Question

日期采用日期时间格式周末的天数必须相同，其余的按数字顺序向下移动。 df1

   Date         Number
0. 12-10-2018   1
1. 13-10-2018   2 
2. 14-10-2018   3
3. 15-10-2018   4
4. 16-10-2018   5
5. 17-10-2018   6
6. 18-10-2018   7
7. 18-10-2018   8
8. 19-10-2018   9

我的代码：

   i = 1
   df1['Number'][0] == 1
   for i range(len(df1[0])):
       if df1[0[i].weekday()== 4:

           df1['Number'].iloc[i+1] = df1['Number'].iloc[i]
       elif df1[0][i].weekday() == 5:
           df1['Number'].iloc[i+1] = df1['Number'].iloc[i]

这给了我输出我的电话号码的信息：

df1

   Date         Number
0. 12-10-2018   1
1. 13-10-2018   1 
2. 14-10-2018   1
3. 15-10-2018   4
4. 16-10-2018   5
5. 17-10-2018   6
6. 18-10-2018   7
7. 19-10-2018   8
8. 20-10-2018   8

我需要输出为：

df1

   Date         Number
0. 12-10-2018   1
1. 13-10-2018   1 
2. 14-10-2018   1
3. 15-10-2018   2
4. 16-10-2018   3
5. 17-10-2018   4
6. 18-10-2018   5
7. 19-10-2018   6
8. 20-10-2018   6

非常感谢

Answer 1

我们可以通过factorize

修复您的输出

df.Number=df.Number.factorize()[0]+1
df
           Date  Number
0.0  12-10-2018       1
1.0  13-10-2018       1
2.0  14-10-2018       1
3.0  15-10-2018       2
4.0  16-10-2018       3
5.0  17-10-2018       4
6.0  18-10-2018       5
7.0  19-10-2018       6
8.0  20-10-2018       6

重新排列周末后将列移位以具有相同的i并遵循数字序列python

1 个答案: