value_counts（）和apply（pd.value_counts）之间的差异

Question

以下字段包含三个值：增量值，完整值和差异值，以及Array[indexes] = 5 和value_counts()

为什么会出现这种差异？

Answer 1

问题是您需要将函数Series.value_counts应用于DataFrame的aomecolumns，因此请使用apply。

与：

相同

df.apply(lambda s: s.value_counts())
#same as
df.apply(pd.value_counts)  

Answer 2

这是一种未记录的方法：

Signature: pd.value_counts(values, sort=True, ascending=False, normalize=False, bins=None, dropna=True)
Docstring:
Compute a histogram of the counts of non-null values.

Parameters
----------
values : ndarray (1-d)
sort : boolean, default True
    Sort by values
ascending : boolean, default False
    Sort in ascending order
normalize: boolean, default False
    If True then compute a relative histogram
bins : integer, optional
    Rather than count values, group them into half-open bins,
    convenience for pd.cut, only works with numeric data
dropna : boolean, default True
    Don't include counts of NaN

Returns
-------
value_counts : Series

如果您将列作为arg传递，则输出与pd.Series.value_counts：

相同

In [8]:
Offline_BackupSchemaIncrementType
df = pd.DataFrame({'Offline_BackupSchemaIncrementType': [0,1,1,2,np.NaN], 'val':np.arange(5)})
df

Out[8]:
   Offline_BackupSchemaIncrementType  val
0                                0.0    0
1                                1.0    1
2                                1.0    2
3                                2.0    3
4                                NaN    4

In [9]:
pd.value_counts(df['Offline_BackupSchemaIncrementType'])

Out[9]:
1.0    2
2.0    1
0.0    1
Name: Offline_BackupSchemaIncrementType, dtype: int64

In [10]:    
df['Offline_BackupSchemaIncrementType'].value_counts()

Out[10]:
1.0    2
2.0    1
0.0    1
Name: Offline_BackupSchemaIncrementType, dtype: int64

然而，当你apply方法时，你正在为每个元素执行此操作，因此返回的Series正在尝试将其与原始df对齐，实际上你得到了一个二维数组：

In [7]:
df['Offline_BackupSchemaIncrementType'].apply(pd.value_counts)

Out[7]:
   0.0  1.0  2.0
0  1.0  NaN  NaN
1  NaN  1.0  NaN
2  NaN  1.0  NaN
3  NaN  NaN  1.0
4  NaN  NaN  NaN

这里的值是列，索引与原始df

相同