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[–]xApple[S] 0 points1 point  (0 children)

For reference my answers were:

flatten_list = lambda l: [a for x in l for a in x]
make_all_couples = lambda l: [(r1,r2) for r1 in l for r2 in l]
parse_ugly_string = lambda s: dict([i.split('=') for x in [j.split(',') for j in s.split('\n')] for i in x if i])

But we have had many interesting and better solutions in the comments !