Lets say I have this list of numbers:
1, 2, 3, 4, 7, 8, 10, 12, 15, 16, 17, 18, 19, 20, 21, 22, 24, 26, 27, 29
And I want to create 5 even groups based on ascending order in the list:
group 1: 1, 2, 3, 4, 7
group 2: 8, 10, 12, 15, 16
group 3: 17, 18, 19, 20, 21
etc...
Can I use python to do this? I know that I can group by 0:4; 5:9... but I have many lists of various lengths that I need to sort into 5 exact groups each.
You can try this:
l = [1, 2, 3, 4, 7, 8, 10, 12, 15, 16, 17, 18, 19, 20, 21, 22, 24, 26, 27, 29]
new_groups = [l[i:i+5] for i in range(0, len(l), 5)]
Output:
[[1, 2, 3, 4, 7], [8, 10, 12, 15, 16], [17, 18, 19, 20, 21], [22, 24, 26, 27, 29]]
If you want to access each group by number, you can build a dictionary:
accessing = {i+1:a for i, a in zip(range(5), new_groups)}
print(acessing[2])
Output:
[8, 10, 12, 15, 16]
Sort your list, and then use the grouper recipe from itertools:
from itertools import zip_longest
l = [1, 2, 3, 4, 7, 8, 10, 12, 15, 16, 17, 18, 19, 20, 21, 22, 24, 26, 27, 29]
l.sort()
def grouper(iterable, n, fillvalue=None):
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
groups = list(grouper(l, 5))
print(groups)
# [(1, 2, 3, 4, 7), (8, 10, 12, 15, 16), (17, 18, 19, 20, 21), (22, 24, 26, 27, 29)]
You can do that based on the length of the list.
l = [...]
groupLength = len(l) // 5
groups = [l[i*groupLength:(i+1)*groupLength] for i in range(5)]
Then either complete the last group with the remaining elements:
groups[-1].extend(l[-len(l)%5:])
Or simply put the remaining elements into a new group:
groups.append(l[-(len(l)%5):])
