正则表达式(\w)*匹配的问题

Question

import
s1 = 'a1b2c3d4e56'
n = re.findall('(\w)*', s1)
print(n)

请教大家，上面这段正则的结果为什么是：['6', '']

Answer 1

正则表达式组的概念

‘(‘’)’ 无命名组

最基本的组是由一对圆括号括起来的正则式。比如上面匹配包夹在字母中间的数字的例子中使用的 (\d+) 

findall 函数只返回了包含在 ’()’ 中的内容, 如果括号内的内容连续时，只返回匹配到的最后一个，

官方文档解释

re.findall(pattern, string, flags=0)
Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found. If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group. Empty matches are included in the result unless they touch the beginning of another match.

返回字符串中匹配规则的所有非重叠匹配，作为字符串列表。

Answer 2

参照如下，括号直接把\w给封闭了

import re
s1 = 'a1b2c3d4e56'
n = re.findall('(\w)*', s1)
print("n---->",n)
#  ||
#  ||
#  ||
# \||/
#  \/
m = re.findall('\w', s1)
for i in m:
    x = re.findall('\w*',i)
print("x---->",x)
print("m---->",m)

结果：

n----> ['6', '']
x----> ['6', '']
m----> ['a', '1', 'b', '2', 'c', '3', 'd', '4', 'e', '5', '6']

Answer 3

findall就是返回一个匹配的字符串列表，模式中包含括号，就是分组的概念;
返回'6'的原因@shanks，返回''的原因是最后的零个字符''也符合(\w)*