python字符串中连续相同字符个数

由字符TF组成的字符串，如：'TTFTTTFFFFTFFTT'
能计算连续相同字符的个数，如返回：T2, F1, T3, F4, T1, F2, T2
另外求得出如下结果的方案['TT', 'F', 'TTT', 'FFFF', 'T', 'FF', 'TT']
请尽量pythonic

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相关建议 2019-12-04

1、打开cmd命令窗口，敲入python命令；

2、编写python代码，先引入itertools包；import itertools

3、再编写字符串计算代码，l = [(k, len(list(g))) for k, g in itertools.groupby('TTFTTTFFFFTFFTT')]

4、查看l的返回值；即为：[('T', 2), ('F', 1), ('T', 3), ('F', 4), ('T', 1), ('F', 2), ('T', 2)]

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其他看法

第1个回答推荐于2017-09-10

def countSeq(TF):
    result = []
    if TF is None or len(TF) == 0:
        return result
    pattern = TF[0]
    count = 1
    for s in TF[1:]:
        if s == pattern:
            count += 1
        else:
            result.append(pattern + str(count))
            pattern = s
            count = 1
    result.append(pattern + str(count))
    return result

def printSeq(TF):
    result = []
    if TF is None or len(TF) == 0:
        return result
    pattern = TF[0]
    result.append(pattern)
    for s in TF[1:]:
        if s == pattern:
            result[-1] += s
        else:
            pattern = s
            result.append(pattern)
    return result

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第2个回答 2018-08-28

import itertools
l = [(k, len(list(g))) for k, g in itertools.groupby('TTFTTTFFFFTFFTT')]
l:[('T', 2), ('F', 1), ('T', 3), ('F', 4), ('T', 1), ('F', 2), ('T', 2))]
不要重复造轮子

第3个回答 2015-10-07

a='TTFTTTFFFFTFFTT'
["%s=%d"%(x,a.count(x)) for x in set(a)]

第4个回答 2017-10-07

下面这个解法应该足够 pythonic 了：

# -*- coding: utf-8 -*-

def split_seq(s):
    import re
    # sep shouldn't conflict with any substring in s, so we make it a little weird ~
    sep = '<[|]>'
    separated_s = re.sub(r'(.)\1*', lambda m: m.group(0)+sep, s)
    return separated_s.split(sep)[:-1]

def count_seq(s):
    lst = split_seq(s)
    return [(s[0],len(s)) for s in lst]

s = 'TTFTTTFFFFTFFTT'
print(split_seq(s))
print(count_seq(s))

# ==================== the result ====================
# ['TT', 'F', 'TTT', 'FFFF', 'T', 'FF', 'TT']
# [('T', 2), ('F', 1), ('T', 3), ('F', 4), ('T', 1), ('F', 2), ('T', 2)]

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