sort list a based on list b, but maintain original order of list a wherever b values are equal

user_10394568 0
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I would like to sort list A based on list B(descending order) such that if two list A,B are

A=["k","a","l","d"]
B=[4,3,7,22]

after sorting, ( sort A based on desc(B) ) A=["d","l","k","a"]

However if B=[0,0,0,0] I would like A to remain A=["k","a","l","d"] after sorting too

so wherever values are equal, the original order must be retained. if B=[0,22,5,0] and A=["k","a","l","d"] then, after sorting, A=["a","l","k","d"]

I have tried using A= [x for _,x in sorted(zip(B,A),reverse=True)],

  • with if A=["k","a","l","d"] and B=[0,0,0,0], this gives A=["a","d","k","l"] but I want A to remain A=["k","a","l","d"]
  • with if A=["k","a","l","d"] and B=[0,22,5,0], this gives A=["a","l","d","k"] but I want A to be A=["a","l","k","d"]

How do I achieve this?

list sorting python
3 Answers

Use the key parameter of sorted

A = ["k", "a", "l", "d"]
Bs = [[4, 3, 7, 22], [0, 0, 0, 0], [0, 22, 5, 0]]

for B in Bs:
    res = [a for _, a in sorted(enumerate(A), key=lambda x: B[x[0]], reverse=True)]
    print(B, res)

Output

[4, 3, 7, 22] ['d', 'l', 'k', 'a']
[0, 0, 0, 0] ['k', 'a', 'l', 'd']
[0, 22, 5, 0] ['a', 'l', 'k', 'd']

As an alternative, you could use:

it_B = iter(B)
res = sorted(A, key=lambda _: next(it_B), reverse=True)

Both ideas work because Timsort (the underlying algorithm of sorted) is stable. Since you are using just one key, when the values are equal, they conserve the appearance order.

user_40015920
user_68909120

Easiest method I've found is to use numpy's Argsort.

index = np.argsort(B)
a = a[index]

Order is kept if they are all the same (i.e. [0,0,0,0]).

user_18382300
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