代码之家 › 专栏 › 技术社区 › Florent

如何用熊猫分割两个不同形状的数据帧?

numpy pandas python

Florent · 技术社区 · 6 年前

我有两个索引相同但形状不同的数据帧,无法将列与数据帧分开 df1 数据框中的列 df2 是的。

预期结果是 df1 / df2 是的。

df1.head()
                           volume  volume        volume         volume  \
timestamp                                                                
2016-07-24 00:00:00+00:00     NaN     NaN           NaN            NaN   
2016-07-25 00:00:00+00:00     NaN     NaN           NaN            NaN   
2016-07-26 00:00:00+00:00     NaN     NaN           NaN  102720.829507   
2016-07-27 00:00:00+00:00     NaN     NaN  3.729644e+05  398346.509801   
2016-07-28 00:00:00+00:00     NaN     NaN  1.326648e+06  244165.794698   

                           volume        volume  volume        volume  
timestamp                                                              
2016-07-24 00:00:00+00:00     NaN           NaN     NaN  1.734943e+07  
2016-07-25 00:00:00+00:00     NaN           NaN     NaN  1.365341e+07  
2016-07-26 00:00:00+00:00     NaN           NaN     NaN  5.199938e+07  
2016-07-27 00:00:00+00:00     NaN  2.471076e+06     NaN  2.558753e+07  
2016-07-28 00:00:00+00:00     NaN  1.642990e+06     NaN  3.118785e+06

df2.head()

timestamp
2016-07-24 00:00:00+00:00    1.734943e+07
2016-07-25 00:00:00+00:00    1.365341e+07
2016-07-26 00:00:00+00:00    5.210210e+07
2016-07-27 00:00:00+00:00    2.882991e+07
2016-07-28 00:00:00+00:00    6.332589e+06
Freq: D, dtype: float64

df1.shape
Out[2126]: (723, 8)

df2.shape
Out[2127]: (723,)

df1.divide(df2, axis= 'index')
ValueError: operands could not be broadcast together with shapes (5784,) (723,)

两个数据帧的结构不同,但索引相同。

type(df1)
Out[2143]: pandas.core.frame.DataFrame

type(df2)
Out[2144]: pandas.core.series.Series

我读到需要重塑其中一个数据帧,所以我尝试了如下操作:

df1.divide(df2.reshape(723,1), axis= 'index')

但它返回一个错误:

ValueError: Unable to coerce to DataFrame, shape must be (723, 8): given (723, 1)

当我皈依 DF2 具有 pd.DataFrame(df2) 然后它抛出一个错误:

TypeError: '<' not supported between instances of 'str' and 'int'

我错过了什么,我该怎么做?

2 回复 | 直到 6 年前

Dtengu 6 年前

试试这个方法我用了一个简单的例子,但如果不起作用,请告诉我。

import pandas as pd
import numpy as np
from IPython.display import display, HTML

CSS = """
.output {
    flex-direction: row;
}
"""

HTML('<style>{}</style>'.format(CSS))


data1 = {"a":[1.,7.,12.],
         "b":[4.,8.,3.],
         "c":[5.,45.,67.]}
data2 = {"a":[3.],
         "b":[2.],
         "c":[8.]}

df1 = pd.DataFrame(data1)
df2 = pd.DataFrame(data2) 
df2 = df2.T
df2 = df2.reset_index()
del df2['index']
display(df1)
display(df2)
display(df1.iloc[:,0:].truediv(df2[0], axis=0)) # this portion of code you want

甲、乙、丙
0 1.0 4.0 5.0
17.0 8.0 45.0条
2 12.0 3.0 67.0条

0个
0 3.0分
12.0条
28.0条

甲、乙、丙
0.333333 1.333333 1.666667
130万4.000000 22.500000
2 150万0.375000 8.375000

gau 6 年前

在使用Delphi(或DIV)函数时,对每个数据框中的相应列/ S进行索引应该是有效的。

df1[['column_1','column_2']].divide(df2[['column_1']], axis= 'index')  

df1[['column_1','column_2']].div(df2[['column_1']], axis= 'index')