Slicing a list and group

Question

For example, if our source list is:

input = [1, 2, 3, 4, 5, 6, 7, 8, 9, ... ]

And I need something like this:

output = {1:[1], 2:[2,3], 3:[4,5,6], 4:[7,8,9,...], ...}

I try like this, but this not work correctly:

groups = {}
N = 1
group = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for i in range(0, len(group)-1):
    groups.update({N:group[i:i+N]})
    N+=1

Answer 1

For completeness sake - you can also write a version which will work on any iterable:

from itertools import islice, count

group = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
res = {k:v for k,v in enumerate(iter(lambda i=iter(group), c=count(1): list(islice(i, next(c))), []), 1)}
# {1: [1], 2: [2, 3], 3: [4, 5, 6], 4: [7, 8, 9, 10]}

Answer 2

You can use one dict comprehension but first you need to find a proper index range that you could split your elements based on them! for this aim you can use a simple mathematical formula that the sum of a sequence from 1...n is n*(n+1)/2 so in this case n*(n+1)/2=len(l) and with solving the equation you can achieve the n with (1+math.sqrt(1+8*len(l)))/2) :

Some examples :

>>> l=[23,12,33,42,5,6,7,8,39,10,11,102]
>>> ind=range(1,int((1+math.sqrt(1+8*len(l)))/2))
>>> {i:l[sum(ind[:i-1]):sum(ind[:i-1])+i] for i in ind}
{1: [23], 2: [12, 33], 3: [42, 5, 6], 4: [7, 8, 39, 10]}

As the length of 11,102 is not 5 so the n will be 4 in this case! but in following it covers all the elements :

>>> l=[23,12,33,42,5,6,7,8,39,10,11,102,4,0,5]
>>> ind=range(1,int((1+math.sqrt(1+8*len(l)))/2))
>>> {i:l[sum(ind[:i-1]):sum(ind[:i-1])+i] for i in ind}
{1: [23], 2: [12, 33], 3: [42, 5, 6], 4: [7, 8, 39, 10], 5: [11, 102, 4, 0, 5]}

And as a better way you can just calculate the sum(ind[:i-1]) one time :

>>> for i in ind:
...    s=sum(ind[:i-1])
...    d[i]=l[s:s+i]
... 
>>> d
{1: [23], 2: [12, 33], 3: [42, 5, 6], 4: [7, 8, 39, 10], 5: [11, 102, 4, 0, 5]}

Last Note as you can see in first example this solution doesn't keep the last elements if their number doesn't match with the corresponding length. if you want to keep the last elements You can use other answers that are good stuffs!