将列表列表的第一个元素与另一个列表进行比较，然后仅添加与新列表匹配的列表

Question

My aim is to compare the 1st element of a list of lists with another list and have the lists added to a new list if the 1st elements of both lists match. So for instance,

list1 = [2,4,7,10]
list2 = [[2,5],[3,7],[1,6],[10,3]]

newlist = [[2,5],[10,3]]

Answer 1

I would use a list comprehension:

[snd for fst, snd in zip(list1, list2) if fst == snd[0]]

This outputs:

[[2, 5], [10, 3]]

Answer 2

You can use zip() :

newlist = []

for a, b in zip(list1, list2):
    if a == b[1]:
        newlist.append(b)

Or with a generator expression:

newlist = [b for a, b in zip(list1, list2) if a == b[1]]

Answer 3

What if the list1 items does NOT have exactly sequence matching list2?

Let's assume list1 has [2, 7, 10, 4], and list2 is the same as posted sample. Using pure zip won't get what you expect!

Alternatively, you could try this approach: (it works w/o assuming there is a matching ordering relationship between the two lists)

L = [2, 7, 4, 10]
M = [[10, 1], [3, 5], [2, 8], [4, 6], [10, 10]]

result = []

for ll in M:
    if ll[0] in set(L):
        result.append(ll)
result.append(ll)

print(result)
# [[10, 1], [2, 8], [4, 6], [10, 10]]