切片列表到有序块

Question

I have dictionary like:

item_count_per_section = {1: 3, 2: 5, 3: 2, 4: 2}

And total count of items retrieved from this dictionary:

total_items = range(sum(item_count_per_section.values()))

Now I want to transform total_items by values of dictionary following way:

items_no_per_section = {1: [0,1,2], 2: [3,4,5,6,7], 3:[8,9], 4:[10,11] }

Ie slice total_items sequencially to sublists which startrs from previous "iteration" index and finished with value from initial dictionary.

Answer 1

You don't need to find total_items at all. You can straightaway use itertools.count , itertools.islice and dictionary comprehension, like this

from itertools import count, islice
item_count_per_section, counter = {1: 3, 2: 5, 3: 2, 4: 2}, count()
print {k:list(islice(counter, v)) for k, v in item_count_per_section.items()}

Output

{1: [0, 1, 2], 2: [3, 4, 5, 6, 7], 3: [8, 9], 4: [10, 11]}

Answer 2

dict comprehension of itertools.islice d iter of total_items :

from itertools import islice
item_count_per_section = {1: 3, 2: 5, 3: 2, 4: 2}
total_items = range(sum(item_count_per_section.values()))

i = iter(total_items)
{key: list(islice(i, value)) for key, value in item_count_per_section.items()}

Outputs:

{1: [0, 1, 2], 2: [3, 4, 5, 6, 7], 3: [8, 9], 4: [10, 11]}

Note: this works for any total_items , not just range(sum(values)) , assuming that was just your sample to keep the question generic. If you do just want the numbers, go with @thefourtheye's answer