Nested list comprehension with two lists

2020-02-08 04:05发布

问题:

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:

[101,202,303]

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

回答1:

The reason it has 9 numbers is because python treats

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

similarly to

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

ie, it is a nested loop



回答2:

List comprehensions are equivalent to for-loops. Therefore, [x + y for x in l2 for y in l1 ] would become:

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

Whereas zip returns tuples containing one element from each list. Therefore [x + y for x,y in zip(l1,l2)] is equivalent to:

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


回答3:

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).

Assuming the length of both lists are the same, you could do:

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


回答4:

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

is equivalent to :

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

So for every element of l you're iterating l2 again and again, as l has 3 elements and l1 has elements so total loops equal 9(len(l)*len(l1)).



回答5:

this sequence

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

is equivalent to

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