新人，第一次玩这个，问问大牛，这个有没更简洁的代码？

比如：201812081000、201812080001、201812080111

要分别修改成
201812081000、20181208001、20181208111


日期搞定位数，后面的低于4位，就格式化成3位，不够补0，如果等于4位，就保持


下面是我写的代码，第一次玩，感觉写的很乱

1. number = str('201812080200')
   
2. numlist = list(number)
   
3. numberlist = [0 for i in range(8)]
   
4. for i in range(8):
   
5.         numberlist[i] = numlist[i]
   
6. numberA = "".join(numberlist)
   
7. numberA = str(numberA)
   
8. numberlist = [0 for i in range(4)]
   
9. for i in range(4):
   
10.         numberlist[i] = numlist[i+8]
    
11. numberB = "".join(numberlist)
    
12. numberB = str(int(numberB))
    
13. if len(numberB) < 4:
    
14.         numberB = str(numberB.zfill(3))
    
15. number = str('%s%s'%(numberA, numberB))
    
16. print(number)

复制代码

number = str('201812080200')

numberA = number[:-4]
numberB = number[-4:].lstrip('0').zfill(3)

number = numberA + numberB

uitb 发表于 2018-12-9 10:52
number = str('201812080200')

numberA = number[:-4]

谢谢大神，已经学会了

1. drawIssue = "2018120900002"
   
2. drawIssue = '%s%s'%(drawIssue[0:8],str(int(drawIssue[8:])).zfill(3))
   
3. print(drawIssue)

复制代码