列表中的所有列表是否相等

Question

Is there a preferred general solution for testing equality of a list of lists. I'm attempting the apply this example more generally to a list of lists. I currently have the following solution, where _list is a generic list of unknown length/number of elements.

all(x == y for x, y in zip(_list, _list[1:]))

or if the order doesn't matter.

all([sorted(x) == sorted(y) for x, y in zip(_list, _list[1:])])

Beyond checking neighbouring lists for equality, how can/should this approach my improved?

Answer 1

I believe, simple and fast enough way would be

all(a[0] == ax for ax in a)

You don't need to compare all elements pairwise, short route is to compare 1 vs all others. And in reality, better to compare against all ( ax in a ) instead of "against others" ( ax in a[1:] ), as slicing is most probably more computation-intensive that single comparison. Depends on your data, tho, but in general I'd suggest not bothering.

Answer 2

use sorted build in function: for example if you have two list x and y then compare like: sorted(x)==sorted(y)

list1 = [7,9,8]
list2 = [8,7,9]
sorted(list1)==sorted(list2)