CODEWITHSUNDEEP
python
a.lex.n
a.lex.n

Reputation: 39

How can I find groups of consecutive items in a python list?

I have a sorted list like this [1,2,3,4,6,7,8,9,10,12,14]

I have looked up different similar solutions but they dont provide help in my case

I want this list to output like this [ [1,4], [6,10], [12], [14] ]
so basically a list of lists with a start and an end of a sequence. Honestly looks very easy but i am kind of stuck on it right now. Any help would be greatly appreciated !

Upvotes: 3

Views: 1872

Answers (5)

Hass786123
Hass786123

Reputation: 676

Using numpy

import numpy as np

myarray = [1,2,3,4,6,7,8,9,10,12,14]
sequences = np.split(myarray, np.array(np.where(np.diff(myarray) > 1)[0]) + 1)
l = []
for s in sequences:
    if len(s) > 1:
        l.append((np.min(s), np.max(s)))
    else:
        l.append(s[0])
print(l)

Output:

[(1, 4), (6, 10), 12, 14]

Upvotes: 1

sakost
sakost

Reputation: 266

Solution can look like this

def make_ranges(l: list):
    prev = l[0]
    start = l[1]
    res = []
    for v in l[1:]:
        if v - 1 != prev:
            if start == prev:
                res.append([start])
            else:
                res.append([start, prev])
            start = v
        prev = v
    if l[-1] - 1 == l[-2]:
        res.append([start, l[-1])
    return res

For example:

print(make_ranges(list(range(10)) + list(range(13, 20)) + [22]))

This code will print [[0, 9], [13, 19], [22]]

Upvotes: 1

rafaelc
rafaelc

Reputation: 59274

Using pandas

import pandas as pd

s = pd.Series([1,2,3,4,6,7,8,9,10,12,14])
s.groupby(s.diff().ne(1).cumsum()).apply(lambda x: [x.iloc[0], x.iloc[-1]] if len(x) >= 2 else [x.iloc[0]]).tolist()

Outputs

[[1, 4], [6, 10], [12], [14]]

Upvotes: 1

Austin
Austin

Reputation: 26039

Use groupby from the standard itertools:

from itertools import groupby

lst = [1,2,3,4,6,7,8,9,10,12,14]

result = []
for k, g in groupby(enumerate(lst), lambda x: x[0] - x[1]):
    g = list(map(lambda x: x[1], g))
    if len(g) > 1:
        result.append([g[0], g[-1]])
    else:
        result.append([g[0]])

print(result)
# [[1, 4], [6, 10], [12], [14]]

Upvotes: 1

Devesh Kumar Singh
Devesh Kumar Singh

Reputation: 20500

You can use more_itertools.consecutive_groups for this available at https://pypi.org/project/more-itertools/

from more_itertools import consecutive_groups

#Get the groupings of consecutive items
li = [list(item) for item in consecutive_groups([1,2,3,4,6,7,8,9,10,12,14])]
#[[1, 2, 3, 4], [6, 7, 8, 9, 10], [12], [14]]

#Use the result to get range groupings
result = [ [item[0],item[-1]] if len(item) > 1 else [item[0]] for item in li]

print(result)
#[[1, 4], [6, 10], [12], [14]]

Upvotes: 8

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