Sorting of lists in number ranges - python-3.x

Question

list = [1,2,,3,4,5,6,1,2,56,78,45,90,34]
range = ["0-25","25-50","50-75","75-100"]
I am coding in python. I want to sort a list of integers in range of numbers and store them in differrent lists.How can i do it?
I have specified my ranges in the the range list.

Answer 1

Create a dictionary with max-value of each bin as key. Iterate through your numbers and append them to the list that's the value of each bin-key:
l = [1,2,3,4,5,6,1,2,56,78,45,90,34]
# your range covers 25 a piece - and share start/endvalues.
# I presume [0-25[ ranges
def inRanges(data,maxValues):
"""Sorts elements of data into bins that have a max-value. Max-values are
given by the list maxValues which holds the exclusive upper bound of the bins."""
d = {k:[] for k in maxValues} # init all keys to empty lists
for n in data:
key = min(x for x in maxValues if x>n) # get key
d[key].append(n) # add number
return d
sortEm = inRanges(l,[25,50,75,100])
print(sortEm)
print([ x for x in sortEm.values()])
Output:
{25: [1, 2, 3, 4, 5, 6, 1, 2], 50: [25, 45, 34],
75: [56], 100: [78, 90]}
[[1, 2, 3, 4, 5, 6, 1, 2], [25, 45, 34], [56], [78, 90]]

Answer 2

Another stable bin approach for your special case (regular intervaled bins) would be to use a calculated key - this would get rid of the key-search in each step.
Stable search means the order of numbers in the list is the same as in the input data:
def inRegularIntervals(data, interval):
"""Sorts elements of data into bins of regular sizes.
The size of each bin is given by 'interval'."""
# init dict so keys are ordered - collection.defaultdict(list)
# would be faster - but this works for lists of a couple of
# thousand numbers if you have a quarter up to one second ...
# if random key order is ok, shorten this to d = {}
d = {k:[] for k in range(0, max(data), interval)}
for n in data:
key = n // interval # get key
key *= interval
d.setdefault(key, [])
d[key ].append(n) # add number
return d
Use on random data:
from random import choices
data = choices(range(100), k = 50)
data.append(135) # add a bigger value to see the gapped keys
binned = inRegularIntervals(data, 25)
print(binned)
Output (\n and spaces added):
{ 0: [19, 9, 1, 0, 15, 22, 4, 9, 12, 7, 12, 9, 16, 2, 7],
25: [25, 31, 37, 45, 30, 48, 44, 44, 31, 39, 27, 36],
50: [50, 50, 58, 60, 70, 69, 53, 53, 67, 59, 52, 64],
75: [86, 93, 78, 93, 99, 98, 95, 75, 88, 82, 79],
100: [],
125: [135], }
To sort the binned lists in place, use
for k in binned:
binned[k].sort()
to get:
{ 0: [0, 1, 2, 4, 7, 7, 9, 9, 9, 12, 12, 15, 16, 19, 22],
25: [25, 27, 30, 31, 31, 36, 37, 39, 44, 44, 45, 48],
50: [50, 50, 52, 53, 53, 58, 59, 60, 64, 67, 69, 70],
75: [75, 78, 79, 82, 86, 88, 93, 93, 95, 98, 99],
100: [],
125: [135]}