Binning a list in groups python

Question

I have a list:

 l = [2.0, 4.0, 5.0, 6.0, 7.0, 8.0, 10.0, 12.0,96.0, 192.0, 480.0, 360.0, 504.0, 300.0]

I want to group the elements in list in group size difference of 10. (i.e, 0-10,10-20,20-30,30-40...etc)

For eg:

Output that I'm looking for is:

[ [2,4,5,6,7,8,10],[12],[96],[192],[300],[360],[480],[504] ]

I tried using:

list(zip(*[iter(l)] * 10))

But getting wrong answer.

try [rounding](https://stackoverflow.com/a/8866125/7540911) – Nullman Mar 18 '19 at 13:04 — Nullman, Mar 18 '19 at 13:04

score 5 · Accepted Answer · answered Mar 18 '19 at 13:09

5

Use itertools.groupby to group together after dividing(//) it by 10

from itertools import groupby
l = [2.0, 4.0, 5.0, 6.0, 7.0, 8.0, 10.0, 12.0,96.0, 192.0, 480.0, 360.0, 504.0, 300.0]

groups = []
for _, g in groupby(l, lambda x: (x-1)//10):
    groups.append(list(g))      # Store group iterator as a list

print(groups)

Output:

[[2.0, 4.0, 5.0, 6.0, 7.0, 8.0, 10.0], [12.0], [96.0], [192.0], [480.0], [360.0], [504.0], [300.0]]

answered Mar 18 '19 at 13:09

Sociopath

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It looks like op wants ordered output. so you can sort initial list `l` to achieve this. – Manuel Fedele Mar 18 '19 at 13:17

score 1 · Answer 2 · edited Mar 18 '19 at 14:04

1

A defaultdict might not be bad for this, it's not in one pass, but you can sort the keys to keep everything in place. The integer divide by 10 will bin everything for you

groups = defaultdict(list)

for i in l:
    groups[int((i-1)//10)].append(i)

groups_list = sorted(groups.values())
groups_list[[2.0, 4.0, 5.0, 6.0, 7.0, 8.0, 10.0], [12.0], [96.0], [192.0], [300.0], [360.0], [480.0], [504.0]]

edited Mar 18 '19 at 14:04

Ena

3,481
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answered Mar 18 '19 at 13:10

C.Nivs

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you could have just done `sorted(groups.values())` :). I like your answer the best – han solo Mar 18 '19 at 13:46
@hansolo thanks, I was thinking about that after I submitted and then the edits rolled through – C.Nivs Mar 18 '19 at 14:08

score 0 · Answer 3 · answered Mar 18 '19 at 13:36

Even though, an answer is accepted, here is another way :

l = [2.0, 4.0, 5.0, 6.0, 7.0, 8.0, 10.0, 12.0,96.0, 192.0, 480.0, 360.0, 504.0, 300.0]
l1 = [int(k) for k in l]
l2 = list(list([k for k in l1 if len(str(k))==j]) for j in range(1,len(str(max(l1))) +1))

OUTPUT :

l2 = [[2, 4, 5, 6, 7, 8], [10, 12, 96], [192, 480, 360, 504, 300]]

score 0 · Answer 4 · answered Mar 18 '19 at 13:52

0

It can be sub listed using dictionary : the key for dict will be value-1/10 if same key comes value will be appended:

gd={}
for i in l:
    k=int((i-1)//10)
    if k in gd:
        gd[k].append(i)
    else:
        gd[k]=[i]
print(gd.values())

answered Mar 18 '19 at 13:52

Pradeep Pandey

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score 0 · Answer 5 · answered Mar 18 '19 at 14:10

You can loop over you list l and create a new list using extend and an if condition:

smaller_list = []
larger_list = []
desired_result_list = []

for element in l:
    if element <= 10:
    smaller_list.extend([element])
else:
    larger_list.append([element])

desired_result_list.extend(larger_list + [smaller_list])

Binning a list in groups python

5 Answers5