[英]Nested list comprehension with two lists
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
它有 9 個數字的原因是因為 python 對待
[x + y for x in l2 for y in l1 ]
類似於
for x in l2: for y in l1: x + y
即,它是一個嵌套循環
列表推導等同於 for 循環。 因此, [x + y for x in l2 for y in l1 ]
將變為:
new_list = [] for x in l2: for y in l1: new_list.append(x + y)
而zip
返回包含每個列表中一個元素的元組。 因此[x + y for x,y in zip(l1,l2)]
等價於:
new_list = [] assert len(l1) == len(l2) for index in xrange(len(l1)): new_list.append(l1[index] + l2[index])
以上答案足以解決您的問題,但我想為您提供一個列表理解解決方案以供參考(因為那是您的初始代碼以及您想要理解的內容)。
假設兩個列表的長度相同,您可以這樣做:
[l1[i] + l2[i] for i in range(0, len(l1))]
[x + y for x in l2 for y in l1 ]
相當於:
lis = [] for x in l: for y in l1: lis.append(x+y)
因此,對於l
的每個元素,您一次又一次地迭代l2
,因為l
有 3 個元素,而l1
有元素,所以總循環等於 9( len(l)*len(l1)
)。
這個序列
res = [x + y for x in l2 for y in l1 ]
相當於
res =[] for x in l2: for y in l1: res.append(x+y)
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