Python | Insert the string at the beginning of all items in a list

Given a list, write a Python program to insert some string at the beginning of all items in that list. 

Examples:

Input : list = [1, 2, 3, 4], str = 'Geek'
Output : list = ['Geek1', 'Geek2', 'Geek3', 'Geek4']

Input : list = ['A', 'B', 'C'], str = 'Team'
Output : list = ['TeamA', 'TeamB', 'TeamC']

There are multiple ways to insert the string at the beginning of all items in a list. 

Approach #1: Using list comprehension List comprehension is an elegant way to define and create list. It can also be used to apply an expression to each element in a sequence. We can use format() function which allows multiple substitutions and value formatting. 

Python3

# Python3 program to insert the string
# at the beginning of all items in a list


def prepend(list, str):

    # Using format()
    str += '{0}'
    list = [str.format(i) for i in list]
    return(list)


# Driver function
list = [1, 2, 3, 4]
str = 'Geek'
print(prepend(list, str))

Output

['Geek1', 'Geek2', 'Geek3', 'Geek4']

Time Complexity: O(n)
Auxiliary Space: O(n), where n is length of list.

Another method in list comprehension is to use ‘%’ instead of format() function 

Python3

# Using '% s'
def prepend(list, str):
    str += '% s'
    list = [str % i for i in list]


list = [1, 2, 3, 4]
str = 'Geek'
print(prepend(list, str))

Approach #2 : Using in-built map() function

Another approach is to use map() function. The function maps the beginning of all items in the list to the string. 

Python3

# Python3 program to insert the string
# at the beginning of all items in a list


def prepend(List, str):

    # Using format()
    str += '{0}'
    List = ((map(str.format, List)))
    return List


# Driver function
list = [1, 2, 3, 4]
str = 'Geek'
print(prepend(list, str))

Output

<map object at 0x7f9f83334dd0>

They are as follows: 

['Geek1', 'Geek2', 'Geek3', 'Geek4']

Approach #3 :  Using a for loop and + operator.

Here is another approach using a for loop to iterate through the elements in the list and concatenating the string and each element using the + operator.

Here is an example of how this approach could be implemented:

Python3

def prepend(lst, s):
    # Create an empty result list
    res = []
    # Iterate through the elements in the list
    for x in lst:
        # Concatenate s and x and append the result to the result list
        res.append(s + str(x))
    # Return the result list
    return res

# Test the function
lst = [1, 2, 3, 4]
s = 'Geek'
print(prepend(lst, s))  # Output: ['Geek1', 'Geek2', 'Geek3', 'Geek4']
#This code is contributed by Edula Vinay Kumar Reddy

Output

['Geek1', 'Geek2', 'Geek3', 'Geek4']

The time complexity is O(n), where n is the length of the list. This is because approach involve iterating through the elements in the list once. The space complexity is O(n), because it only creates a single list of size n to store the result.