 # Python | time.get_clock_info () method

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Method # 1: Using a Loop
This is a brute force method to accomplish this particular task. In this we repeat each element, check for its presence in another list, if so, then increase its number and put it in a tuple.

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``` # Python3 code to demonstrate how it works # Frequency of grouping list items # using a loop   # initialize the list test_list = [ 1 , 3 , 3 , 1 , 4 , 4 ] < p>   # print original list print ( "The original list:" + str (test_list))   # Frequency of grouping list items # using a loop res = [] temp = dict () for ele in test_list : if ele in temp: temp [ele] = temp [ele] + 1   else :  temp [ele] = 1 for key in temp:   res.append ((key, temp [key]))   # print result print ( "Frequency of list elements:" + str (res)) ```

` ` Output:

` The original list: [1, 3, 3, 1, 4, 4] Frequency of list elements: [(1, 2), (3, 2), (4, 2)] `

Method # 2: Using ` Counter () + items () `
A combination of the two functions can be used to accomplish this task. They accomplish this task using built-in constructs and are shorthand for this task.

 ` # Python3 code to demonstrate how it works ` ` # Frequency of grouping list items ` ` # using Counter () + items () ` ` from ` ` collections ` ` import ` ` Counter `   ` # initialize the list ` ` test_list ` ` = ` ` [` ` 1 ` `, ` ` 3 ` `, ` ` 3 ` `, ` ` 1 ` , ` 4 ` `, ` ` 4 ] ````   # print the original list print ( "The original list: " + str (test_list))   # Frequency of grouping list items # using Counter () + items () res = list (Counter (test_list) .items ())    # print result print (< / code> "Frequency of list elements:" + str (res)) ```

Output:

``` The original list: [1, 3, 3, 1, 4, 4] Frequency of list elements: [(1, 2), (3, 2), (4, 2)]

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