Sorting multiple lists together in place

Question

I have lists a,b,c,... of equal length. I'd like to sort all of them the order obtained by sorting a, i.e., I could do the decorate-sort-undecorate pattern

a, b, c = map(list, zip(*sorted(zip(a, b, c))))

or something like that. However, I'd like that the lists are sorted in place (I assume that sorted pulls everything from the temporary iterator passed to it to a temporary list, and then zip stuff into three output lists, so every datum in the input is copied twice unnecessarily) without creating temporary objects. So what I don't mean is:

a_sorted, b_sorted, c_sorted = map(list, zip(*sorted(zip(a, b, c))))
a[:] = a_sorted
b[:] = b_sorted
c[:] = c_sorted

How can I achieve that?

Kelly Bundy · Accepted Answer

I think "without creating temporary objects" is impossible, especially since "everything is an object" in Python.

You could get O(1) space / number of objects if you implement some sorting algorithm yourself, though if you want O(n log n) time and stability, it's difficult. If you don't care about stability (seems likely, since you say you want to sort by a but then actually sort by a, b and c), heapsort is reasonably easy:

def sort_together_heapsort(a, b, c):
    n = len(a)
    def swap(i, j):
        a[i], a[j] = a[j], a[i]
        b[i], b[j] = b[j], b[i]
        c[i], c[j] = c[j], c[i]
    def siftdown(i):
        while (kid := 2*i+1) < n:
            imax = kid if a[kid] > a[i] else i
            kid += 1
            if kid < n and a[kid] > a[imax]:
                imax = kid
            if imax == i:
                return
            swap(i, imax)
            i = imax
    for i in range(n // 2)[::-1]:
        siftdown(i)
    while n := n - 1:
        swap(0, n)
        siftdown(0)

Anyway, if someone's interested in just saving some amount of memory, that can be done by decorating in-place (building tuples and storing them in a):

def sort_together_decorate_in_a(a, b, c):
    for i, a[i] in enumerate(zip(a, b, c)):
        pass
    a.sort()
    for i, [a[i], b[i], c[i]] in enumerate(a):
        pass

Or if you trust that list.sort will ask for keys for the elements in order (at least in CPython it does, already did so when the key parameter was introduced 18 years ago, and I suspect will keep doing so):

def sort_together_iter_key(a, b, c):
    it = iter(a)
    b.sort(key=lambda _: next(it))
    it = iter(a)
    c.sort(key=lambda _: next(it))
    a.sort()

Testing memory and time with three lists of 100,000 elements:

15,072,520 bytes   152 ms  sort_together_sorted_zip
15,072,320 bytes   166 ms  sort_together_sorted_zip_2
14,272,576 bytes   152 ms  sort_together_sorted_zip_X
 6,670,708 bytes   126 ms  sort_together_decorate_in_a
 6,670,772 bytes   177 ms  sort_together_decorate_in_first_X
 5,190,212 bytes   342 ms  sort_multi_by_a_guest_X
 1,597,400 bytes   100 ms  sort_together_iter_key
 1,597,448 bytes   102 ms  sort_together_iter_key_X
       744 bytes  1584 ms  sort_together_heapsort
       704 bytes  1663 ms  sort_together_heapsort_X
       168 bytes  1326 ms  sort_together_heapsort_opti
       188 bytes  1512 ms  sort_together_heapsort_opti_X

Note:

The second solution is a shortened/improved version of yours, no need for temporary variables and conversions to lists.
The solutions with _X suffix are versions that take arbitrarily many lists as parameters.
The @a_guest is from their answer. Runtime-wise it currently benefits from my data being random, as that doesn't expose that solution's worst case complexity O(m * n²), where m is the number of lists and n is the length of each list.

Testing memory and time with ten lists of 100,000 elements:

19,760,808 bytes   388 ms  sort_together_sorted_zip_X
12,159,100 bytes   425 ms  sort_together_decorate_in_first_X
 5,190,292 bytes  1249 ms  sort_multi_by_a_guest_X
 1,597,528 bytes   393 ms  sort_together_iter_key_X
       704 bytes  4186 ms  sort_together_heapsort_X
       188 bytes  4032 ms  sort_together_heapsort_opti_X

The whole code (Try it online!):

import tracemalloc as tm
from random import random
from timeit import timeit

def sort_together_sorted_zip(a, b, c):
    a_sorted, b_sorted, c_sorted = map(list, zip(*sorted(zip(a, b, c))))
    a[:] = a_sorted
    b[:] = b_sorted
    c[:] = c_sorted

def sort_together_sorted_zip_2(a, b, c):
    a[:], b[:], c[:] = zip(*sorted(zip(a, b, c)))

def sort_together_sorted_zip_X(*lists):
    sorteds = zip(*sorted(zip(*lists)))
    for lst, lst[:] in zip(lists, sorteds):
        pass

def sort_together_decorate_in_a(a, b, c):
    for i, a[i] in enumerate(zip(a, b, c)):
        pass
    a.sort()
    for i, [a[i], b[i], c[i]] in enumerate(a):
        pass

def sort_together_decorate_in_first_X(*lists):
    first = lists[0]
    for i, first[i] in enumerate(zip(*lists)):
        pass
    first.sort()
    for i, values in enumerate(first):
        for lst, lst[i] in zip(lists, values):
            pass

def sort_together_iter_key(a, b, c):
    it = iter(a)
    b.sort(key=lambda _: next(it))
    it = iter(a)
    c.sort(key=lambda _: next(it))
    a.sort()

def sort_together_iter_key_X(*lists):
    for lst in lists[1:]:
        it = iter(lists[0])
        lst.sort(key=lambda _: next(it))
    lists[0].sort()

def sort_together_heapsort(a, b, c):
    n = len(a)
    def swap(i, j):
        a[i], a[j] = a[j], a[i]
        b[i], b[j] = b[j], b[i]
        c[i], c[j] = c[j], c[i]
    def siftdown(i):
        while (kid := 2*i+1) < n:
            imax = kid if a[kid] > a[i] else i
            kid += 1
            if kid < n and a[kid] > a[imax]:
                imax = kid
            if imax == i:
                return
            swap(i, imax)
            i = imax
    for i in range(n // 2)[::-1]:
        siftdown(i)
    while n := n - 1:
        swap(0, n)
        siftdown(0)

def sort_together_heapsort_X(*lists):
    a = lists[0]
    n = len(a)
    def swap(i, j):
        for lst in lists:
            lst[i], lst[j] = lst[j], lst[i]
    def siftdown(i):
        while (kid := 2*i+1) < n:
            imax = kid if a[kid] > a[i] else i
            kid += 1
            if kid < n and a[kid] > a[imax]:
                imax = kid
            if imax == i:
                return
            swap(i, imax)
            i = imax
    for i in range(n // 2)[::-1]:
        siftdown(i)
    while n := n - 1:
        swap(0, n)
        siftdown(0)

def sort_together_heapsort_opti(a, b, c):
    # Avoid inner functions and range-loop to minimize memory.
    # Makes it faster, too. But duplicates code. Not recommended.
    n = len(a)
    i0 = n // 2 - 1
    while i0 >= 0:
        i = i0
        while (kid := 2*i+1) < n:
            imax = kid if a[kid] > a[i] else i
            kid += 1
            if kid < n and a[kid] > a[imax]:
                imax = kid
            if imax == i:
                break
            a[i], a[imax] = a[imax], a[i]
            b[i], b[imax] = b[imax], b[i]
            c[i], c[imax] = c[imax], c[i]
            i = imax
        i0 -= 1
    while n := n - 1:
        a[0], a[n] = a[n], a[0]
        b[0], b[n] = b[n], b[0]
        c[0], c[n] = c[n], c[0]
        i = 0
        while (kid := 2*i+1) < n:
            imax = kid if a[kid] > a[i] else i
            kid += 1
            if kid < n and a[kid] > a[imax]:
                imax = kid
            if imax == i:
                break
            a[i], a[imax] = a[imax], a[i]
            b[i], b[imax] = b[imax], b[i]
            c[i], c[imax] = c[imax], c[i]
            i = imax

def sort_together_heapsort_opti_X(*lists):
    # Avoid inner functions and range-loop to minimize memory.
    # Makes it faster, too. But duplicates code. Not recommended.
    a = lists[0]
    n = len(a)
    i0 = n // 2 - 1
    while i0 >= 0:
        i = i0
        while (kid := 2*i+1) < n:
            imax = kid if a[kid] > a[i] else i
            kid += 1
            if kid < n and a[kid] > a[imax]:
                imax = kid
            if imax == i:
                break
            for lst in lists:
                lst[i], lst[imax] = lst[imax], lst[i]
            i = imax
        i0 -= 1
    while n := n - 1:
        for lst in lists:
            lst[0], lst[n] = lst[n], lst[0]
        i = 0
        while (kid := 2*i+1) < n:
            imax = kid if a[kid] > a[i] else i
            kid += 1
            if kid < n and a[kid] > a[imax]:
                imax = kid
            if imax == i:
                break
            for lst in lists:
                lst[i], lst[imax] = lst[imax], lst[i]
            i = imax

def sort_multi_by_a_guest_X(a, *lists):
    indices = list(range(len(a)))
    indices.sort(key=lambda i: a[i])
    a.sort()
    for lst in lists:
        for i, j in enumerate(indices):
            while j < i:
                j = indices[j]
            lst[i], lst[j] = lst[j], lst[i]

funcs = [
    sort_together_sorted_zip,
    sort_together_sorted_zip_2,
    sort_together_sorted_zip_X,
    sort_together_decorate_in_a,
    sort_together_decorate_in_first_X,
    sort_multi_by_a_guest_X,
    sort_together_iter_key,
    sort_together_iter_key_X,
    sort_together_heapsort,
    sort_together_heapsort_X,
    sort_together_heapsort_opti,
    sort_together_heapsort_opti_X,
]

n = 100000
a0 = [random() for _ in range(n)]
b0 = [x + 1 for x in a0]
c0 = [x + 2 for x in a0]

for _ in range(3):
    for func in funcs:

        a, b, c = a0[:], b0[:], c0[:]
        time = timeit(lambda: func(a, b, c), number=1)
        assert a == sorted(a0)
        assert b == sorted(b0)
        assert c == sorted(c0)

        a, b, c = a0[:], b0[:], c0[:]
        tm.start()
        func(a, b, c)
        memory = tm.get_traced_memory()[1] 
        tm.stop()

        print(f'{memory:10,} bytes  {int(time * 1e3):4} ms  {func.__name__}')
    print()

Sorting multiple lists together in place

Answers (2)

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