In python given an array like so:
a = [ 0, 1, 3, 4, 6, 7, 8, 10, 14 ]
I would like to split this into three uneven groups, such that I end up with something like this:
b = [0, 1, 3, 4]
c = [6, 7, 8]
d = [10, 14]
I want to group numbers by multiples of 5. So any integers from 0 - 4 would end up in the first array, 5 - 9 in the second, and so on.
Itertools.groupby is always the answer!
Here we round each number down to the nearest 5, and then group by equal numbers:
>>> for n, g in itertools.groupby(a, lambda x: round(x/5)*5):
print list(g)
[0, 1, 3, 4]
[6, 7, 8]
[10, 14]
We can be more or less time efficient if we know something about the numbers we're working with. We could also come up with a very quick one that's terribly memory inefficient, but consider this, if it fits your purposes:
#something to store our new lists in
range = 5 #you said bounds of 5, right?
s = [ [] ]
for number in a:
foundit = false
for list in s:
#deal with first number
if len( list ) == 0:
list.append( number )
else:
#if our number is within the same range as the other number, add it
if list[0] / range == number / range:
foundit = true
list.append( number )
if foundit == false:
s.append( [ number ] )
Now that I understand your definition of groups better, I think this relatively simple answer will not only work, it should also be very fast:
from collections import defaultdict
a = [0, 1, 3, 4, 6, 7, 8, 10, 14]
chunk_size = 5
buckets = defaultdict(list)
for n in a:
buckets[n/chunk_size].append(n)
for bucket,values in sorted(buckets.iteritems()):
print '{}: {}'.format(bucket, values)
Output:
0: [0, 1, 3, 4]
1: [6, 7, 8]
2: [10, 14]
bandcto be? In other words, what properties do you want the unevenness to have?