Question

我有一个数据集，其中包含客户的ID以及名为“ WEEK1”，“ WEEK2”等的指标。如果在该特定星期登记了客户，则值= 1，否则为0，如下所示：

ID WEEK1 WEEK2 WEEK3 WEEK4 WEEK5
1   0     0     1     0     1
2   0     0     0     0     1
3   1     0     1     0     1
4   0     0     0     0     0
5   1     1     1     1     1
6   1     0     0     0     0
7   0     1     1     1     0

我想做的是搜索客户注册的第一周，保持该周的指标= 1，然后将该客户ID的所有其他周指标值更改为0，即O / P：-

ID WEEK1 WEEK2 WEEK3 WEEK4 WEEK5
1   0     0     1     0     0  ## WEEK5 is changed to 0 here
2   0     0     0     0     1  ## nothing changed
3   1     0     0     0     0  ## WEEK3 and WEEK5 is changed to 0
4   0     0     0     0     0
5   1     0     0     0     0
6   1     0     0     0     0
7   0     1     0     0     0

因此，对于每个客户ID，我们找到第一个WEEK，其值= 1，然后将所有下一个WEEK值= 0。

现在我已经使用if-else尝试过，将每个条件一个一个地放置，如下所示：

if df['WEEK1'] == 1:
    df['WEEK2'] = 0
    df['WEEK3'] = 0
    df['WEEK4'] = 0
    df['WEEK5'] = 0
elif df['WEEK2'] == 1:
    df['WEEK3'] = 0
    df['WEEK4'] = 0
    df['WEEK5'] = 0
... and so on

当只有5个WEEK列时，使用if-else对我有用，但是现在我获得了52个WEEK列的数据，除了使用if-else之外，我找不到其他选择。

因此，任何适用于在这5列上强加层次结构并且还可以扩展到可变数量的列（如52、104等）的东西都会很有帮助。

Answer 1

使用：

#if first column is not index
df = df.set_index('ID')
df = df.where(df.shift(axis=1).eq(1).cumsum(axis=1).eq(0), 0)
print (df)
    WEEK1  WEEK2  WEEK3  WEEK4  WEEK5
ID                                   
1       0      0      1      0      0
2       0      0      0      0      1
3       1      0      0      0      0
4       0      0      0      0      0
5       1      0      0      0      0
6       1      0      0      0      0
7       0      1      0      0      0

详细信息和说明：

第一个DataFrame.shift右边的值：

print (df.shift(axis=1))
    WEEK1  WEEK2  WEEK3  WEEK4  WEEK5
ID                                   
1     NaN    0.0    0.0    1.0    0.0
2     NaN    0.0    0.0    0.0    0.0
3     NaN    1.0    0.0    1.0    0.0
4     NaN    0.0    0.0    0.0    0.0
5     NaN    1.0    1.0    1.0    1.0
6     NaN    1.0    0.0    0.0    0.0
7     NaN    0.0    1.0    1.0    1.0

如果可能，用1比较另一个值，例如1或0，否则省略此步骤：

print (df.shift(axis=1).eq(1))
    WEEK1  WEEK2  WEEK3  WEEK4  WEEK5
ID                                   
1   False  False  False   True  False
2   False  False  False  False  False
3   False   True  False   True  False
4   False  False  False  False  False
5   False   True   True   True   True
6   False   True  False  False  False
7   False  False   True   True   True

通过DataFrame.cumsum获取每行的累积总和：

print (df.shift(axis=1).eq(1).cumsum(axis=1))
    WEEK1  WEEK2  WEEK3  WEEK4  WEEK5
ID                                   
1       0      0      0      1      1
2       0      0      0      0      0
3       0      1      1      2      2
4       0      0      0      0      0
5       0      1      2      3      4
6       0      1      1      1      1
7       0      0      1      2      3

按0比较：

print (df.shift(axis=1).eq(1).cumsum(axis=1).eq(0))
    WEEK1  WEEK2  WEEK3  WEEK4  WEEK5
ID                                   
1    True   True   True  False  False
2    True   True   True   True   True
3    True  False  False  False  False
4    True   True   True   True   True
5    True  False  False  False  False
6    True  False  False  False  False
7    True   True  False  False  False

上次由掩码False设置为DataFrame.where到0的值：

print (df.where(df.shift(axis=1).eq(1).cumsum(axis=1).eq(0), 0))
    WEEK1  WEEK2  WEEK3  WEEK4  WEEK5
ID                                   
1       0      0      1      0      0
2       0      0      0      0      1
3       1      0      0      0      0
4       0      0      0      0      0
5       1      0      0      0      0
6       1      0      0      0      0
7       0      1      0      0      0

在多列上强加层次结构，基于其他列更改列值的随机数

1 个答案: