How do I sort a list in the same order as another list in Python?-CodePudding

Let's say I have two lists with the first representing total volume and the second representing the respective unique item IDs whose volume is equal to the total volume:

a = [50, 45, 90, 75]

b = [[1,2,3], [5,6,7], [4], [8,9]]

For example, 50 would be the total volume of the items with the IDs of 1,2, and 3.

I am currently sorting list a to be from smallest to largest, but I also want to arrange the order of list b to match its total volume. Based on the previously made lists, I want both lists to sort like this:

a = [45, 50, 75, 90]

b = [[5,6,7], [1,2,3], [8,9], [4]]

To sort list a, I am writing:

a.sort(key=lambda x: x, reverse = False)

But I'm not sure write the code in Python for list b to sort based on the same pattern. I have to keep the same data structure, so both lists a and b must stay a list and list of lists, respectively. Could anyone tell me how to solve this issue?

CodePudding user response：

You can zip the two lists together, sort the tuples based on the a value, and then unzip:

a = [45, 50, 75, 90]
b = [[5,6,7], [1,2,3], [8,9], [4]]

a,b = map(list, zip(*sorted(zip(a, b), key = lambda x: x[0])))

print(a, b, sep='\n')

Output:

[45, 50, 75, 90]
[[5, 6, 7], [1, 2, 3], [8, 9], [4]]

CodePudding user response：

You can use zip() and two list comprehensions:

a = [50, 45, 90, 75]
b = [[1,2,3], [5,6,7], [4], [8,9]]

a = [x for x, _ in sorted(zip(a, b))]
b = [y for _, y in sorted(zip(a, b))]
print(a)
print(b)

This outputs:

[45, 50, 75, 90]
[[5, 6, 7], [1, 2, 3], [8, 9], [4]]