# Python – Group first elements by second elements in Tuple list

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: }

```

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: }

```

Attention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.

To begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.