1,将False视为0,而后者将逻辑为True的值视为True,并将逻辑为False的值删除.作为示范：

import numpy as np
import pandas as pd

arr = np.array([1,2,3,4,5])
arr                           # this is a numpy array 
array([1, 2, 3, 4, 5])
arr[[True, False, True]]
array([2, 1, 2])              # check here how it is actually picking the value at position 
                              # 1 and 0 alternatively;

ser = pd.Series([1,2,3,4,5])
ser                           # this is a pandas Series
0    1
1    2
2    3
3    4
4    5
dtype: int64
ser[[True, False, True]]      # in pandas Series, it will pick up values where the logic is True;
0    1
2    3
dtype: int64

你会看到他们的行为方式不同.由于您的数组是一个numpy数组,我们不能使用逻辑索引来获取值.为了得到你想要的结果,我们可以尝试从I中提取真值的索引,然后在你的数组上使用它：

array[I[I == True].index,0] = 3
array[I[I == True].index,1] = 7
array[I[I == True].index,2] = 2
array[I[I == True].index,3] = 5
print(array)


[[ nan  nan  nan  nan]
 [  3.   7.   2.   5.]
 [ nan  nan  nan  nan]
 [  3.   7.   2.   5.]
 [ nan  nan  nan  nan]]
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