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Python – Cross tuple summation grouping
• Last Updated : 22 Jun, 2020

Sometimes, while working with Python tuple records, we can have a problem in which we need to perform summation grouping of 1st element of tuple pair w.r.t similar 2nd element of tuple. This kind of problem can have application in day-day programming. Let’s discuss certain ways in which this task can be performed.

Input : test_list = [(1, 5), (7, 4), (9, 6), (11, 6)]
Output : [(7, 4), (1, 5), (20, 6)]

Input : test_list = [(1, 8)]
Output : [(1, 8)]

Method #1 : Using loop
This is brute force way in which this task can be performed. In this we check for similar second elements and perform summation till then and perform the accumulated grouping.

 `# Python3 code to demonstrate working of ``# Cross tuple summation grouping``# Using loop`` ` `# initializing list``test_list ``=` `[(``4``, ``5``), (``7``, ``5``), (``8``, ``6``), (``10``, ``6``), (``10``, ``4``), (``6``, ``7``), (``3``, ``7``)]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# Concatenate Similar Key values``# Using loop``temp ``=` `dict``()``for` `ele1, ele2 ``in` `test_list:``    ``temp[ele2] ``=` `temp.get(ele2, ``0``) ``+` `ele1``res ``=` `[(ele2, ele1) ``for` `(ele1, ele2) ``in` `temp.items()]`` ` `# printing result ``print``(``"The grouped records are : "` `+` `str``(res)) `
Output :

```The original list is : [(4, 5), (7, 5), (8, 6), (10, 6), (10, 4), (6, 7), (3, 7)]
The grouped records are : [(10, 4), (11, 5), (18, 6), (9, 7)]
```

Method #2 : Using `groupby() + sum() + zip()` + list comprehension
The combination of above functions can be used to solve this problem. In this, we perform the task of finding sum using sum(). The task of grouping is done by groupby().

 `# Python3 code to demonstrate working of ``# Cross tuple summation grouping``# Using groupby() + sum() + zip() + list comprehension``from` `itertools ``import` `groupby`` ` `# initializing list``test_list ``=` `[(``4``, ``5``), (``7``, ``5``), (``8``, ``6``), (``10``, ``6``), (``10``, ``4``), (``6``, ``7``), (``3``, ``7``)]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# Concatenate Similar Key values``# Using groupby() + sum() + zip() + list comprehension``res ``=` `[(``sum``(``next``(``zip``(``*``ele))), key) ``for` `key, ele ``in` `groupby(``                       ``test_list, key ``=` `lambda` `tup:tup[``1``])]`` ` `# printing result ``print``(``"The grouped records are : "` `+` `str``(res)) `
Output :
```The original list is : [(4, 5), (7, 5), (8, 6), (10, 6), (10, 4), (6, 7), (3, 7)]
The grouped records are : [(10, 4), (11, 5), (18, 6), (9, 7)]
```

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