python3.5 如何用map做出和zip同樣的效果?
問題描述
如下面這段代碼,我想把map和zip做出同樣的效果
name=[’a’,’b’,’c’]age=[10,11,12]nation=[’中國’,’にほん’,’Deutsch’]U1=list(zip(name,age,nation))print(U1)U2=map(None,name,age,nation)print(list(U2))
可是顯示:
[(’a’, 10, ’中國’), (’b’, 11, ’にほん’), (’c’, 12, ’Deutsch’)]Traceback (most recent call last): File 'F:/python/PT/program/nine neijian3.py', line 8, in <module> print(list(U2))TypeError: ’NoneType’ object is not callable
但是我去掉map里面的None:
U2=map(name,age,nation)print(list(U2))
顯示:
print(list(U2))TypeError: ’list’ object is not callable`
請各位大神賜教。
問題解答
回答1:map(lambda a,b,c: (a,b,c), name, age, nation)
回答2:name=[’a’,’b’,’c’]age=[10,11,12]nation=[’中國’,’にほん’,’Deutsch’]U1=list(zip(name,age,nation))print(U1)U2 = map(lambda a,b,c: (a,b,c), name, age, nation)print(list(U2))
第一個報錯NoneType ,是用于None object,所以不能輸出很正常。NoneType is the type for the None object, which is an object that indicates no value. You cannot add it to strings or other objects.
Python map()方法好像不是你這么用的,據我了解是應該是這個樣子的。描述很簡單,第一個參數接收一個函數名,第二個參數接收一個可迭代對象。語法map(f, iterable)基本上等于:[f(x) for x in iterable]實例
>>> def add100(x):... return x+100... >>> hh = [11,22,33]>>> map(add100,hh)[111, 122, 133]
http://stackoverflow.com/ques...
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