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Python – Group first elements by second elements in Tuple list
  • Last Updated : 02 Sep, 2020

Given List of tuples, group 1st elements on basis of 2nd elements.

Input : test_list = [(6, 5), (2, 7), (2, 5), (8, 7), (3, 7)]
Output : {5: [6, 2], 7: [2, 8, 3]}
Explanation : 5 occurs along with 6 and 2 in tuple list, hence grouping.

Input : test_list = [(6, 5), (2, 7), (2, 5), (8, 7)]
Output : {5: [6, 2], 7: [2, 8]}
Explanation : 5 occurs along with 6 and 2 in tuple list, hence grouping.

Method #1 : Using loop + groupby() + sorted() + list comprehension + lambda

In this, the elements are sorted for grouping, function provided by lambda, then only first elements are extracted using list comprehension out of result. And final dictionary formation using loop.



Python3




# Python3 code to demonstrate working of 
# Group first elements by second elements in Tuple list
# Using loop + groupby() + sorted() + list comprehension + lambda
from itertools import groupby
  
# initializing list
test_list = [(6, 5), (2, 7), (2, 5), (8, 7), (9, 8), (3, 7)]
  
# printing original list
print("The original list is : " + str(test_list))
  
res = dict()
  
# forming equal groups
for key, val in groupby(sorted(test_list, key = lambda ele: ele[1]), key = lambda ele: ele[1]):
    res[key] = [ele[0] for ele in val] 
  
# printing results
print("Grouped Dictionary : " + str(res))
Output
The original list is : [(6, 5), (2, 7), (2, 5), (8, 7), (9, 8), (3, 7)]
Grouped Dictionary : {5: [6, 2], 7: [2, 8, 3], 8: [9]}

Method #2 : Using dictionary comprehension 

This is method similar to above, just a one-liner shorthand handled using dictionary comprehension.

Python3




# Python3 code to demonstrate working of 
# Group first elements by second elements in Tuple list
# Using dictionary comprehension 
from itertools import groupby
  
# initializing list
test_list = [(6, 5), (2, 7), (2, 5), (8, 7), (9, 8), (3, 7)]
  
# printing original list
print("The original list is : " + str(test_list))
  
# shorthand to solve problem
res = {key : [v[0] for v in val] for key, val in groupby(sorted(test_list, key = lambda ele: ele[1]), key = lambda ele: ele[1])}
  
# printing results
print("Grouped Dictionary : " + str(res))
Output
The original list is : [(6, 5), (2, 7), (2, 5), (8, 7), (9, 8), (3, 7)]
Grouped Dictionary : {5: [6, 2], 7: [2, 8, 3], 8: [9]}

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