Given a list of tuples, the task is to convert it into list of all digits which exists in elements of list. Let’s discuss certain ways in which this task is performed.
Method #1: Using re The most concise and readable way to convert list of tuple into list of all digits which exists in elements of list is by using re.
Python3
# Python code to convert list of tuples into # list of all digits which exists # in elements of list. # Importing import re # Input list initialization lst = [( 11 , 100 ), ( 22 , 200 ), ( 33 , 300 ), ( 44 , 400 ), ( 88 , 800 )] # Using re temp = re.sub(r '[\[\]\(\), ]' , '', str (lst)) # Using set Output = [ int (i) for i in set (temp)] # Printing output print ( "Initial List is :" , lst) print ( "Output list is :" , Output) |
Initial List is : [(11, 100), (22, 200), (33, 300), (44, 400), (88, 800)] Output list is : [2, 1, 8, 4, 3, 0]
Method #2: Using itertools.chain() and lambda() This is yet another way to perform this particular task using lambda().
Python3
# Python code to convert list of tuples into # list of all digits which exists # in elements of list. # Importing from itertools import chain # Input list initialization lst = [( 11 , 100 ), ( 22 , 200 ), ( 33 , 300 ), ( 44 , 400 ), ( 88 , 800 )] # using lambda temp = map ( lambda x: str (x), chain.from_iterable(lst)) # Output list initialization Output = set () # Adding element in Output for x in temp: for elem in x: Output.add(elem) # Printing output print ( "Initial List is :" , lst) print ( "Output list is :" , Output) |
Initial List is : [(11, 100), (22, 200), (33, 300), (44, 400), (88, 800)] Output list is : {'4', '3', '2', '1', '8', '0'}
Method #3: Using list(),map(),join() and set() methods
Initially we convert list containing tuple of integers to list containing tuple of strings.Later we will concatenate the string(obtained by joining tuple strings) to an empty string.And then use set to remove duplicates.Finally convert them to integer type and print output list
Python3
# Python code to convert list of tuples into # list of all digits which exists # in elements of list. # Input list initialization lst = [( 11 , 100 ), ( 22 , 200 ), ( 33 , 300 ), ( 44 , 400 ), ( 88 , 800 )] p = "" for i in lst: x = list ( map ( str ,i)) p + = "".join(x) p = list ( map ( int , set (p))) # Printing output print ( "Initial List is :" , lst) print ( "Output list is :" , p) |
Initial List is : [(11, 100), (22, 200), (33, 300), (44, 400), (88, 800)] Output list is : [8, 3, 2, 4, 0, 1]
Time complexity: O(n), where n is the length of list
Auxiliary Space: O(m), where m is the number of digits.
Method #4: Using list comprehension and the isdigit method
You can use a list comprehension and the isdigit method to extract all the digits from the tuples in the list. The isdigit method returns True if a character is a digit, and False otherwise.
Python3
# Initialize the list lst = [( 11 , 100 ), ( 22 , 200 ), ( 33 , 300 ), ( 44 , 400 ), ( 88 , 800 )] # Extract the digits using a list comprehension and the isdigit method digits = set () # Print the resulting list print ( "The list of digits:" , digits) #This code is contributed by Edula Vinay Kumar Reddy |
The list of digits: {'0', '8', '1', '2', '3', '4'}
Time complexity is O(n)
Auxiliary Space is O(m), where n is the length of the list and m is the number of digits.
Method #5: Using nested loops and string conversion
Step-by-step approach:
- Initialize the input list lst with the given values.
- Initialize an empty list digits to store the extracted digits.
- Loop through each tuple tup in the input list.
- Loop through each item in the current tuple.
- Loop through each character char in the string representation of the current item.
- Check if the current character is a digit using the isdigit() method.
- If it is a digit, convert it to an integer using the int() function and append it to the digits list.
- Check if the current character is a digit using the isdigit() method.
- Use the set() function to remove duplicates from the digits list and convert it back to a list.
- Print the initial list and the output list.
Below is the implementation of the above approach:
Python3
# Input list initialization lst = [( 11 , 100 ), ( 22 , 200 ), ( 33 , 300 ), ( 44 , 400 ), ( 88 , 800 )] # Using nested loops and string conversion digits = [] for tup in lst: for item in tup: for char in str (item): if char.isdigit(): digits.append( int (char)) # Using set to remove duplicates Output = list ( set (digits)) # Printing output print ( "Initial List is :" , lst) print ( "Output list is :" , Output) |
Initial List is : [(11, 100), (22, 200), (33, 300), (44, 400), (88, 800)] Output list is : [0, 1, 2, 3, 4, 8]
Time complexity: O(n*m), where n is the length of the input list and m is the maximum length of the items in the tuples (assuming integer items only).
Auxiliary space: O(k), where k is the total number of digits in all the items in the input list.
Method #6: Using numpy.ravel():
- Import the numpy library.
- Initialize a list of tuples called last.
- Use numpy’s ravel() method to convert the list of tuples to a 1-D numpy array called arr.
- Use a list comprehension and the isdigit() method to extract the digits from the array and convert them to integers.
- Use set() to remove any duplicates in the list of digits.
- Convert the set back to a list called digits.
- Print the original list and the resulting list.
Python3
import numpy as np # Initialize the list lst = [( 11 , 100 ), ( 22 , 200 ), ( 33 , 300 ), ( 44 , 400 ), ( 88 , 800 )] # Convert the list of tuples to a 1-D numpy array using ravel() method arr = np.array(lst).ravel() # Extract the digits using a list comprehension and the isdigit method digits = [ int (d) for d in ''.join( map ( str , arr)) if d.isdigit()] # Remove duplicates using set digits = list ( set (digits)) # Print the resulting list print ( "Initial List is :" , lst) print ( "Output list is :" , digits) |
Output
Initial List is : [(11, 100), (22, 200), (33, 300), (44, 400), (88, 800)] Output list is : [0, 1, 2, 3, 4, 8]
The time complexity of this code is O(n), where n is the total number of digits.
The space complexity of this code is O(n), where n is the total number of digits.