Say I have one list,
list1 = ['Dog', 'Cat', 'Monkey', 'Parakeet', 'Zebra']
And another,
list2 = [[True, True], [False], [True], [False]]
(You can imagine that the second list was created with an itertools.groupby on an animal being a house pet.)
Now say I want to give the first list the same sublist structure as the second.
list3 = [['Dog', 'Cat'], ['Monkey'], ['Parakeet'], ['Zebra']]
I might do something like this:
list3 = []
lengths = [len(x) for x in list2]
count = 0
for leng in lengths:
templist = []
for i in range(leng):
templist.append(list1[count])
count += 1
list3.append(templist)
Which I haven't totally debugged but I think should work. My question is if there is a more pythonic or simple way to do this? This seems kind of convoluted and I have to imagine there is a more graceful way to do it (maybe in itertools which never ceases to impress me).
You can simply use next for a no-import solution:
list1 = ['Dog', 'Cat', 'Monkey', 'Parakeet', 'Zebra']
list_1 = iter(list1)
list2 = [[True, True], [False], [True], [False]]
new_list2 = [[next(list_1) for _ in i] for i in list2]
Output:
[['Dog', 'Cat'], ['Monkey'], ['Parakeet'], ['Zebra']]
However, for input n levels deep, you can use recursion:
list1 = ['Dog', 'Cat', 'Monkey', 'Parakeet', 'Zebra']
list_1 = iter(list1)
list2 = [[True, [True, False]], [False], [[True]]]
def place_data(current):
return [next(list_1) if isinstance(i, int) else place_data(i) for i in current]
print(place_data(list2))
Output:
[['Dog', ['Cat', 'Monkey']], ['Parakeet'], [['Zebra']]]
Make an iter of list1 and islice the necessary N elements of each list2,e g:
from itertools import islice
list1 = ['Dog', 'Cat', 'Monkey', 'Parakeet', 'Zebra']
list2 = [[True, True], [False], [True], [False]]
it = iter(list1)
list3 = [list(islice(it, len(el))) for el in list2]
# [['Dog', 'Cat'], ['Monkey'], ['Parakeet'], ['Zebra']]
This is one way:
from itertools import accumulate
idx_acc = [0] + list(accumulate(map(len, list2)))
list1 = [list1[i:j] for i, j in zip(idx_acc, idx_acc[1:])]
# [['Dog', 'Cat'], ['Monkey'], ['Parakeet'], ['Zebra']]