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                ## 概述 lambda 是一個表達式,而不是一個語句。 `lambda args:expression` ## 嵌套作用域和lambda ~~~python def func(): x = 4 action = (lambda n, y=x: y ** n) return action x = func() print(x(2)) --> 16 ~~~ 參考之前的作用域介紹,需要使用默認參數**從上層作用域將值傳遞給**lambda ```python def func(): x = 4 action = (lambda n, y=x: y ** n) return action x = func() print(x(2)) --> 16 ``` ## 作用域與帶有循環變量的默認參數相比較 >[danger]特例 如果lambda或者def在函數中定義,嵌套在一個循環之中,并且嵌套的函數引用了上層作用域的變量,該變量被循環所改變,所有這個循環中產生的函數將有相同的值,即最后一個循環完成時被引用變量的值 ```python def makeActions(): acts = [] for i in range(5): acts.append(lambda x: i ** x) return acts ``` 分析,i從0迭代到4,最終i = 4 測試結果 ``` acts = makeActions() acts[0](2) acts[2](2) acts[4](2) 都等于16 ``` 為了讓這類代碼能夠工作,必須使用默認參數把當前值傳遞給嵌套作用域中的變量,@p432 ```python def makeActions(): acts = [] for i in range(5): acts.append(lambda x, y=i: y ** x) return acts ``` 測試結果 ``` acts = makeActions() acts[0](2) --> 0 acts[2](2) --> 4 acts[4](2) --> 16 ```
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