value_counts()和apply(pd.value_counts)之间的区别

Question

以下字段包含三个值: incremental、full和differential以及value_counts()和apply(pd.value_counts)

为什么会有这样的差异？

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Answer 1

有问题，你需要将函数Series.value_counts应用到DataFrame的一个列中，所以请使用apply。

它与以下内容相同：

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，当您对方法执行apply时，您将对每个元素执行此操作，因此返回的系列将尝试将其与原始df对齐，实际上您将获得一个二维数组：

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