本文介绍了用逻辑(布尔)表达式切片Pandas Dataframe的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

在尝试用逻辑表达式对Pandas数据框进行切片时,我遇到了异常.

I am getting an exception as I try to slice with a logical expression my Pandas dataframe.

我的数据具有以下形式:

My data have the following form:

df
    GDP_norm    SP500_Index_deflated_norm
Year        
1980    2.121190    0.769400
1981    2.176224    0.843933
1982    2.134638    0.700833
1983    2.233525    0.829402
1984    2.395658    0.923654
1985    2.497204    0.922986
1986    2.584896    1.09770

df.info()
<class 'pandas.core.frame.DataFrame'>
Int64Index: 38 entries, 1980 to 2017
Data columns (total 2 columns):
GDP_norm                     38 non-null float64
SP500_Index_deflated_norm    38 non-null float64
dtypes: float64(2)
memory usage: 912.0 bytes

命令如下:

df[((df['GDP_norm'] >=3.5 & df['GDP_norm'] <= 4.5) & (df['SP500_Index_deflated_norm'] > 3)) | (

   (df['GDP_norm'] >= 4.0 & df['GDP_norm'] <= 5.0) & (df['SP500_Index_deflated_norm'] < 3.5))]

错误消息如下:

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

推荐答案

我建议分别创建布尔型掩码,以提高可读性并简化错误处理.

I suggest create boolean masks separately for better readibility and also easier error handling.

m1m2代码中缺少(),问题出在运算符优先级上:

Here are missing () in m1 and m2 code, problem is in operator precedence:

文档-6.16.参见&的运算符优先级具有更高的优先级,例如>=:

docs - 6.16. Operator precedence where see & have higher priority as >=:

Operator                                Description

lambda                                  Lambda expression
if – else                               Conditional expression
or                                      Boolean OR
and                                     Boolean AND
not x                                   Boolean NOT
in, not in, is, is not,                 Comparisons, including membership tests    
<, <=, >, >=, !=, ==                    and identity tests
|                                       Bitwise OR
^                                       Bitwise XOR
&                                       Bitwise AND

(expressions...), [expressions...],     Binding or tuple display, list display,       
{key: value...}, {expressions...}       dictionary display, set display


m1 = (df['GDP_norm'] >=3.5) & (df['GDP_norm'] <= 4.5)
m2 = (df['GDP_norm'] >= 4.0) & (df['GDP_norm'] <= 5.0)

m3 = m1 & (df['SP500_Index_deflated_norm'] > 3)
m4 = m2 & (df['SP500_Index_deflated_norm'] < 3.5)

df[m3 | m4]

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09-25 00:41