以下内容在Python 2中有效,但在3中无效。是否可以在Python 3中访问局部变量?还是这些情况的替代解决方案?
[('{name_var}_{i:02d}of{maxpg:02d}.{date_var}').format(i, **locals())
for i in range(start, end)]
Python 3中的错误
KeyError:'local_var'
下面是上面的一个更简单的玩具示例(在Python 2中有效,但在3中无效)
local_var = 'hello'
['{local_var}'.format(**locals()) for i in range(1)]
Python 3中的错误
KeyError:'local_var'
答案 0 :(得分:2)
如@ user2357112在评论中所解释,列表理解在Python 3中具有其自身的局部范围(因此是locals()
字典)。
比较:
>>> var=1
>>> [locals() for _ in range(1)]
[{'_': 0, '.0': <range_iterator object at 0x7f5b65cb7270>}]
使用
>>> [l for l in (locals(), )]
[{'__name__': '__main__', '__doc__': None, '__package__': None, '__loader__': <class '_frozen_importlib.BuiltinImporter'>, '__spec__': None, '__annotations__': {}, '__builtins__': <module 'builtins' (built-in)>, 'var': 1}]
在第一种情况下,函数locals
在列表理解代码内被调用,而在第二种情况下,函数调用的结果作为参数传递给列表理解。
dis
模块确认:
>>> from dis import dis
>>> def f(): return [locals() for _ in range(1)]
...
>>> dis(f)
1 0 LOAD_CONST 1 (<code object <listcomp> at 0x7fc8173bd9c0, file "<stdin>", line 1>)
2 LOAD_CONST 2 ('f.<locals>.<listcomp>')
4 MAKE_FUNCTION 0
6 LOAD_GLOBAL 0 (range)
8 LOAD_CONST 3 (1)
10 CALL_FUNCTION 1
12 GET_ITER
14 CALL_FUNCTION 1
16 RETURN_VALUE
未调用locals
函数。您会在列表理解代码中看到该调用:
>>> dis(f.__code__.co_consts[1])
1 0 BUILD_LIST 0
2 LOAD_FAST 0 (.0)
>> 4 FOR_ITER 10 (to 16)
6 STORE_FAST 1 (_)
8 LOAD_GLOBAL 0 (locals)
10 CALL_FUNCTION 0
12 LIST_APPEND 2
14 JUMP_ABSOLUTE 4
>> 16 RETURN_VALUE
而
>>> def g(): return [l for l in (locals(),)]
...
>>> dis(g)
1 0 LOAD_CONST 1 (<code object <listcomp> at 0x7f5b65cb8930, file "<stdin>", line 1>)
2 LOAD_CONST 2 ('g.<locals>.<listcomp>')
4 MAKE_FUNCTION 0
6 LOAD_GLOBAL 0 (locals)
8 CALL_FUNCTION 0
10 BUILD_TUPLE 1
12 GET_ITER
14 CALL_FUNCTION 1
16 RETURN_VALUE
locals
函数在列表理解执行之前被调用,迭代器被构建并传递给列表理解。
关于您的特定问题,您可以将locals
的求值强制超出列表理解范围(请注意i=i
:这不是位置参数):
>>> d = locals()
>>> ['{name_var}_{i:02d}of{maxpg:02d}.{date_var}'.format(i=i, **d) for i in range(start, end)]
['VAR_00of01.2019-01-01', 'VAR_01of01.2019-01-01', 'VAR_02of01.2019-01-01', 'VAR_03of01.2019-01-01', 'VAR_04of01.2019-01-01', 'VAR_05of01.2019-01-01', 'VAR_06of01.2019-01-01', 'VAR_07of01.2019-01-01', 'VAR_08of01.2019-01-01', 'VAR_09of01.2019-01-01']
如果您的Python版本是3.6或更高版本,则可以使用(f个字符串)[https://docs.python.org/3/whatsnew/3.6.html#pep-498-formatted-string-literals]
>>> [f'{name_var}_{i:02d}of{maxpg:02d}.{date_var}' for i in range(start, end)]
['VAR_00of01.2019-01-01', 'VAR_01of01.2019-01-01', 'VAR_02of01.2019-01-01', 'VAR_03of01.2019-01-01', 'VAR_04of01.2019-01-01', 'VAR_05of01.2019-01-01', 'VAR_06of01.2019-01-01', 'VAR_07of01.2019-01-01', 'VAR_08of01.2019-01-01', 'VAR_09of01.2019-01-01']
但是,我认为在每次迭代中在locals()
中进行查找不是一个好主意。您可以一次构建format_string
并将其用于列表理解:
>>> format_string = '{name_var}_{{i:02d}}of{maxpg:02d}.{date_var}'.format(**locals())
>>> format_string
'VAR_{i:02d}of01.2019-01-01'
或(> = 3.6):
>>> format_string = f'{name_var}_{{i:02d}}of{maxpg:02d}.{date_var}'
然后您拥有:
>>> [format_string.format(i=i) for i in range(start, end)]
['VAR_00of01.2019-01-01', 'VAR_01of01.2019-01-01', 'VAR_02of01.2019-01-01', 'VAR_03of01.2019-01-01', 'VAR_04of01.2019-01-01', 'VAR_05of01.2019-01-01', 'VAR_06of01.2019-01-01', 'VAR_07of01.2019-01-01', 'VAR_08of01.2019-01-01', 'VAR_09of01.2019-01-01']