创建运行时长度的python列表的pythonic方法

Question

The f=[] in the below code seems a waste of a line, but I don't know how to get around it.

1)

f=[]
for x in X:
   f.append(foo(x))

2)

f=[]
[f.append(foo(x)) for x in X]

I was just wondering what the most "pythonic" way to do this is. The f=[] line seems unpythonic.

Answer 1

您最好阅读有关python中的列表推导的信息

f = [foo(x) for x in X]

Answer 2

f = [None]*len(X)

creates a list of None elements in the same size as X, and is O(n)

f = list(X) 

copies X to f and has the same time complexity as the above according to https://wiki.python.org/moin/TimeComplexity

If you meant to specifically create a list where every element is foo(x), then yeah:

f = [foo(x) for x in X]