# Python | Count the sublists containing given element in a list

• Last Updated : 16 Mar, 2019

Given a list of lists, write a Python program to count the number of sublists containing the given element x.

Examples:

```Input : lst = [1, 3, 5], [1, 3, 5, 7], [1, 3, 5, 7, 9]]
x = 1
Output : 3

Input : lst = (['a'], ['a', 'c', 'b'], ['d'])
x = 'a'
Output : 2
```

Approach #1 : Naive Approach

Count the number of lists containing x. Initialize count to 0, then start a for loop and check if x exists in each list or not. If yes, increment count.

 `# Python3 Program to count number of``# list containing a certain element ``# in a list of lists`` ` `def` `countList(lst, x):``    ``count ``=` `0``    ``for` `i ``in` `range``(``len``(lst)):``        ``if` `x ``in` `lst[i]:``            ``count``+``=` `1``         ` `    ``return` `count``     ` `# Driver Code``lst ``=` `([``'a'``], [``'a'``, ``'c'``, ``'b'``], [``'d'``]) ``x ``=` `'a'``print``(countList(lst, x))`

Output:

```2
```

Approach #2 : List comprehension (Alternative for naive)
A simple one-liner list comprehension can also do the job by simply converting the above mentioned Naive approach into one-liner for loop.

 `# Python3 Program to count number of``# list containing a certain element ``# in a list of lists`` ` `def` `countList(lst, x):``     ` `    ``return` `sum``(x ``in` `item ``for` `item ``in` `lst)``     ` `# Driver Code``lst ``=` `([``'a'``], [``'a'``, ``'c'``, ``'b'``], [``'d'``]) ``x ``=` `'a'``print``(countList(lst, x))`

Output:

```2
```

Approach #3 : Using `chain.from_iterable()` and `Counter`

We can use Counter to count how many lists ‘x’ occurs in. Since we don’t want to count ‘x’ for more than once for each inner list, we’ll convert each inner list to sets. After this, join those sets of elements into one sequence using `chain.from_iterable()`.

 `# Python3 Program to count number of``# list containing a certain element ``# in a list of lists``from` `itertools ``import` `chain``from` `collections ``import` `Counter`` ` `def` `countList(lst, x):``     ` `    ``return` `Counter(chain.from_iterable(``set``(i) ``for` `i ``in` `lst))[x]``     ` `# Driver Code``lst ``=` `([``'a'``], [``'a'``, ``'c'``, ``'b'``], [``'d'``]) ``x ``=` `'a'``print``(countList(lst, x))`

Output:

```2
```

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