Binary element list grouping in Python

Suppose we have a list of lists in which each sublist has two elements. One element of each sublist is common across many other subjects of the list. We need to create a final list which will show sublists grouped by common elements.

With set and map

In the given list the first element is a string and the second element is a number. So we create a temp list which will hold the second element of each sublist. Then we compare is sublist with each of the element in the temp list and designer follow to group them.

Example

 Live Demo

listA = [['Mon', 2], ['Tue', 3], ['Wed', 3],
["Thu", 1], ['Fri', 2], ['Sat', 3],
['Sun', 1]]
# With set and map
temp = set(map(lambda i: i[1], listA))
res = [[j[0] for j in listA if j[1] == i] for i in temp]
# Result
print("The list with grouped elements is : \n" ,res)

Output

Running the above code gives us the following result −

The list with grouped elements is :
[['Thu', 'Sun'], ['Mon', 'Fri'], ['Tue', 'Wed', 'Sat']]

With groupby and itemgetter

The itemgetter function is used to get the second element of each sublist. Then a group by function is applied based on the result from itemgetter function.

Example

 Live Demo

from itertools import groupby
from operator import itemgetter
listA = [['Mon', 2], ['Tue', 3], ['Wed', 3],["Thu", 1], ['Fri', 2], ['Sat', 3],['Sun', 1]]
# With groupby
listA.sort(key = itemgetter(1))
groups = groupby(listA, itemgetter(1))
res = [[i[0] for i in val] for (key, val) in groups]
# Result
print("The list with grouped elements is : \n" ,res)

Output

Running the above code gives us the following result −

The list with grouped elements is :
[['Thu', 'Sun'], ['Mon', 'Fri'], ['Tue', 'Wed', 'Sat']]

With defaultdict

We Apply the defaultdict function to get the second elements of the sublist as the keys of the dictionary. Then in the resulting list we append the values from the first element of the sublist.

Example

 Live Demo

import collections
listA = [['Mon', 2], ['Tue', 3], ['Wed', 3],["Thu", 1], ['Fri', 2], ['Sat', 3],['Sun', 1]]
# With defaultdict
res = collections.defaultdict(list)
for val in listA:
   res[val[1]].append(val[0])
# Result
print("The list with grouped elements is : \n" ,res)

Output

Running the above code gives us the following result −

The list with grouped elements is :
defaultdict(<class 'list'>, {2: ['Mon', 'Fri'], 3: ['Tue', 'Wed', 'Sat'], 1: ['Thu', 'Sun']})

Pradeep Elance

Published on 09-Jul-2020 12:46:10

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