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# Python – Similar Consecutive elements frequency

• Last Updated : 01 Mar, 2020

Sometimes, while working with Python, we can have a problem in which we have to find the occurrences of elements that are present consecutively. This problem have usage in school programming and data engineering. Let’s discuss certain ways in which this task can be performed.

Method #1 : Using loop
This is brute force method by which this problem can be solved. In this, we iterate loop and count till we get other number.

 `# Python3 code to demonstrate ``# Similar Consecutive elements frequency``# using loop`` ` `# initializing list ``test_list ``=` `[``2``, ``2``, ``3``, ``3``, ``3``, ``3``, ``4``, ``4``, ``4``]`` ` `# printing original list``print` `(``"The original list is : "` `+` `str``(test_list))`` ` `# Similar Consecutive elements frequency``# using loop``res ``=` `[]``count ``=` `1``for` `ele ``in` `range``(``0``, ``len``(test_list) ``-``1``):``    ``if` `test_list[ele] !``=` `test_list[ele ``+` `1``]:``        ``res.append((test_list[ele], count))``        ``count ``=` `1``    ``else` `:``        ``count ``=` `count ``+` `1``res.append((test_list[``len``(test_list) ``-``1``], count))`` ` `# printing result ``print` `(``"The consecutive element frequency is : "` `+` `str``(res))`
Output :
```The original list is : [2, 2, 3, 3, 3, 3, 4, 4, 4]
The consecutive element frequency is : [(2, 2), (3, 4), (4, 3)]
```

Method #2 : Using `groupby() + len()` + list comprehension
The combination of above methods can be used to perform this task. In this, we group the consecutive elements and extract the count using len(). List comprehension is used to bind both the logics together.

 `# Python3 code to demonstrate ``# Similar Consecutive elements frequency``# using groupby() + len() + list comprehension``from` `itertools ``import` `groupby`` ` `# initializing list ``test_list ``=` `[``2``, ``2``, ``3``, ``3``, ``3``, ``3``, ``4``, ``4``, ``4``]`` ` `# printing original list``print` `(``"The original list is : "` `+` `str``(test_list))`` ` `# Similar Consecutive elements frequency``# using groupby() + len() + list comprehension``res ``=` `[(k, ``len``(``list``(j))) ``for` `k, j ``in` `groupby(test_list)]`` ` `# printing result ``print` `(``"The consecutive element frequency is : "` `+` `str``(res))`
Output :
```The original list is : [2, 2, 3, 3, 3, 3, 4, 4, 4]
The consecutive element frequency is : [(2, 2), (3, 4), (4, 3)]
```

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