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
test_list = [( 'g' , 4 ), ( 'f' , 3 ), ( 'g' , 2 )]
print ( "The original list : " + str (test_list))
res = ''
for char, freq in test_list:
res = res + char * freq
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
test_list = [( 'g' , 4 ), ( 'f' , 3 ), ( 'g' , 2 )]
print ( "The original list : " + str (test_list))
res = ''.join(char * freq for char, freq in test_list)
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
import functools
test_list = [( 'g' , 4 ), ( 'f' , 3 ), ( 'g' , 2 )]
print ( "The original list : " + str (test_list))
res = functools. reduce ( lambda x, y : x + y[ 0 ] * y[ 1 ], test_list, '')
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 #4: Using the map() function and lambda function
Python3
char_freq = (( 'g' , 4 ), ( 'f' , 3 ), ( 'g' , 2 ))
my_list = list ( map ( lambda x: x[ 0 ] * x[ 1 ], char_freq))
my_string = "".join(my_list)
print (my_string)
|
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
char_freq = (( 'g' , 4 ), ( 'f' , 3 ), ( 'g' , 2 ))
my_iterable = itertools.chain.from_iterable(itertools.repeat(c, f) for c, f in char_freq)
my_string = ''.join(my_iterable)
print (my_string)
|
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 ))
char_freq_arr = np.array(char_freq)
chars = char_freq_arr[:, 0 ]
freqs = char_freq_arr[:, 1 ].astype( int )
repeated_chars = np.repeat(chars, freqs)
concatenated_chars = np.concatenate([repeated_chars])
my_string = ''.join(concatenated_chars)
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.
Share your thoughts in the comments
Please Login to comment...