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# Python | Count the occurrence of all list items in a tuple

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Examples :

`  Input:  tuple = (’a’,’ a’, ’c’,’ b’, ’d’ ) list = [’a’,’ b’]  Output:  3  Input:  tuple = (1, 2, 3, 1, 4, 6, 7, 1, 4) list = [1, 4, 7]  Output:  6 `

Approach # 1: Naive Approach

First Approach — naive approach. Use a for loop and traverse the given list and count the occurrences of each element of the tuple in the list. Finally, return the score.

 ` # Python3 Program for counting cases ` ` # of all list items in tuple ` ` from ` ` collections ` ` import ` ` Counter `   ` def ` ` countOccurrence (tup, lst): ` ` count ` ` = ` ` 0 ` ` for ` ` item ` ` in ` ` tup : ` ` if ` ` item ` ` in ` ` lst: ` ` ` ` count ` ` + ` ` = ` ` 1 `   ` return ` ` count `   ` Driver code ` ` tup ` ` = ` ` (` `’ a’ ` `, ` ` ’a’ ` `, ` ` ’c’ ` `, ` `’ b’ ` `, ` ` ’d’ ` `) ` ` lst ` ` = ` ` [` ` ’a’ ` `, ` ` ’b’ ` `] ` ` print ` ` (countOccurrence (tup, lst)) `

Exit :

` 3 `

Approach # 2 Using a counter
From in the Python Collections module, import the counter to solve this problem. Counter — it is a container that keeps track of how many times equivalent values ​​are added. After storing the result in & # 39; counts & # 39;, we use a for loop and count how many times each item in the list occurs in & # 39; counts & # 39; and sum it up to get the final result.

` `

` # Python3 Case Counting Program # of all list items in a tuple from collections import Counter   def countOccurrence (tup, lst):   counts = Counter (tup) return sum (counts [i] for i in lst)   Driver code tup = ( ’a’ , ’ a ’ , ’ c’ , ’b’ , ’ d’ ) lst = [ ’a’ , ’ b’ ] print (countOccurrence (tup, lst)) `

` ` Exit:

` 3 `

Approach # 3. Using Set
Another method to solve this problem — using the given data structure. Just convert the given list into a set that removes all duplicates. Now, for each element of the list, count its occurrence in a tuple and sum them.

 ` # Python3 Counting program ` ` # of all list items in a tuple `   ` def ` ` countOccurrence (tup, lst): ` ` lst ` ` = ` ` set ` ` (lst) ` ` return ` ` sum ` ` (` ` 1 ` ` for ` ` x ` ` in ` ` tup ` ` if ` ` x ` ` in ` ` lst) ` ` `  ` Driver code ` ` tup ` ` = ` ` (` ` ’a’ ` `, ` ` ’a’ ` `, ` `’ c’ ` `, ` ` ’b’ ` `, ` `’ d’ ` `) ` ` lst ` ` = ` ` [` ` ’a’ ` `, ` `’ b’ ` `] ` ` print ` ` (countOccurrence (tup, lst)) `

Exit :

` 3 `

Approach # 4 Used Python dictionary definition
Get each tuple element and its frequency as a key: value pair in the Python dictionary, then using a for loop, for each list element, count its occurrences in the tuple and sum them.

` `

` # Python3 Case Counting Program # of all list items in a tuple   def countOccurrence (tup, lst): dct = {} for i in tup:   if not dct.get (i): dct [i] = 0 dct [i] + = 1   return sum (dct.get (i, 0 ) for i in lst)   Driver code tup = ( ’a’ , ’ a’  , ’c’ , ’ b’ , ’d’ ) lst = [ ’a’ , ’ b’ ] print (countOccurrence (tup, lst)) `

` ` Exit :

` 3 `

Approach # 5 : Python ` numpy.in1d ​​() `
Python numpy gives us a direct method to find a solution to a given problem, and this is ` numpy.in1d ​​() `. This method checks if each element of the one-dimensional array is present in the second array. Since the list is also a one-dimensional array, this method can be applied here.

 ` # Python3 Case Counting Program ` ` # of all list items in a tuple ` ` import ` ` numpy as np `   ` def ` ` countOccurrence (tup , lst): ` ` return ` ` np .in1d ​​(tup, lst). ` ` sum ` ` () `   ` Driver code ` ` tup ` ` = ` ` (` ` ’a’ ` `, < / code> ’a’ , ’ c’ , ’b’ , ’ d’ ) `` lst = [ ’a’ , ’ b’ ] print (countOccurrence (tup, lst)) `

Exit:

` 3 `

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