# Python | Count occurrences of an element in a Tuple

• Difficulty Level : Medium
• Last Updated : 19 Sep, 2022

In this program, we need to accept a tuple and then find the number of times an item is present in the tuple. This can be done in various ways, but in this article, we will see how this can be done using a simple approach and how inbuilt functions can be used to solve this problem.

Examples:

`Tuple: (10, 8, 5, 2, 10, 15, 10, 8, 5, 8, 8, 2)`
```Input : 4
Output : 0 times

Input : 10
Output : 3 times

Input : 8
Output : 4 times```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1(Simple Approach):

We keep a counter that keeps on increasing if the required element is found in the tuple.

## Python3

 `# Program to count the number of times an element``# Present in the list`  `def` `countX(tup, x):``    ``count ``=` `0``    ``for` `ele ``in` `tup:``        ``if` `(ele ``=``=` `x):``            ``count ``=` `count ``+` `1``    ``return` `count`  `# Driver Code``tup ``=` `(``10``, ``8``, ``5``, ``2``, ``10``, ``15``, ``10``, ``8``, ``5``, ``8``, ``8``, ``2``)``enq ``=` `4``enq1 ``=` `10``enq2 ``=` `8` `print``(countX(tup, enq))``print``(countX(tup, enq1))``print``(countX(tup, enq2))`

Output

```0
3
4```

Method 2(Using count()):

The idea is to use method count() to count number of occurrences.

## Python3

 `# Program to count the number of times an element``# Present in the list``# Count function is used`  `def` `Count(tup, en):``    ``return` `tup.count(en)`  `# Driver Code``tup ``=` `(``10``, ``8``, ``5``, ``2``, ``10``, ``15``, ``10``, ``8``, ``5``, ``8``, ``8``, ``2``)``enq ``=` `4``enq1 ``=` `10``enq2 ``=` `8``print``(Count(tup, enq), ``"times"``)``print``(Count(tup, enq1), ``"times"``)``print``(Count(tup, enq2), ``"times"``)`

Output

```0 times
3 times
4 times```

Method 3: Using list comprehension

## Python3

 `tuple1 ``=` `(``10``, ``8``, ``5``, ``2``, ``10``, ``15``, ``10``, ``8``, ``5``, ``8``, ``8``, ``2``)``x ``=` `[i ``for` `i ``in` `tuple1 ``if` `i ``=``=` `10``]``print``(``"the element 10 occures"``, ``len``(x), ``"times"``)`

Output

`the element 10 occures 3 times`

Method #4: Using enumerate function

## Python3

 `tuple1``=``(``10``, ``8``, ``5``, ``2``, ``10``, ``15``, ``10``, ``8``, ``5``, ``8``, ``8``, ``2``)``x``=``[i ``for` `a,i ``in` `enumerate``(tuple1) ``if` `i``=``=``10``]``print``(``"the element 10 occures"``,``len``(x),``"times"``)`

Output

`the element 10 occures 3 times`

Method #5: Using lambda function

## Python3

 `tuple1``=``(``10``, ``8``, ``5``, ``2``, ``10``, ``15``, ``10``, ``8``, ``5``, ``8``, ``8``, ``2``)``x``=``list``(``filter``(``lambda` `i: (i``=``=``10``),tuple1))``print``(``"the element 10 occures"``,``len``(x),``"times"``)`

Output

`the element 10 occures 3 times`

Method #6: Using counter function

## Python3

 `from` `collections ``import` `Counter``tuple1``=``(``"10"``, ``"8"``, ``"5"``, ``"2"``, ``"10"``, ``"15"``, ``"10"``, ``"8"``, ``"5"``, ``"8"``, ``"8"``, ``"2"``)``x``=``Counter(tuple1)``print``(``"the element 10 occures"``,x[``"10"``],``"times"``)`

Output

`the element 10 occures 3 times`

Method#7: Using Countof function

## Python3

 `tuple1``=``(``"10"``, ``"8"``, ``"5"``, ``"2"``, ``"10"``, ``"15"``, ``"10"``, ``"8"``, ``"5"``, ``"8"``, ``"8"``, ``"2"``)` `import` `operator as op` `print``(op.countOf(tuple1,``"10"``))`

Output

`3`

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