对列表中的所有元素进行平方
问题描述
我被告知
编写一个函数 square(a),它接受一个数字数组 a 并返回一个包含平方的每个值的数组.
Write a function, square(a), that takes an array, a, of numbers and returns an array containing each of the values of a squared.
一开始我有
def square(a):
for i in a: print i**2
但这不起作用,因为我正在打印,并且没有像我被问到的那样返回.所以我尝试了
But this does not work since I'm printing, and not returning like I was asked. So I tried
def square(a):
for i in a: return i**2
但这只是我数组的最后一个数字的平方.我怎样才能让它对整个列表进行平方?
But this only squares the last number of my array. How can I get it to square the whole list?
解决方案
你可以使用列表推导:
def square(list):
return [i ** 2 for i in list]
或者你可以map
它:
def square(list):
return map(lambda x: x ** 2, list)
或者您可以使用生成器.它不会返回列表,但您仍然可以遍历它,并且由于您不必分配整个新列表,因此它可能比其他选项更节省空间:
Or you could use a generator. It won't return a list, but you can still iterate through it, and since you don't have to allocate an entire new list, it is possibly more space-efficient than the other options:
def square(list):
for i in list:
yield i ** 2
或者您可以执行无聊的旧 for
-循环,尽管这不像某些 Python 程序员所希望的那样惯用:
Or you can do the boring old for
-loop, though this is not as idiomatic as some Python programmers would prefer:
def square(list):
ret = []
for i in list:
ret.append(i ** 2)
return ret
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