How can I find same values in a list and group together a new list?

From this list:

N = [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5]

I’m trying to create:

L = [[1],[2,2],[3,3,3],[4,4,4,4],[5,5,5,5,5]]

Any value which is found to be the same is grouped into it’s own sublist.
Here is my attempt so far, I’m thinking I should use a while loop?

global n

n = [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5] #Sorted list
l = [] #Empty list to append values to

def compare(val):
   """ This function receives index values
   from the n list (n[0] etc) """
   
   global valin
   valin = val

   global count
   count = 0

    for i in xrange(len(n)):
        if valin == n[count]: # If the input value i.e. n[x] == n[iteration]
            temp = valin, n[count]
             l.append(temp) #append the values to a new list
             count +=1
        else:
          count +=1
    

for x in xrange (len(n)):
    compare(n[x]) #pass the n[x] to compare function

Answers:

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Method 1

Use itertools.groupby:

from itertools import groupby

N = [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5]

print([list(j) for i, j in groupby(N)])

Output:

[[1], [2, 2], [3, 3, 3], [4, 4, 4, 4], [5, 5, 5, 5, 5]]

Side note: Prevent from using global variable when you don’t need to.

Method 2

Someone mentions for N=[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 1] it will get [[1], [2, 2], [3, 3, 3], [4, 4, 4, 4], [5, 5, 5, 5, 5], [1]]

In other words, when numbers of the list isn’t in order or it is a mess list, it’s not available.

So I have better answer to solve this problem.

from collections import Counter

N = [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5]
C = Counter(N)

print [ [k,]*v for k,v in C.items()]

Method 3

You can use itertools.groupby along with a list comprehension

>>> l =  [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5]
>>> [list(v) for k,v in itertools.groupby(l)]
[[1], [2, 2], [3, 3, 3], [4, 4, 4, 4], [5, 5, 5, 5, 5]]

This can be assigned to the variable L as in

L = [list(v) for k,v in itertools.groupby(l)]

Method 4

You’re overcomplicating this.

What you want to do is: for each value, if it’s the same as the last value, just append it to the list of last values; otherwise, create a new list. You can translate that English directly to Python:

new_list = []
for value in old_list:
    if new_list and new_list[-1][0] == value:
        new_list[-1].append(value)
    else:
        new_list.append([value])

There are even simpler ways to do this if you’re willing to get a bit more abstract, e.g., by using the grouping functions in itertools. But this should be easy to understand.

If you really need to do this with a while loop, you can translate any for loop into a while loop like this:

for value in iterable:
    do_stuff(value)

iterator = iter(iterable)
while True:
    try:
        value = next(iterator)
    except StopIteration:
        break
    do_stuff(value)

Or, if you know the iterable is a sequence, you can use a slightly simpler while loop:

index = 0
while index < len(sequence):
    value = sequence[index]
    do_stuff(value)
    index += 1

But both of these make your code less readable, less Pythonic, more complicated, less efficient, easier to get wrong, etc.

Method 5

You can do that using numpy too:

import numpy as np

N = np.array([1,2,2,3,3,3,4,4,4,4,5,5,5,5,5])
counter = np.arange(1, np.alen(N))
L = np.split(N, counter[N[1:]!=N[:-1]])

The advantage of this method is when you have another list which is related to N and you want to split it in the same way.

Method 6

Another slightly different solution that doesn’t rely on itertools:

#!/usr/bin/env python

def group(items):
    """
    groups a sorted list of integers into sublists based on the integer key
    """
    if len(items) == 0:
        return []

    grouped_items = []
    prev_item, rest_items = items[0], items[1:]

    subgroup = [prev_item]
    for item in rest_items:
        if item != prev_item:
            grouped_items.append(subgroup)
            subgroup = []
        subgroup.append(item)
        prev_item = item

    grouped_items.append(subgroup)
    return grouped_items

print group([1,2,2,3,3,3,4,4,4,4,5,5,5,5,5])
# [[1], [2, 2], [3, 3, 3], [4, 4, 4, 4], [5, 5, 5, 5, 5]]

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