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Python | Check if element exists in list of lists
• Last Updated : 20 Mar, 2019

Given a list of lists, the task is to determine whether the given element exists in any sublist or not. Given below are a few methods to solve the given task.

Method #1: Using any()

`any()` method return true whenever a particular element is present in a given iterator.

 `# Python code to demonstrate ` `# finding whether element ` `# exists in listof list ` ` `  `# initialising nested lists ` `ini_list ``=` `[[``1``, ``2``, ``5``, ``10``, ``7``], ` `            ``[``4``, ``3``, ``4``, ``3``, ``21``], ` `            ``[``45``, ``65``, ``8``, ``8``, ``9``, ``9``]] ` ` `  `elem_to_find ``=` `8` `elem_to_find1 ``=` `0` ` `  `# element exists in listof listor not? ` `res1 ``=` `any``(elem_to_find ``in` `sublist ``for` `sublist ``in` `ini_list) ` `res2 ``=` `any``(elem_to_find1 ``in` `sublist ``for` `sublist ``in` `ini_list) ` ` `  `# printing result ` `print``(``str``(res1), ``"\n"``, ``str``(res2)) `

Output:

```True
False
```

Method #2: Using operator in

The ‘in’ operator is used to check if a value exists in a sequence or not. Evaluates to true if it finds a variable in the specified sequence and false otherwise.

 `# Python code to demonstrate ` `# finding whether element ` `# exists in listof list ` ` `  `# initialising nested lists ` `ini_list ``=` `[[``1``, ``2``, ``5``, ``10``, ``7``], ` `            ``[``4``, ``3``, ``4``, ``3``, ``21``], ` `            ``[``45``, ``65``, ``8``, ``8``, ``9``, ``9``]] ` ` `  `elem ``=` `8` `elem1 ``=` `0` ` `  `# element exists in listof listor not? ` `res1 ``=` `elem ``in` `(item ``for` `sublist ``in` `ini_list ``for` `item ``in` `sublist) ` `res2 ``=` `elem1 ``in` `(item ``for` `sublist ``in` `ini_list ``for` `item ``in` `sublist) ` ` `  `# printing result ` `print``(``str``(res1), ``"\n"``, ``str``(res2)) `

Output:

```True
False
```

Method #3: Using `itertools.chain()`

Make an iterator that returns elements from the first iterable until it is exhausted, then proceeds to the next iterable, until all of the iterables are exhausted.

 `# Python code to demonstrate ` `# finding whether element ` `# exists in listof list ` `from`  `itertools ``import` `chain ` ` `  `# initialising nested lists ` `ini_list ``=` `[[``1``, ``2``, ``5``, ``10``, ``7``],  ` `            ``[``4``, ``3``, ``4``, ``3``, ``21``], ` `            ``[``45``, ``65``, ``8``, ``8``, ``9``, ``9``]] ` ` `  `elem_to_find ``=` `8` `elem_to_find1 ``=` `0` ` `  `# element exists in listof listor not? ` `res1 ``=` `elem_to_find ``in` `chain(``*``ini_list) ` `res2 ``=` `elem_to_find1 ``in` `chain(``*``ini_list) ` ` `  `# printing result ` `print``(``str``(res1), ``"\n"``, ``str``(res2)) `

Output:

```True
False
```

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