如果元素包含在列表中，则从列表中删除元组

Question

I have a list of players and a list of tuples

players = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
shortlist = [('a', 'b'), ('c', 'd'), ('e',)]

There are two pairs and a single tuple. I want each player to choose one tuple from the shortlist, however they cannot choose the tuple which they are in, if they are in the shortlist.

My current attempt:

for person in players:
    for item in shortlist:
        if person in item:
            new_list = shortlist.remove(item)
            their_choice = random.choices(new_list, k=1)
            df[person] = their_choice

But I am just getting empty values in the dataframe. I think there must be a way to do this using list comprehension but I can't quite figure it out. Thanks in advance

Answer 1

You wouldn't want to modify the list whilst looping over it (eg with remove ).

Perhaps you're looking for something like this?

import random
choices = []
for person in players:
    eligible_items = [item for item in shortlist if person not in item]
    their_choice = random.choices(eligible_items)[0]
    choices.append(their_choice)

Or in one big list comprehension?

[
    random.choices([item for item in shortlist if person not in item])[0]
    for person in players
]

Answer 2

modifying your code a little the second loop can be done in a list comprehension, so something like this should work:

for person in players:
    df[person] = random.choices(shortlist, [person not in choice for choice in shortlist])

what is happening here is we assign a weight to each choice in the second argument of choices. The weights are calculated using list comprehension for every person and they are False for the choice containing the person and True for the others. Luckily random.choices() accepts True False list as weights or conversion to 0 and 1 would be needed.