排列元组列表的元素

Question

I have a list 'l' of tuples.

l = [('apple',4), ('carrot',2), ('apple',1), ('carrot',7)]

I want to arrange the first elements of tuples according to the values in ascending order.

The expected result is:

result = [('apple', (1,4)), ('carrot', (2,7))]

I tried as:

for x in l:
  variables = list(set(x[0]))

I suppose that there is more better way of doing it. Any ideas please.

Answer 1

You could use a defaultdict to collect those values, and then get the items from the dictionary to get the desired result:

>>> l = [('apple',4), ('carrot',2), ('apple',1), ('carrot',7)]
>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> for k, v in l:
        d[k].append(v)

>>> dict(d)
{'carrot': [2, 7], 'apple': [4, 1]}
>>> list(d.items())
[('carrot', [2, 7]), ('apple', [4, 1])]

In order to sort those sublists then, you could use a list comprehension:

>>> [(k, tuple(sorted(v))) for k, v in d.items()]
[('carrot', (2, 7)), ('apple', (1, 4))]

And if you want to sort that also by the “key”, just sort that resulting list using list.sort() .

Answer 2

Here's a one liner for you:

>>> a = [('apple',4), ('carrot',2), ('apple',1), ('carrot',7)]
>>> sorted([(n, tuple(sorted([e[1] for e in a if e[0] == n]))) for n in set(e for e,f in a)])
[('apple', (1, 4)), ('carrot', (2, 7))]

This sorts both the first element ( apple , carrot , ...), and each second element ( (1,4) (2,7) ).

Note that @poke's solution does not sort it.

Answer 3

Here is my solution:

from collections import defaultdict

l = [('apple',4), ('carrot',2), ('apple',1), ('carrot',7)]

d = defaultdict(list)
for i, j in l:
    d[i].append(j)

result = sorted([tuple([x, tuple(sorted(y))]) for x, y in d.items()])

print(result)

And here is the result:

[('apple', (1, 4)), ('carrot', (2, 7))]