Given Strings with words, the task is to write a Python program to split each word into two halves on the basis of assigned percentages according to the given values.
Example:
Input : test_str = ‘neveropen is best for all Lazyroar and cs students’, per_splt = 50
Output : Lazyroarf orLazyroar i s be st f or a ll ge eks a nd c s stud ents
Explanation : Each word after splitting by 50 percent, result is output.Input : test_str = ‘neveropen is best for all Lazyroar and cs students’, per_splt = 70
Output : Lazyroarforg eeks i s be st fo r al l gee ks an d c s stude nts
Explanation : Each word after splitting by 70 percent, result is output.
Method 1: Using split() + len() + slice + join()
In this, we split each word and perform a percentage split of each word using len() and slicing. The result is joined in an intermediate fashion using a loop.
Python3
# Python3 code to demonstrate working of# Split each word into percent segment in list# Using split() + len() + slice + loop# initializing stringtest_str = 'neveropen is best for all Lazyroar and cs students'# printing original stringprint("The original string is : " + str(test_str))# initializing percent split per_splt = 50test_str = test_str.split()res = ''for ele in test_str: prop = int((per_splt/100) * len(ele)) new_str1 = ele[:prop] new_str2 = ele[prop:] res += new_str1 + " " + new_str2 + " " # printing resultprint("Segmented words : " + str(res)) |
Output:
The original string is : neveropen is best for all Lazyroar and cs students
Segmented words : Lazyroarf orLazyroar i s be st f or a ll ge eks a nd c s stud ents
Method 2: Using join()
Similar to the above method, the difference is that join() is used to concatenate the resultant string.
Python3
# Python3 code to demonstrate working of# Split each word into percent segment in list# Using join()# initializing stringtest_str = 'neveropen is best for all Lazyroar and cs students'# printing original stringprint("The original string is : " + str(test_str))# initializing percent split per_splt = 50test_str = test_str.split()# one liner solution using join()res = ' '.join([ele[:int((per_splt/100) * len(ele))] + " " + ele[int((per_splt/100) * len(ele)):] for ele in test_str])# printing resultprint("Segmented words : " + str(res)) |
Output:
The original string is : neveropen is best for all Lazyroar and cs students
Segmented words : Lazyroarf orLazyroar i s be st f or a ll ge eks a nd c s stud ents
Time Complexity: O(n)
Auxiliary Space: O(n)
Approach 3: Using regular expressions
In this approach, we use the power of regular expressions to split the words.
Python3
# Python3 code to demonstrate working of# Split each word into percent segment in list# Using regular expressions # importing moduleimport re # initializing stringtest_str = 'neveropen is best for all Lazyroar and cs students' # printing original stringprint("The original string is : " + str(test_str)) # initializing percent splitper_splt = 50 # one liner solution using join()res = ' '.join([re.sub(r'(\w{' + str(int((per_splt/100) * len(ele))) + r'})(\w+)', r'\1 \2', ele) for ele in test_str.split()]) # printing resultprint("Segmented words : " + str(res))#This code is contributed by Edula Vinay Kumar Reddy |
The original string is : neveropen is best for all Lazyroar and cs students Segmented words : Lazyroarf orLazyroar i s be st f or a ll ge eks a nd c s stud ents
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 4: Using the itertools module.
Step-by-step approach:
- Import the itertools module.
- Split the input string into words using the split() method and store it in a variable.
- Loop through each word in the variable created in step 2.
- Find the length of the word using the len() method and calculate the number of characters required for the segment using the percent split.
- Use the itertools module to split the word into segments of the desired length.
- Convert the iterator to a list and join the segments with a space in between.
- Append the segmented word to a list.
- Join the segmented words in the list with a space and return the segmented string.
Python3
# import itertools moduleimport itertools# initializing stringtest_str = 'neveropen is best for all Lazyroar and cs students'# initializing percent splitper_splt = 50# function to split each word into percent segmentdef segment_words(test_str, per_splt): # initialize list to store segmented words segmented_list = [] # split string into words words = test_str.split() # loop through each word for word in words: # calculate number of characters required for segment seg_length = int((per_splt / 100) * len(word)) # split word into segments of desired length using itertools segments = itertools.zip_longest(*[iter(word)]*seg_length, fillvalue='') # join segments with space in between segmented_word = ' '.join([''.join(segment) for segment in segments]) # append segmented word to list segmented_list.append(segmented_word) # join segmented words in list and return segmented string return ' '.join(segmented_list)# call function and print resultprint("Segmented words : " + segment_words(test_str, per_splt)) |
Segmented words : Lazyroarf orgeek s i s be st f o r a l l ge ek s a n d c s stud ents
Time complexity: O(nm), where n is the number of words in the input string and m is the length of the longest word.
Auxiliary space: O(nm), where n is the number of words in the input string and m is the length of the longest word.
Method 5: Use the textwrap module:
Step-by-step approach:
- Import the textwrap module.
- Define the segment_words function that takes test_str and per_splt as input.
- Initialize an empty list segmented_list to store the segmented words.
- Split the input string test_str into individual words and store them in words.
- Loop through each word in words.
- Calculate the desired segment length seg_length as a percentage of the length of the current word.
- Use the textwrap.wrap function to split the word into segments of desired length.
- Join the segments with a space in between to form the segmented word.
- Append the segmented word to segmented_list.
- Join the segmented words in segmented_list with a space in between and return the resulting segmented string.
Python3
import textwrapdef segment_words(test_str, per_splt): segmented_list = [] words = test_str.split() for word in words: seg_length = int((per_splt / 100) * len(word)) segments = textwrap.wrap(word, seg_length) segmented_word = ' '.join(segments) segmented_list.append(segmented_word) return ' '.join(segmented_list)test_str = 'neveropen is best for all Lazyroar and cs students'per_splt = 50segmented_str = segment_words(test_str, per_splt)print(segmented_str) |
Lazyroarf orgeek s i s be st f o r a l l ge ek s a n d c s stud ents
Time complexity: O(n*k), where n is the number of words in the input string and k is the length of the longest word in the input string.
Auxiliary space: O(n*k), where n is the number of words in the input string and k is the length of the longest word in the input string.
