Python | Construct string from character frequency tuple

Last Updated : 09 Apr, 2023

Sometimes, while working with data, we can have a problem in which we need to perform construction of string in a way that we have a list of tuple having character and it's corresponding frequency and we require to construct a new string from that. Let's discuss certain ways in which this task can be performed.

Method #1 : Using loop This is brute force method in which this task can be performed. In this, we iterate the list and perform string concatenation using * operator and keep building string this way.

Python3

# Python3 code to demonstrate working of
# String construction from character frequency
# using loop

# initialize list
test_list = [('g', 4), ('f', 3), ('g', 2)]

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

# String construction from character frequency
# using loop
res = ''
for char, freq in test_list:
    res = res + char * freq

# printing result
print("The constructed string is : " + str(res))

Output :

The original list : [('g', 4), ('f', 3), ('g', 2)]
The constructed string is : ggggfffgg

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #2 : Using join() + list comprehension The combination of above functionalities can be used to perform this task. In this, we perform the task of extraction using list comprehension and making string using join().

Python3

# Python3 code to demonstrate working of
# String construction from character frequency
# using join() + list comprehension

# initialize list
test_list = [('g', 4), ('f', 3), ('g', 2)]

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

# String construction from character frequency
# using join() + list comprehension
res = ''.join(char * freq for char, freq in test_list)

# printing result
print("The constructed string is : " + str(res))

Output :

The original list : [('g', 4), ('f', 3), ('g', 2)]
The constructed string is : ggggfffgg

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #3 : Using reduce()

Python3

# Python3 code to demonstrate working of
# String construction from character frequency
# using reduce()
 
# importing functools for reduce()
import functools
 
# initialize list
test_list = [('g', 4), ('f', 3), ('g', 2)]
 
# printing original list
print("The original list : " + str(test_list))
 
# String construction from character frequency
# using reduce()
res = functools.reduce(lambda x, y : x + y[0] * y[1], test_list, '')
 
# printing result
print("The constructed string is : " + str(res))
#this code is contributed by edula vinay kumar reddy

Output

The original list : [('g', 4), ('f', 3), ('g', 2)]
The constructed string is : ggggfffgg

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #4: Using the map() function and lambda function

Python3

# Initialize the character frequency tuple
char_freq = (('g', 4), ('f', 3), ('g', 2))

# Use the map function to repeat the characters
my_list = list(map(lambda x: x[0]*x[1], char_freq))

# Use the join method to convert the list to a string
my_string = "".join(my_list)

# Print the constructed string
print(my_string)
# This code is contributed by Vinay Pinjala.

Output

ggggfffgg

Time Complexity: O(n)
Auxiliary Space: O(n)

Method 5: Use the itertools module.

Step-by-step approach:

Import the itertools module.
Use a list comprehension with the itertools.repeat() function to repeat each character by the given frequency.
Use the itertools.chain.from_iterable() function to combine the repeated character strings into a single iterable.
Use the str.join() method to convert the iterable into a string.
Print the constructed string.

Below is the implementation of the above approach:

Python3

import itertools

# Initialize the character frequency tuple
char_freq = (('g', 4), ('f', 3), ('g', 2))

# Use itertools to repeat the characters and combine them into a single iterable
my_iterable = itertools.chain.from_iterable(itertools.repeat(c, f) for c, f in char_freq)

# Use join to convert the iterable into a string
my_string = ''.join(my_iterable)

# Print the constructed string
print(my_string)

Output

ggggfffgg

Time complexity: O(n), where n is the total number of characters to be repeated.
Auxiliary space: O(n), where n is the total number of characters to be repeated, since we need to store the repeated character strings in memory before joining them into a single iterable.

Method 6: Using the numpy.repeat() function along with the join() function:

Steps:

Import the numpy module.
Convert the char_freq tuple into a numpy array using the numpy.array() function.
Extract the characters and their frequencies into separate arrays using indexing.
Use the numpy.repeat() function to repeat each character by its frequency.
Concatenate the repeated characters using the numpy.concatenate() function.
Convert the concatenated array into a string using the join() function.
Print the constructed string.

Below is the implementation of the above approach:

Python3

import numpy as np

char_freq = (('g', 4), ('f', 3), ('g', 2))

# Convert the char_freq tuple
# into a numpy array
char_freq_arr = np.array(char_freq)

# Extract the characters and their
# frequencies into separate arrays
chars = char_freq_arr[:, 0]
freqs = char_freq_arr[:, 1].astype(int)

# Repeat each character by its frequency
# using numpy.repeat()
repeated_chars = np.repeat(chars, freqs)

# Concatenate the repeated characters
# using numpy.concatenate()
concatenated_chars = np.concatenate([repeated_chars])

# Convert the concatenated array into
# a string using join()
my_string = ''.join(concatenated_chars)

# Print the constructed string
print(my_string)

Output:

ggggfffgg

Time Complexity: O(N), where N is the total number of characters in the constructed string.
Auxiliary Space: O(N), where N is the total number of characters in the constructed string.