如何替换与正则表达式模式匹配的字符串？

Question

i have list with numbers. I want to extract all the numbers with consecutive digits like 444,888,1111 etc.. My following python code with regex is working exactly the way I wanted.

import re 
numbers = ['44444', '123', '444', '454', '76587', '888', '9090', '1111', '321', '4321', '4563', '3333', '543', '765', '4567', '555', '99999', '11211','11']
conList = [] 

for num in numbers:
  if re.search(r'^(\d)\1+$',num):
    conList.append(num)

print('conList :',conList) 

Result:

conList : ['44444', '444', '888', '1111', '3333', '555', '99999', '11']

Now I am trying to achieve the same result without using a regex pattern. How can I replace that regex pattern with some python code? (I am trying this only for educational purpose. I will post my answer if I got one.)

Answer 1

This is my attempt to solve this problem.

for num in numbers:
  if len(set(num)) == 1 and len(num) >1:
    conList.append(num)
print('conList :',conList)

---

Or, using list comprehension

conList = [ num for num in numbers if len(num) > 1 and len(set(num)) == 1 ]
print('conList : %s', conList)

---

Or, using filter

conList = filter(lambda x: len(x) > 1 and len(set(x)) == 1, numbers)

Answer 2

使用pythonic和快速的列表理解方法：

[i for i in numbers if len(set(i)) == 1]

Answer 3

You can do this in one line using list comprehensions as the following:

numbers = ['44444', '123', '444', '454', '76587', '888', '9090', '1111', '321', '4321', '4563', '3333', '543', '765', '4567',
     '555', '99999', '11211', '11']
conList = [x for x in numbers if x.count(x[0]) == len(x)]
print conList

output:

['44444', '444', '888', '1111', '3333', '555', '99999', '11']