如何在Python中将列表中的每个项目附加到列表中的列表？

Question

Sorry for the puzzling title. For example, I have a list: [a, b, c, d]

I want to generate a ranked list with different combinations in this format:

[a]
[b]
[c]
[d]
[a, b]
[a, c]
[a, d]
[a, b, c]
[a, b, d]
[a, b, c, d]

I'm having trouble in generating this list. So far what I did first was generate each list by adding an element per iteration:

[]
[a]
[a, b]
[a, b, c]

Then I generated the lengths of how the ranked list should look like:

[]
[]
[]
[]
[a]
[a]
[a]
[a, b]
[a, b]
[a, b, c]

Now I'm stuck from here. Is there a library in Python that allows me to do this or can you only do this manually in code? The last thing I have to do is to do a one-to-one list appending from the original list that I generated at the top.

Here's the code of what I have attempted, assume original_list is the original list I made at the top and new_list is the list that I have generated right above this text:

for x in range(0, len(original_list)):
    new_list[x].append(original_list[x])

This doesn't work apparently since it appends every item from original_list to the first 4 items in new_list .

EDIT: The elements should be alphabetical with only the last element having different combinations with no repeating element since I'm attempting this on a list with 21 items.

Answer 1

Using the powerset recipe from itertools recipes , you could do:

from itertools import chain, combinations


def powerset(iterable):
    s = list(iterable)
    it = chain.from_iterable(combinations(s, r) for r in range(len(s) + 1))
    return map(list, (e for e in it if e))


result = sorted(powerset(['a', 'b', 'c', 'd']), key=lambda x: (len(x), ''.join(x)))
for s in result:
    print(s)

Output

['a']
['b']
['c']
['d']
['a', 'b']
['a', 'c']
['a', 'd']
['b', 'c']
['b', 'd']
['c', 'd']
['a', 'b', 'c']
['a', 'b', 'd']
['a', 'c', 'd']
['b', 'c', 'd']
['a', 'b', 'c', 'd']

UPDATE

Given the updated requirements you could do:

lst = ['a', 'b', 'c', 'd']
length = len(lst)


def patterns(l):
    for i in range(length):
        for c in l[i:]:
            yield l[:i] + [c]


for pattern in sorted(patterns(lst), key=lambda x: (len(x), ''.join(x))):
    print(pattern)

Output

['a']
['b']
['c']
['d']
['a', 'b']
['a', 'c']
['a', 'd']
['a', 'b', 'c']
['a', 'b', 'd']
['a', 'b', 'c', 'd']

Answer 2

Use simple iteration through list appending required to a new list:

lst = ['a', 'b', 'c', 'd', 'e']

nlst = []
for i in range(len(lst)):
    for y in lst[i:]:
        nlst.append(lst[:i] + list(y))

for x in nlst:
    print(x)

# ['a']
# ['b']
# ['c']
# ['d']
# ['e']
# ['a', 'b']
# ['a', 'c']
# ['a', 'd']
# ['a', 'e']
# ['a', 'b', 'c']
# ['a', 'b', 'd']
# ['a', 'b', 'e']
# ['a', 'b', 'c', 'd']
# ['a', 'b', 'c', 'e']
# ['a', 'b', 'c', 'd', 'e']

Answer 3

Try this:

from itertools import combinations

a = ['a', 'b', 'c', 'd']
result = [list(combinations(a,i))for i in range(1,len(a)+1)]

and for printing it like that:

for i in result:
    print(*list(i), sep='\n')