How to efficiently scramble the tuples generated by itertools.combinations()?

Question content

I am using itertools.combinations() to generate a list of two-item tuples based on a list of non-repeating elements. Then I shuffle the resulting list. However, the contents of the tuple itself are organized chronologically. For example, run the following code:

import random
import itertools
items = ["a","b","c","d","e"]
item_combos = list(itertools.combinations(items, 2))
random.shuffle(item_combos)
print(item_combos)

Output:

['a', 'b', 'c', 'd', 'e']
[('b', 'd'), ('a', 'e'), ('b', 'c'), ('a', 'd'), ('a', 'b'), ('a', 'c'), ('c', 'e'), ('c', 'd'), ('b', 'e'), ('d', 'e')]

The

characters are sorted in the tuple by the time they appeared in the input list (not alphabetically, the input list just happens to be in alphabetical order. Shuffling the list doesn't solve the problem, it just hides it). "c" will always appear to the left of "d" and "a" will always appear to the left of everything else.

My solution was to simply replace all tuples with scrambled tuples (shown below). This works, but proves to be very slow, especially on larger lists.

for i in range(len(item_combos)):
    item_combos[i] = tuple(random.sample(item_combos[i], 2))

Is there a faster way to produce similar output?

Correct answer

Selecting random tuples to reverse instead of shuffling each tuple works much faster. This has the same result, since each tuple in item_combos contains only two items.

New tuple "scrambling" code:

for i in range(len(item_combos)):
    if random.random()<.5:
        item_combos[i] = item_combos[i][::-1]

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