Question

我正在努力在两个数据帧之间进行基本的条件逐元素乘法。假设我有以下两个数据框：

df1 = pd.DataFrame({'A': [-0.1,0.3,-0.4, 0.8,-0.5,-0.1,0.3,-0.4, 0.8,-1.2],'B': [-0.2,0.5,0.3,-0.5,0.1,-0.2,0.5,0.3,-0.5,0.9]},index=[0, 1, 2, 3,4,5,6,7,8,9])
df2=pd.DataFrame({'C': [-0.003,0.03848,-0.04404, 0.018,-0.1515,-0.02181,0.233,-0.0044, 0.01458,-0.015],'D': [-0.0152,0.0155,0.03,-0.0155,0.0151,-0.012,0.035,0.0013,-0.0005,0.009]},index=[0, 1, 2, 3,4,5,6,7,8,9])

这个想法是根据df1的值乘以df1 and df2.shift(-1)（从本质上讲，不是矩阵乘法）。如果(df1>=0.50 or df1<=-0.50)，则将df1和df2.shift（-1）相乘。否则，我只输入0。

此示例中的期望结果应为以下内容（列名称是df1以及df1索引的列名称）：

df3=pd.DataFrame({'A': [0,0,0, -0.1212,0.010905,0,0,0, -0.012,'NaN'],'B': [0,0.015,0,-0.00755,0,0,0.00065,0,-0.0045,'NaN']},index=[0, 1, 2, 3,4,5,6,7,8,9])

我尝试了以下代码：

import numpy as np
import pandas as pd
df3=np.where((df1>=0.50 or df1 <=-0.50),df1*df2.shift(-1),0)

然后我得到一个DataFrame的真值是模棱两可的。使用a.empty，a.bool（），a.item（），a.any（）或a.all（）。谢谢。

Answer 1

将|与OR构造函数一起按位DataFrame使用：

arr = np.where((df1>=0.50) | (df1 <=-0.50),df1*df2.shift(-1),0)
df3 = pd.DataFrame(arr, index=df1.index, columns=df1.columns)
print (df3)
          A        B
0  0.000000  0.00000
1  0.000000  0.01500
2  0.000000  0.00000
3 -0.121200 -0.00755
4  0.010905  0.00000
5  0.000000  0.00000
6  0.000000  0.00065
7  0.000000  0.00000
8 -0.012000 -0.00450
9       NaN      NaN

Numpy解决方案应该更快：

arr2 = np.concatenate([df2.values[1:, ], 
                       np.repeat(np.nan, len(df2.columns))[None, :]])

arr = np.where((df1.values>=0.50) | (df1.values <=-0.50),df1.values*arr2,0)
df3 = pd.DataFrame(arr, index=df1.index, columns=df1.columns)
print (df3)
          A        B
0  0.000000  0.00000
1  0.000000  0.01500
2  0.000000  0.00000
3 -0.121200 -0.00755
4  0.010905  0.00000
5  0.000000  0.00000
6  0.000000  0.00065
7  0.000000  0.00000
8 -0.012000 -0.00450
9       NaN      NaN

两个熊猫数据帧的条件逐元素乘法

1 个答案: