Python 带有两个列表的嵌套列表理解
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Nested list comprehension with two lists
提问by André Caldas
I understand how the simple list comprehension works eg.:
我了解简单列表理解的工作原理，例如：
[x*2 for x in range(5)] # returns [0,2,4,6,8]
and also I understand how the nested list comprehesion works:
而且我理解嵌套列表理解是如何工作的：
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
所以，当我尝试这样做时
l1 = [100,200,300]
l2 = [0,1,2]
[x + y for x in l2 for y in l1 ]
I expected this:
我期待这个：
[101,202,303]
but I got this:
但我得到了这个：
[100,200,300,101,201,301,102,202,302]
so I got a better way solve the problem, which gave me what I want
所以我找到了解决问题的更好方法，这给了我我想要的
[x + y for x,y in zip(l1,l2)]
but I didn't understood the return of 9 elements on the first code
但我没有理解第一个代码上9个元素的返回
采纳答案by rspencer
The reason it has 9 numbers is because python treats
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 ]
similarly to
类似于
for x in l2:
for y in l1:
x + y
ie, it is a nested loop
即，它是一个嵌套循环
回答by Ashwini Chaudhary
[x + y for x in l2 for y in l1 ]
is equivalent to :
相当于：
lis = []
for x in l:
for y in l1:
lis.append(x+y)
So for every element of lyou're iterating l2again and again, as lhas 3 elements and l1has elements so total loops equal 9(len(l)*len(l1)).
因此，对于l您l2一次又一次地迭代的每个元素，因为l有 3 个元素并且l1有元素所以总循环等于 9( len(l)*len(l1))。
回答by Wesley Baugh
List comprehensions are equivalent to for-loops. Therefore, [x + y for x in l2 for y in l1 ]would become:
列表推导式等效于 for 循环。因此，[x + y for x in l2 for y in l1 ]会变成：
new_list = []
for x in l2:
for y in l1:
new_list.append(x + y)
Whereas zipreturns tuples containing one element from each list. Therefore [x + y for x,y in zip(l1,l2)]is equivalent to:
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)]相当于：
new_list = []
assert len(l1) == len(l2)
for index in xrange(len(l1)):
new_list.append(l1[index] + l2[index])
回答by oleg
this sequence
这个序列
res = [x + y for x in l2 for y in l1 ]
res = [x + y for x in l2 for y in l1 ]
is equivalent to
相当于
res =[]
for x in l2:
for y in l1:
res.append(x+y)
回答by Sticky
The above answers will suffice for your question but I wanted to provide you with a list comprehension solution for reference (seeing as that was your initial code and what you're trying to understand).
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（因为这是您的初始代码以及您试图理解的内容）。
Assuming the length of both lists are the same, you could do:
假设两个列表的长度相同，你可以这样做：
[l1[i] + l2[i] for i in range(0, len(l1))]
