Sort list using benchmark list

Question

I have a list with benchmark order of items:

benchmark = ['Wind', 'Sun', 'Tree', 'Human', 'Cat']

Also I have many small lists with two items:

list1 = ['Cat', 'Wind']
list2 = ['Tree', 'Sun']
list3 = ['Wind', 'Human']

I want to sort them using order in benchmark. So expected output is:

list1 = ['Wind', 'Cat']
list2 = ['Sun', 'Tree']
list3 = ['Wind', 'Human']

How can I do it in most efficient way?

Answer 1

Try this,

benchmark = ['Wind', 'Sun', 'Tree', 'Human', 'Cat']
list1 = ['Cat', 'Wind']
list2 = ['Tree', 'Sun']
list3 = ['Wind', 'Human']

print(sorted(list1, key=benchmark.index))
print(sorted(list2, key=benchmark.index))
print(sorted(list3, key=benchmark.index))

Answer 2

you can use zip function

sorted_list1 = [item for _,item in sorted(zip(benchmark,list1))]