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Python – Equable Minimial Records
• Last Updated : 22 Jun, 2020

Sometimes, while working with Python records, we can have a problem in which we need to extract one of the records that are equal to certain index, which is mimimal of other index. This kind of problem occurs in domains such as web development. Let’s discuss certain ways in which this task can be performed.

Input :
test_list = [(‘Gfg’, 13, 5), (‘is’, 13, 6), (‘best’, 13, 2), (‘CS’, 13, 2)]
eq_idx = 2, min_idx = 3
Output : [(‘best’, 13, 2)]

Input :
test_list = [(‘Gfg’, 12, 5), (‘is’, 12, 6), (‘best’, 13, 2), (‘CS’, 13, 3)]
eq_idx = 2, min_idx = 3
Output : [(‘Gfg’, 12, 5), (‘best’, 13, 2)]

Method #1 : Using list comprehension + `min() + lambda`
The combination of above functions can be used to solve this problem. In this, we use min() to extract the minimum index value, and list comprehension and lambda functions are used to perform the task of grouping elements.

 `# Python3 code to demonstrate working of ``# Equable Minimial Records``# Using min() + list comprehension + lambda`` ` `# initializing list``test_list ``=` `[(``'Gfg'``, ``12``, ``5``), (``'is'``, ``13``, ``6``), (``'best'``, ``12``, ``2``), (``'CS'``, ``13``, ``2``)]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# initializing Equate index ``eq_idx ``=` `2`` ` `# initializing min index ``min_idx ``=` `3`` ` `# Equable Minimial Records``# Using min() + list comprehension + lambda``res ``=` `[``min``((ele ``for` `ele ``in` `test_list ``if` `ele[eq_idx ``-` `1``] ``=``=` `sub),``      ``key ``=` `lambda` `a: ``int``(a[min_idx ``-` `1``]))``      ``for` `sub ``in` `{b[eq_idx ``-` `1``] ``for` `b ``in` `test_list}]`` ` `         ` `# printing result ``print``(``"Equable Minimal Records : "` `+` `str``(res)) `
Output :

```The original list is : [('Gfg', 12, 5), ('is', 13, 6), ('best', 12, 2), ('CS', 13, 2)]
Equable Minimal Records : [('best', 12, 2), ('CS', 13, 2)]
```

Method #2 : Using `groupby() + filter()` + lambda
The combination of above functions can be used to solve this problem. In this, we perform the task of grouping using groupby() and filter() + lambda are used for task of matching conditions.

 `# Python3 code to demonstrate working of ``# Equable Minimial Records``# Using groupby() + filter() + lambda``from` `itertools ``import` `groupby`` ` `# initializing list``test_list ``=` `[(``'Gfg'``, ``12``, ``5``), (``'is'``, ``13``, ``6``), (``'best'``, ``12``, ``2``), (``'CS'``, ``13``, ``2``)]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# initializing Equate index ``eq_idx ``=` `2`` ` `# initializing min index ``min_idx ``=` `3`` ` `# Equable Minimial Records``# Using groupby() + filter() + lambda``res ``=` `[]``for` `k, val ``in` `groupby(test_list, ``lambda` `sub: sub[eq_idx ``-` `1``]):``    ``res.append(``min``(``filter``(``lambda` `sub : sub[eq_idx ``-` `1``] ``=``=` `k, test_list),``                                   ``key ``=` `lambda` `sub : sub[min_idx ``-` `1``]))``res ``=` `list``(``set``(res))`` ` `         ` `# printing result ``print``(``"Equable Minimal Records : "` `+` `str``(res)) `
Output :
```The original list is : [('Gfg', 12, 5), ('is', 13, 6), ('best', 12, 2), ('CS', 13, 2)]
Equable Minimal Records : [('best', 12, 2), ('CS', 13, 2)]
```

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