Nested list comprehension with two lists

Question

I understand how the simple list comprehension works eg.:

[x*2 for x in range(5)] # returns [0,2,4,6,8]

and also I understand how the nested list comprehesion works:

w_list = ["i_have_a_doubt", "with_the","nested_lists_comprehensions"]

# returns the list of strings without underscore and capitalized
print [replaced.title() for replaced in [el.replace("_"," ")for el in w_list]]

so, when I tried do this

l1 = [100,200,300]
l2 = [0,1,2]
[x + y for x in l2 for y in l1 ]

I expected this:

[100,201,302]

but I got this:

[100,200,300,101,201,301,102,202,302]

so I got a better way solve the problem, which gave me what I want

[x + y for x,y in zip(l1,l2)]

but I didn't understood the return of 9 elements on the first code

Answer 1

它有 9 个数字的原因是因为 python 对待

[x + y for x in l2 for y in l1 ]

类似于

for x in l2: for y in l1: x + y

即，它是一个嵌套循环

Answer 2

列表推导等同于 for 循环。 因此， [x + y for x in l2 for y in l1 ]将变为：

 new_list = [] for x in l2: for y in l1: new_list.append(x + y)

而zip返回包含每个列表中一个元素的元组。 因此[x + y for x,y in zip(l1,l2)]等价于：

 new_list = [] assert len(l1) == len(l2) for index in xrange(len(l1)): new_list.append(l1[index] + l2[index])

Answer 3

以上答案足以解决您的问题，但我想为您提供一个列表理解解决方案以供参考（因为那是您的初始代码以及您想要理解的内容）。

假设两个列表的长度相同，您可以这样做：

 [l1[i] + l2[i] for i in range(0, len(l1))]

Answer 4

[x + y for x in l2 for y in l1 ]

相当于：

 lis = [] for x in l: for y in l1: lis.append(x+y)

因此，对于l的每个元素，您一次又一次地迭代l2 ，因为l有 3 个元素，而l1有元素，所以总循环等于 9( len(l)*len(l1) )。

Answer 5

这个序列

res = [x + y for x in l2 for y in l1 ]

相当于

res =[] for x in l2: for y in l1: res.append(x+y)