从python中的19个列表中制作一个列表

Question

I have 19 big lists of numbers in python and want to make a new list from all of them. so at the end I would have one list. I made the following code for two lists and works perfectly:

a2 = [x for x in l2 if x+3 in l1] 

this code looks for the elements in two lists only if the difference between them is 3. which means it finds any element in l1 if there is an element in l2 which is 3 units bigger than that element in l1 and the new list would have the mentioned elements of l1

for example if l1 = [1,2,3,6,5,7,8,9] and l2 = [4, 6,10,11,13,14]

new = [1,3,7,8]

but I don't know how to include the rest of lists. so, instead of looking for the elements in l2 , it looks for the "3 units bigger elements than l1 " in l 2 .... l19 and returns the 3 units smaller element in l1 (the the new list). do you know how to do that in a single line?

Answer 1

Put all of your other lists into a larger list and combine into one set using set.union :

superlist = [l3, ..., l19]
s = set(l2).union(*superlist)
a2 = [x for x in l1 if x+3 in s]

You can, of course, decide to pass the lists directly to set.union .

Membership lookup with sets - O(1) complexity - is way faster than that for lists - O(n).

Answer 2

a2 = [x for x in l if x+3 in l1+l2+l3+l4+l5+...+l19]

For the sake of completeness, here also another one-liner inspired by the other answers you can use in case of more than 19 lists:

a2=[x for x in l if x+3 in set().union(*[eval('l'+str(i+1)) for i in range(19)])]

but ...

ATTENTION

such kind of one-liner look very nice, but are deadly in case of really large lists ( they can unnecessary take all of your CPU-power for extremely looooong time ), so you will be glad to use a multi-liner instead:

bigL = l1+l2+l3+l4+l5+...+l19
setBigL = set(bigL)
a2 = [x for x in l if x+3 in setBigL]

For the sake of completeness here how to get bigL without the need of writing all the list names:

noOfLists = 19; strListAddition = ''
for i in range(19): strListAddition+='l'+str(i+1)+'+'
bigL = eval(strListAddition[0:-1])

Answer 3

If I don't misunderstand your meaning, you can try this, and if you hava 19 list, just change the range part:

l1 = [4, 6,10,11,13,14]
l2 = [1,2,3,6,5,7,8,9]
l3=[3,7,10,11]

print([x for i in range(2,4) for x in eval('l'+str(i)) if x+3 in l1])

Output:

[1, 3, 7, 8, 3, 7, 10, 11]