字符串分割到數組與許多焦炭親物品

Question

我有一個字符串，我想STR的例如數組：

"hello world" 
["hel", "lo ", "wor", "ld"] 

或

["hell", "o wo", "rld"] 

我看到list(message)將是確定的，但只是爲

["h", "e", "l", "l", "o", " ", "w", "o", "r", "l", "d", ] 

任何想法？

Answer 1

>>> s = 'hello world' 
>>> [s[i:i+3] for i in range(len(s)) if not i % 3] 
['hel', 'lo ', 'wor', 'ld'] 

對於更通用的解決方案（即自定義的分割），試試這個功能：

def split_on_parts(s, *parts): 
    total = 0 
    buildstr = [] 
    for p in parts: 
     buildstr.append(s[total:total+p]) 
     total += p 
    return buildstr 

s = 'hello world' 
print split_on_parts(s, 3, 3, 3, 3) 
print split_on_parts(s, 4, 3, 4) 

它產生的輸出：

['hel', 'lo ', 'wor', 'ld'] 
['hell', 'o w', 'orld'] 

OR如果你真的在一個班輪的心情：

def split_on_parts(s, *parts): 
    return [s[sum(parts[:p]):sum(parts[:p+1])] for p in range(len(parts))] 

Answer 2

>>> lst = ['he', 'llo', ' wo', 'rld'] 
>>> ''.join(lst) 
'hello world' 
>>> s = 'hello world' 
>>> list(s) 
['h', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd'] 

這些是基礎;如果您有任何具體要求，請對此帖發表評論，我會更新我的答案。

Answer 3

>>> def split_length(s, l): 
...  return [s[i:i+l] for i in range(0, len(s), l)] 
... 
>>> split_length("hello world", 3) 
['hel', 'lo ', 'wor', 'ld'] 
>>> split_length("hello world", 4) 
['hell', 'o wo', 'rld'] 

Answer 4

>>> s 
'hello world' 
>>> filter(lambda n: n != ' ', re.split("(o|wo)", s)) 
['hell', 'o', 'wo', 'rld'] 
>>> filter(lambda n: n != ' ', re.split("(lo|wor)", s)) 
['hel', 'lo', 'wor', 'ld'] 

不確定它是如何（按什麼標準）準確地被分開。

Answer 5

`list` is a python key word. You can use list and indexing power of list to manipulate your result. 

In [5]: s = 'hello world' 

In [6]: s.split() 
Out[6]: ['hello', 'world'] 

In [7]: list(s) 
Out[7]: ['h', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd']