【每日一坑 3】 找数字

crossin先生

FERRYMAN 发表于 2021-2-11 09:46
老师，看了答案之后我发现正则表达式是'[\d|.]+'

但是‘|’不是‘or’ 的意思吗？‘【】’不是表示里面的 ...

这里错了，后面楼层里有解释
https://bbs.crossincode.com/foru ... =2301&fromuid=3

不考虑小数点的话，直接 \d+ 就可以

FERRYMAN

crossin先生 发表于 2021-2-11 15:17
这里错了，后面楼层里有解释
https://bbs.crossincode.com/forum.php?mod=redirect&goto=findpost&ptid=8 ...

好的，谢谢老师

ZZZZZ

1. text = “aAsmr3idd4bgs7Dlsf9eAF”
   
2. b = “”
   
3. c = “”
   
4. for a in text:
   
5.     if a in ['0','1','2','3','4','5','6','7','8','9']:
   
6.         b += a
   
7.     else:
   
8.         c += str(a)
   
9. print(b,c)
   
10.         

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ZZZZZ

text = input()
new_text = []
for i in text:
  if i in list(range(0, 9)):
    new_text.append(i)
for i in new_text:
  print(i)

chinaso1111

text = "aAsmr3idd4bgs7Dlsf9eAF"
num = ''
for i in range(len(text)):
        if text[i].isdigit():
                print("'{}'".format(text[i]),end="")

chinaso1111

1. for i in str1:
   
2.     if i.isdigit() == True:
   
3.         print(i,end="")

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if i.isdigit() == True:
print(i)

Danieldepython

1. import re
   
2. text = "aAsmr3idd4bgs7Dlsf9eAF"
   
3. text_new = re.findall("\d",text)

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xiaolx29

1. from functools import reduce
   
2. 
   
3. text = "aAsmr3idd4bgs7Dlsf9eAF"
   
4. text2 = reduce(lambda a, b: ''.join((a, b)), (map(lambda i: i if i.isdigit() else '', text)))

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我直接上 mapreduce 大法了