轉換Python字符串列出

Question

我這裏顯示格式化爲一個列表：

>>> x = "[26, 25]" 
>>> list(x) 
['[', '2', '6', ',', ' ', '2', '5', ']'] 

如何轉換與元素的列表如下所示：

>>> x 
[25, 26] 

Answer 1

使用ast.literal_eval()：

import ast 

ast.literal_eval(x) 

或將其視爲JSON，使用json.loads()：

import json 

json.loads(x) 

ast.literal_eval()以Python文字作爲輸入（因此Python源代碼中的文本會給你一個值），json.loads()需要JSON輸入。

演示：

>>> import ast, json 
>>> x = "[26, 25]" 
>>> ast.literal_eval(x) 
[26, 25] 
>>> json.loads(x) 
[26, 25] 

兩者之間的區別僅進場時，你有比你輸入字符串整數列表以上; JSON字符串包含unicode，Python 2上的ast.literal_eval()不能正確解釋，而JSON類型只是ast.literal_eval()支持的子集。

>>> x = '["List", "with", "strings", "including", "snowman", "\u2603"]' 
>>> ast.literal_eval(x) 
['List', 'with', 'strings', 'including', 'snowman', '\\u2603'] 
>>> json.loads(x) 
[u'List', u'with', u'strings', u'including', u'snowman', u'\u2603'] 
>>> x = "(1, 2, 3, 'a tuple is Python syntax, not JSON')" 
>>> ast.literal_eval(x) 
(1, 2, 3, 'a tuple is Python syntax, not JSON') 
>>> json.loads(x) 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
    File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python2.7/json/__init__.py", line 338, in loads 
    return _default_decoder.decode(s) 
    File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python2.7/json/decoder.py", line 365, in decode 
    obj, end = self.raw_decode(s, idx=_w(s, 0).end()) 
    File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python2.7/json/decoder.py", line 383, in raw_decode 
    raise ValueError("No JSON object could be decoded") 
ValueError: No JSON object could be decoded 

Answer 2

比「EVAL」更好的（從安全和速度（可能）的觀點）將是：

[int(v) for v in x[1:-1].replace(" ", "").split(",")] 

Answer 3

另一種選擇，如果你知道自己在做正確的事

x = "[26, 25]" 
y = eval(x) 
print y