问题
I need to seek the "matchingpoint" of two list where List "a" is bigger than List "b". So far I've found this earlier post: Check if a list is part of another list while preserving the list sequence
That helps a lot. But I need to now, where the lists fit.
a = [3, 4, 1, 2, 4, 1, 5]
b = [4, 1, 2]
"b" fits in "a" but instead of a TRUE value I would like to have a[1] as first matchingpoint
回答1:
You could use next to fetch the first matching point:
a = [3, 4, 1, 2, 4, 1, 5]
b = [4, 1, 2]
starting_point = next((a[i] for i in range(len(a)) if b == a[i:i + len(b)]), -1)
print(starting_point)
Output
4
UPDATE
If you need both the index and the value at the index, return the index instead of the value, for example:
position = next((i for i in range(len(a)) if b == a[i:i + len(b)]), -1)
print("position", position)
print("starting point", a[position])
Output
position 1
starting point 4
Note the change, now is i instead of a[i]
来源：https://stackoverflow.com/questions/64907723/check-if-list-is-part-of-list-keeping-order-and-find-postion