1

我需要分析每个用户最后一次活跃的最后 60 天。

我的数据框包含每个用户（'DataSourceId'）处于活动状态的日期（'CalendarDate'）（'Activity' 一个整数） - 每个日期一行。我已经按 DataSourceId 对数据框进行了分组，所以我在列中有日期，并且我抓住了每个用户活动的最后一天“max_date”：

df['max_date'] = df.groupby('DataSourceId')['CalendarDate'].transform('max')

datetime64[ns]尽管 'CalendarDate' 和 'max_date' 实际上是格式（活动值是） ，但数据看起来像这样 float64：

ID    Jan1    Jan2    Jan3    Jan4    Jan5...  max_date
1               8              15      10        Jan5
2       2              13                        Jan3
3       6      11                                Jan2

现在，我想将日历日期中的列重新排列为每一行的“最后 x 天”。像这样：

ID    Last    Last-1    Last-2    Last-3  ...  Last-x
1      10       15                   8  
2      13                  2           
3      11        6

我还没有找到任何类似转换的例子，我真的被困在这里。

编辑： 在调整了 jezrael 的解决方案后，我注意到它在某些情况下失败了。

我认为问题与jezrael解决方案中的这段代码有关：r = data_wide.bfill().isna().sum(axis=1).values

示例：此数据失败（和r = [0 3]）：

CalendarDate                         2017-07-02 2017-07-03 2017-07-06 2017-07-07 2017-07-08 2017-07-09
DataSourceId                                                                                          
1000648                                     NaN     188.37     178.37        NaN     128.37      18.37
1004507                                   51.19        NaN      52.19      53.19        NaN        NaN

具体来说，重新对齐的数据框如下所示：

              Last-0  Last-1  Last-2  Last-3  Last-4  Last-5
DataSourceId                                                
1000648        18.37  128.37     NaN  178.37  188.37     NaN
1004507        52.19     NaN   51.19     NaN     NaN   53.19

如果我通过将 ID 1000648 更改为 1100648（使其成为第二行）来更改数据框中的顺序，则结果是（r = [0 2]）：

              Last-0  Last-1  Last-2  Last-3  Last-4  Last-5
DataSourceId                                                
1004507          NaN     NaN   53.19   52.19     NaN   51.19
1100648          NaN  178.37  188.37     NaN   18.37  128.37
4

3 回答 3

0

请尝试以下代码，如果这有帮助，请告诉我。

df = df.iloc[:,list(range(len(df.columns)-1,0,-1))]
print(df)
于 2018-09-13T10:48:36.157 回答
0

您可以使用此代码首先查找最后一个连续的空值，然后通过每个系列的计数移位，它将起作用。

df1 = df[df.columns.difference(['ID'])]
df1 = df1.apply(lambda x:x.shift(x[::-1].isnull().cumprod().sum())[::-1],axis=1)
df1.columns = ['Last-'+str(i) for i in range(df1.columns.shape[0])]
df1['ID'] = df['ID']

出去：

   Last-0   Last-1  Last-2  Last-3  Last-4  ID
0   10.0    15.0    NaN     8.0     NaN     1
1   13.0    NaN     2.0     NaN     NaN     2
2   11.0    6.0     NaN     NaN     NaN     3
于 2018-09-13T11:15:09.730 回答
0

如果性能很重要，请稍作改动numpy solution：

#select all columns without last
A = df.iloc[:, 1:-1].values
print (A)
[[nan  8. nan 15. 10.]
 [ 2. nan 13. nan nan]
 [ 6. 11. nan nan nan]]

#count NaNs values
r = df.bfill(axis=1).isna().sum(axis=1).values
#oldier pandas versions
#r = df.bfill(axis=1).isnull().sum(axis=1).values
#boost solution by https://stackoverflow.com/a/30428192
#r = A.shape[1] - (~np.isnan(A)).cumsum(axis=1).argmax(axis=1) - 1
print (r)
[0 2 3]

rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]

# Use always a negative shift, so that column_indices are valid.
# (could also use module operation)
r[r < 0] += A.shape[1]
column_indices = np.flip(column_indices - r[:,np.newaxis], axis=1)
print (column_indices)
[[ 4  3  2  1  0]
 [ 2  1  0 -1 -2]
 [ 1  0 -1 -2 -3]]

result = A[rows, column_indices]
#https://stackoverflow.com/a/51613442
#result = strided_indexing_roll(A,r)
print (result)
[[10. 15. nan  8. nan]
 [13. nan  2. nan nan]
 [11.  6. nan nan nan]]

c = [f'Last-{x}' for x in np.arange(result.shape[1])]
df1 = pd.DataFrame(result, columns=c)
df1.insert(0, 'ID', df['ID'])
print (df1)
   ID  Last-0  Last-1  Last-2  Last-3  Last-4
0   1    10.0    15.0     NaN     8.0     NaN
1   2    13.0     NaN     2.0     NaN     NaN
2   3    11.0     6.0     NaN     NaN     NaN

编辑：

如果ID是索引，则解决方案会有所更改-仅删除第一列.iloc[:, :-1]和最后使用DataFrame构造函数：

A = df.iloc[:, :-1].values
print (A)
[[nan  8. nan 15. 10.]
 [ 2. nan 13. nan nan]
 [ 6. 11. nan nan nan]]

r = df.bfill(axis=1).isna().sum(axis=1).values
print (r)
[0 2 3]

rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]

# Use always a negative shift, so that column_indices are valid.
# (could also use module operation)
r[r < 0] += A.shape[1]
column_indices = np.flip(column_indices - r[:,np.newaxis], axis=1)
print (column_indices)
[[ 4  3  2  1  0]
 [ 2  1  0 -1 -2]
 [ 1  0 -1 -2 -3]]

result = A[rows, column_indices]
print (result)
[[10. 15. nan  8. nan]
 [13. nan  2. nan nan]
 [11.  6. nan nan nan]]

c = [f'Last-{x}' for x in np.arange(result.shape[1])]
#use DataFrame constructor
df1 = pd.DataFrame(result, columns=c, index=df.index)
print (df1)
    Last-0  Last-1  Last-2  Last-3  Last-4
ID                                        
1     10.0    15.0     NaN     8.0     NaN
2     13.0     NaN     2.0     NaN     NaN
3     11.0     6.0     NaN     NaN     NaN
于 2018-09-13T11:33:49.383 回答