Question

DataFrame是：

A = pd.DataFrame({    'key':['II','I','I','III','II'],
                       'Z':['a', 'b', 'c', 'd', 'e'],
                       'd':[1,2,0,2,0],
                       'e':[0,2,0,3,0],
                       'f':[0,3,0,4,0],})

我有一个功能（它只是在features中编码data_aggregate）：

def Encode(data_aggregate, features):
    for feature in features:       
        l = len(data_aggregate)
        groups = data_aggregate.groupby(by=feature)
        groups_sizes = groups.size()
        for key in groups_sizes.keys():
            ratio = groups_sizes[key] / l
            data_aggregate.loc[data_aggregate.feature == key, [feature]] = ratio
    return data_aggregate

当我将A传递给Encode时：

new = Encode(A, ['key'])

它返回错误：

   ---------------------------------------------------------------------------
    AttributeError                            Traceback (most recent call last)
    <ipython-input-31-e896b315a285> in <module>()
    ----> 1 A.loc[A.feature == key, [feature]] = 111

    C:\ProgramData\Anaconda3\lib\site-packages\pandas\core\generic.py in __getattr__(self, name)
       2742             if name in self._info_axis:
       2743                 return self[name]
    -> 2744             return object.__getattribute__(self, name)
       2745 
       2746     def __setattr__(self, name, value):

    AttributeError: 'DataFrame' object has no attribute 'feature'

Answer 1

您不能使用点表示法引用包含字符串的变量名称的列。

试试这个

def Encode(data_aggregate, features):
    for feature in features:       
        l = len(data_aggregate)
        groups = data_aggregate.groupby(by=feature)
        groups_sizes = groups.size()
        for key in groups_sizes.keys():
            ratio = groups_sizes[key] / l
            data_aggregate.loc[data_aggregate[feature] == key, [feature]] = ratio
    return data_aggregate

<强> 溶液
也就是说，你可以用

做同样的事情

A.assign(key=A.key.map(A.key.value_counts(normalize=True)))

   Z  d  e  f  key
0  a  1  0  0  0.4
1  b  2  2  3  0.4
2  c  0  0  0  0.4
3  d  2  3  4  0.2
4  e  0  0  0  0.4

功能化

def Encode(df, features):
    d = {k: df[k].map(df[k].value_counts(normalize=True)) for k in features}
    return df.assign(**d)

Encode(A, ['key'])

   Z  d  e  f  key
0  a  1  0  0  0.4
1  b  2  2  3  0.4
2  c  0  0  0  0.4
3  d  2  3  4  0.2
4  e  0  0  0  0.4

Encode(A, ['key', 'Z'])

     Z  d  e  f  key
0  0.2  1  0  0  0.4
1  0.2  2  2  3  0.4
2  0.2  0  0  0  0.4
3  0.2  2  3  4  0.2
4  0.2  0  0  0  0.4

如何调用大熊猫＆＃39; DataFrame列，传递了一个函数

1 个答案: