替換字符串條紋

Question

計數

比方說，我有以下形式的字符串：

"000000111100011100001011000000001111" 

，我想創建一個包含1條紋的長列表：

[4, 3, 1, 2, 4] 

對此有沒有很好的一行？

Answer 1

如果你不介意的from itertools import groupby ...

>>> from itertools import groupby 
>>> [len(list(g)) for k, g in groupby(s) if k == '1'] 
[4, 3, 1, 2, 4] 

Answer 2

>>> s = "000000111100011100001011000000001111" 
>>> items = set(s) 
>>> counts = [s.count(x) for x in items] 
>>> counts 
[1, 1] 
>>> 

Answer 3

可與正則表達式來完成，雖然不是很優雅的迭代工具解決方案

answer = [len(item) for item in filter(None, re.split(r"[^1]+", test_string))] 

，或者更優雅：

answer = [len(item) for item in re.findall(r"1+", test_string)] 

and mo再優雅依舊（學分喬恩）：

answer = map(len, re.findall("1+", test_string)) 

Answer 4

>>> mystr = "000000111100011100001011000000001111" 
>>> [len(s) for s in re.split("0+", mystr) if s] 
[4, 3, 1, 2, 4] 

Answer 5

沒有正則表達式的需要，只是str.split

>>> mystr = "000000111100011100001011000000001111" 
>>> [len(s) for s in mystr.split('0') if s] 
[4, 3, 1, 2, 4]