[英]Python: why does this code fail? assigning variable to list of itself in for loop
a = 'hello'
b = None
c = None
for x in [a, b, c]:
if isinstance(x, (basestring, dict, int, float)) or x is None:
x = [x]
a
返回“你好”,但預期['你好']。 但是,這有效:
a = 'hello'
a = [a]
a
返回['你好']
要實現這一點,首先你要了解你有兩個不同的結果,a和x(對於每個元素),以及列表[a,b,c]的引用,僅在for循環中使用,而且永遠不會更多。
為了實現目標,您可以這樣做:
a = 'hello'
b = None
c = None
lst = [a, b, c] #here I create a reference for a list containing the three references above
for i,x in enumerate(lst):
if isinstance(x, (str,dict, int, float)) or x is None: #here I used str instead basestring
lst[i] = [x]
print(lst[0]) #here I print the reference inside the list, that's why the value is now showed modified
['你好']
但正如我所說,如果你打印(a)它會再次顯示:
'你好'#here我打印了變量a的值,沒有修改
因為你從來沒有做過任何事。
看一下這個問題,了解更多關於引用的信息如何通過引用傳遞變量?
讓我們一次完成這一行;
a = 'hello' # assigns 'hello' to a.
b = None # etc...
c = None
for x in [a, b, c]: # Creates list from the values in 'a' 'b' and 'c', then iterates over them.
# This if statement is always True. x is always either a string, or None.
if isinstance(x, (basestring, dict, int, float)) or x is None:
x = [x] # On the first iteration, sets variable to the expected ['hello']
# After that though - it replaces it with [None]
a # = 'hello' - you have not assigned anything to the variable except on the first line.
唯一設置為['hello']
變量是x
,用None
快速覆蓋。 如果您將if檢查更改為exclude or x is None
並分配給a
而不是x
,則可以獲得所需的結果。
值得注意的是,當你啟動for循環時會創建列表[a, b, c]
。 改變a
b
或c
在for循環將沒有任何效果-這樣的名單已經產生。
其他答案很棒。 我認為,概括來說,list元素是不可變的/可散列的,因此for循環返回一個副本,而不是對原始對象的引用。 我應該發現這一點,但教我編碼太長時間不睡覺!
我最終使用了這個:
def lst(x):
if isinstance(x, (basestring, dict, int, float)) or x is None:
x = [x]
return x
[a, b, c] = map(lst, [a, b, c])
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