字符串python中重复字符(包括非重复字符)的计数

Question

我需要找到按字母组重复字符的计数例如，如果我有一个字符串，s = "hggdsaajhjhajadj"，那么我需要计数为

h -, g - 2 d 1, s - 1, a, h - j - 1,等等

而不是{a: 4, ' d ': 2,"g":2,"h":3,"j":4,"s":1}

下面的代码按字母给出了计数。

s = "hggdsaajhjhajadj"
def find_repeated(string):
    table = {}
    for char in string.lower():
        if char in table:
            table[char] += 1
        elif char != " ":
            table[char] = 1
        else:
            table[char] = 0
    return table
print find_repeated(s)

{a: 4, ' d ': 2,"g":2,"h":3,"j":4,"s":1}

如果我尝试使用以下命令，

 for c in sorted(set(s)):
       i = 1; 
       while c * i in s:
           i += 1
       print c, "-", i - 1

然后，我得到以下内容:

a - 2 d - 1 g - 2 h - 1 j - 1 s - 1

你能给我一些建议吗?

Answer 1

Python处理连续组的工具是itertools.groupby:

>>> from itertools import groupby
>>> s = "hggdsaajhjhajadj"
>>> [(k, len(list(g))) for k,g in groupby(s)]
[('h', 1), ('g', 2), ('d', 1), ('s', 1), ('a', 2), ('j', 1), ('h', 1), ('j', 1), ('h', 1), ('a', 1), ('j', 1), ('a', 1), ('d', 1), ('j', 1)]

groupby返回一个对象，如果你遍历它，你会得到键和一个遍历组元素的迭代器:

>>> grouped = groupby(s)
>>> for key, group in grouped:
...     print(key, list(group))
...     
h ['h']
g ['g', 'g']
d ['d']
s ['s']
a ['a', 'a']
j ['j']
h ['h']
j ['j']
h ['h']
a ['a']
j ['j']
a ['a']
d ['d']
j ['j']

Answer 2

下面的函数执行您指定的操作:

def mycount(s):
        i = 0
        res = []
        while i<len(s):
                j = i+1
                while j<len(s) and s[i] == s[j]:
                        j += 1
                res.append( (s[i],j-i) )
                i = j
        return res