Question

我有一个pandas数据框，其中有一列包含带有一堆属性的类。我希望将其中一些属性扩展到新列中。我有一些有效的代码，但看起来有点讨厌，并使用一个eval。什么是更蟒蛇的方式做到这一点

import pandas as pd

#Boilerplate for minimal, reproducible example
class cl:
    class inner:
        na1 = "nested atribute one"
        na2 = "nested atribute two"
    def __init__(self, name):
        self.name = name
    a1 = "atribute one"
    a2 = "atribute one"
    inner_atts = inner()


class_object1 = cl("first")
class_object2 = cl("second")
data = [class_object1,class_object2]
data_frame = pd.DataFrame(data,columns=['class object'])
####################
info_to_get = {'name','a1','a2','inner_atts.na1','inner_atts.na2'}

for x in info_to_get:
    sr = 'y.{0}'.format(x)
    data_frame['{0}'.format(x)] = data_frame['class object'].apply(lambda y: eval(sr,{'y':y}))

print(data_frame)

Answer 1

使用operator.attrgetter：

import operator

info_to_get = list(info_to_get)
df[info_to_get] = pd.DataFrame(df['class object'].apply(operator.attrgetter(*info_to_get)).tolist())

输出：

                             class object       inner_atts.na1  \
0  <__main__.cl object at 0x7f08002d27b8>  nexted atribute one   
1  <__main__.cl object at 0x7f08002d2a90>  nexted atribute one   

        inner_atts.na2            a2   name            a1  
0  nexted atribute two  atribute one  first  atribute one  
1  nexted atribute two  atribute one    two  atribute one

Answer 2

关于熊猫的第一件事是，它不适合存储和处理无法向量化的任何东西-开销很大，最好使用列表和循环对其进行迭代。

也就是说，我将使用列表理解来做到这一点。

from operator import attrgetter

f = attrgetter(*info_to_get)
pd.DataFrame([f(c) for c in df['class object']], columns=info_to_get)

        inner_atts.na2    name            a2       inner_atts.na1            a1
0  nexted atribute two   first  atribute one  nexted atribute one  atribute one
1  nexted atribute two  second  atribute one  nexted atribute one  atribute one

Evidence suggests，使用列表组合处理不可矢量化的数据，可以最大程度地提高速度。

如何避免使用“ eval”功能遍历属性列表

2 个答案: