Python | Associating a single value with all list items

Sometimes we come across a utility in which we have a list and we wish to associate with it any one of the given values. This can occur in many phases of programming and knowing the shorthands to it can be useful. Let’s discuss certain ways in which this can be done.

Associating a single value with all list items Using map() + lambda

This task can be done using the map() function which is an inbuilt Python function that is generally used to associate or aggregate values. The lambda function can feed a particular value to the map function for its execution. 

Python3

# Python3 code to demonstrate
# associate value in list
# using map() + lambda

# initializing list
test_list = [1, 4, 5, 8, 3, 10]

# initializing value to associate
val = 'geeks'

# printing the original list
print("The original list is : " + str(test_list))

# printing value
print("The value to be attached to each value : " + str(val))

# using map() + lambda
# associate value in list
res = list(map(lambda i: (i, val), test_list))

# printing result
print("The modified attached list is : " + str(res))

Output

The original list is : [1, 4, 5, 8, 3, 10]
The value to be attached to each value : geeks
The modified attached list is : [(1, 'geeks'), (4, 'geeks'), (5, 'geeks'), (8, 'geeks'), (3, 'geeks'), (10, 'geeks')]

Time complexity: O(n) where n is the length of the list “test_list”.
Auxiliary space: O(n) where n is the length of the list “res”.

Associating a single value with all list items Using zip() + itertools.repeat()

The zip function can be used to attach the required value with the elements in a sequence and repeat function can be used to extend the task to all the list elements in more efficient manner. 

Python3

# Python3 code to demonstrate
# associate value in list
# using zip() + itertools.repeat()
from itertools import repeat

# initializing list
test_list = [1, 4, 5, 8, 3, 10]

# initializing value to associate
val = 'geeks'

# printing the original list
print("The original list is : " + str(test_list))

# printing value
print("The value to be attached to each value : " + str(val))

# using zip() + itertools.repeat()
# associate value in list
res = list(zip(test_list, repeat(val)))

# printing result
print("The modified attached list is : " + str(res))

Output

The original list is : [1, 4, 5, 8, 3, 10]
The value to be attached to each value : geeks
The modified attached list is : [(1, 'geeks'), (4, 'geeks'), (5, 'geeks'), (8, 'geeks'), (3, 'geeks'), (10, 'geeks')]

Time Complexity: O(n) where n is the length of the list ‘test_list’.
Auxiliary Space: O(n) as we are creating a new list ‘res’ of length n.

Associating a single value with all list items Using for loop and tuple() method

Python3

# Python3 code to demonstrate
# associate value in list

# initializing list
test_list = [1, 4, 5, 8, 3, 10]

# initializing value to associate
val = 'geeks'

# printing the original list
print("The original list is : " + str(test_list))

# printing value
print("The value to be attached to each value : " + str(val))
res = []
for i in test_list:
    x = [i, "geeks"]
    res.append(tuple(x))


# printing result
print("The modified attached list is : " + str(res))

Output

The original list is : [1, 4, 5, 8, 3, 10]
The value to be attached to each value : geeks
The modified attached list is : [(1, 'geeks'), (4, 'geeks'), (5, 'geeks'), (8, 'geeks'), (3, 'geeks'), (10, 'geeks')]

The time complexity of this code is O(n), where n is the number of elements in the list “test_list”.

The auxiliary space complexity is O(n), as a new list “res” with n elements is created. 

Python3

test_list = [1, 4, 5, 8, 3, 10]
val = 'geeks'

# Use zip() and a generator expression to create a list of tuples with the elements
# of test_list and a sequence of val values.
res = list(zip(test_list, (val for _ in test_list)))

# Print the original list and value
print("The original list is :", test_list)
print("The value to be attached to each value :", val)

# Print the modified attached list
print("The modified attached list is :", res)
#This code is contributed by Edula Vinay Kumar Reddy

Output

The original list is : [1, 4, 5, 8, 3, 10]
The value to be attached to each value : geeks
The modified attached list is : [(1, 'geeks'), (4, 'geeks'), (5, 'geeks'), (8, 'geeks'), (3, 'geeks'), (10, 'geeks')]

The zip() function combines the elements of two iterables into tuples, and the generator expression (val for _ in test_list) generates a sequence of vals that is as long as the list. The resulting tuples are then collected into a list using the list() function.

Time complexity: O(n), where n is the length of the list. This is because the zip() function and generator expression only need to iterate over the list once.
Auxiliary space: O(n), because the resulting list will have n elements.

Associating a single value with all list items Using list comprehension

Step-by-step approach:

• Initialize the list test_list and the value to associate val.
• Use a list comprehension to iterate over each element i in test_list and create a tuple (i, val).
• The resulting list of tuples is assigned to the variable res.
• Print the modified list res.

Below is the implementation of the above approach:

Python3

test_list = [1, 4, 5, 8, 3, 10]
val = 'geeks'

res = [(i, val) for i in test_list]

print("The modified attached list is : " + str(res))

Output

The modified attached list is : [(1, 'geeks'), (4, 'geeks'), (5, 'geeks'), (8, 'geeks'), (3, 'geeks'), (10, 'geeks')]

Time Complexity: O(n) as we iterate through the list once.
Auxiliary Space: O(n) as we create a new list of tuples.

Associating a single value with all list items Using numpy library

Python3

import numpy as np

test_list = [1, 4, 5, 8, 3, 10]
val = 'geeks'

# convert list to numpy array
arr = np.array(test_list)

# create numpy array with same length as input list, and fill with given string value
str_arr = np.full((len(test_list),), val)

# horizontally stack both arrays
res_arr = np.column_stack((arr, str_arr))

# convert resulting numpy array to list
res_list = res_arr.tolist()

# print the resulting list
print("The modified attached list is : " + str(res_list))

Output

The modified attached list is : [['1', 'geeks'], ['4', 'geeks'], ['5', 'geeks'], ['8', 'geeks'], ['3', 'geeks'], ['10', 'geeks']]

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, because we are creating a new numpy array and a new list.

Associating a single value with all list items Using dictionary comprehension

Python3

# initializing list
test_list = [1, 4, 5, 8, 3, 10]

# initializing value to associate
val = 'geeks'

# printing the original list
print ("The original list is : " + str(test_list))

# printing value
print ("The value to be attached to each value : " +  str(val))

# using dictionary comprehension
# associate value in list
res = {i: val for i in test_list}

# converting dictionary to list of tuples
res = list(res.items())

# printing result
print ("The modified attached list is : " + str(res))

Output

The original list is : [1, 4, 5, 8, 3, 10]
The value to be attached to each value : geeks
The modified attached list is : [(1, 'geeks'), (4, 'geeks'), (5, 'geeks'), (8, 'geeks'), (3, 'geeks'), (10, 'geeks')]

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)