Python – Extract digits from Tuple list
Sometimes, while working with Python lists, we can have a problem in which we need to perform extraction of all the digits from tuple list. This kind of problem can find its application in data domains and day-day programming. Let’s discuss certain ways in which this task can be performed.
Input : test_list = [(15, 3), (3, 9)]
Output : [9, 5, 3, 1]Input : test_list = [(15, 3)]
Output : [5, 3, 1]
Method #1: Using map() + chain.from_iterable() + set() + loop
The combination of above functions can be used to solve this problem. In this, we perform the task of flattening list using chain.from_iterable(), and then the digits are extracted using brute method. set() is used to remove duplicate digits.
Python3
# Python3 code to demonstrate working of # Extract digits from Tuple list # Using map() + chain.from_iterable() + set() + loop from itertools import chain # initializing list test_list = [( 15 , 3 ), ( 3 , 9 ), ( 1 , 10 ), ( 99 , 2 )] # printing original list print ( "The original list is : " + str (test_list)) # Extract digits from Tuple list # Using map() + chain.from_iterable() + set() + loop temp = map ( lambda ele: str (ele), chain.from_iterable(test_list)) res = set () for sub in temp: for ele in sub: res.add(ele) # printing result print ( "The extracted digits : " + str (res)) |
The original list is : [(15, 3), (3, 9), (1, 10), (99, 2)] The extracted digits : {'1', '0', '3', '2', '9', '5'}
Time Complexity: O(n*m), where n is the length of the input list and m is the maximum length of any tuple element in the list.
Auxiliary Space: O(k), where k is the number of unique digits in the input list. This is because the set data structure is used to store the extracted digits.
Method #2: Using regex expression
This is yet another way in which this task can be performed. In this, an appropriate regex expression is used to extract the required unique digits.
Python3
# Python3 code to demonstrate working of # Extract digits from Tuple list # Using regex expression import re # initializing list test_list = [( 15 , 3 ), ( 3 , 9 ), ( 1 , 10 ), ( 99 , 2 )] # printing original list print ( "The original list is : " + str (test_list)) # Extract digits from Tuple list # Using regex expression temp = re.sub(r '[\[\]\(\), ]' , '', str (test_list)) res = [ int (ele) for ele in set (temp)] # printing result print ( "The extracted digits : " + str (res)) |
The original list is : [(15, 3), (3, 9), (1, 10), (99, 2)] The extracted digits : [5, 9, 2, 0, 1, 3]
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(n), where n is the length of the input list.
Method #3: Using list(),str(),map(),set() methods .
Initially converted all elements of tuple to string and concatenated them.Later used set() method to remove the duplicates, converted string elements to integer elements and finally converted them to list datatype.
Python3
# Python3 code to demonstrate working of # Extract digits from Tuple list # initializing list test_list = [( 15 , 3 ), ( 3 , 9 ), ( 1 , 10 ), ( 99 , 2 )] # printing original list print ( "The original list is : " + str (test_list)) x = "" # Extract digits from Tuple list for i in test_list: for j in i: x + = str (j) res = list ( map ( int , set (x))) # printing result print ( "The extracted digits : " + str (res)) |
The original list is : [(15, 3), (3, 9), (1, 10), (99, 2)] The extracted digits : [2, 3, 0, 1, 9, 5]
Method#4: Using list Comprehension
Python3
# Initializing list test_list = [( 15 , 3 ), ( 3 , 9 ), ( 1 , 10 ), ( 99 , 2 )] # Printing original list print ( "The original list is : " + str (test_list)) # Extracting digits from Tuple list using list comprehensions temp = ''.join([ str (i) for sublist in test_list for i in sublist]) result = set (temp) result = [ int (i) for i in result] # Printing result print ( "The extracted digits : " + str ( list (result))) #This code is contributed by Vinay Pinjala. |
The original list is : [(15, 3), (3, 9), (1, 10), (99, 2)] The extracted digits : ['0', '9', '1', '3', '5', '2']
Time complexity : O(n)
Auxiliary Space : O(n)
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