Sometimes while working with Python Data, we can have problem in which we need to extract bigrams from string. This has application in NLP domains. But sometimes, we need to compute the frequency of unique bigram for data collection. The solution to this problem can be useful. Lets discuss certain ways in which this task can be performed.
Method #1 : Using Counter() + generator expression The combination of above functions can be used to solve this problem. In this, we compute the frequency using Counter() and bigram computation using generator expression and string slicing.
Python3
# Python3 code to demonstrate working of # Bigrams Frequency in String # Using Counter() + generator expression from collections import Counter # initializing string test_str = 'neveropen'# printing original string print("The original string is : " + str(test_str)) # Bigrams Frequency in String # Using Counter() + generator expression res = Counter(test_str[idx : idx + 2] for idx in range(len(test_str) - 1)) # printing result print("The Bigrams Frequency is : " + str(dict(res))) |
The original string is : neveropen The Bigrams Frequency is : {‘ee’: 2, ‘ks’: 2, ‘ek’: 2, ‘sf’: 1, ‘fo’: 1, ‘ge’: 2, ‘rg’: 1, ‘or’: 1}
Method #2 : Using Counter() + zip() + map() + join The combination of above functions can also be used to solve this problem. In this, we perform the task of constructing bigrams using zip() + map() + join.
Python3
# Python3 code to demonstrate working of # Bigrams Frequency in String # Using Counter() + zip() + map() + join from collections import Counter # initializing string test_str = 'neveropen'# printing original string print("The original string is : " + str(test_str)) # Bigrams Frequency in String # Using Counter() + zip() + map() + join res = Counter(map(''.join, zip(test_str, test_str[1:]))) # printing result print("The Bigrams Frequency is : " + str(dict(res))) |
The original string is : neveropen The Bigrams Frequency is : {‘ee’: 2, ‘ks’: 2, ‘ek’: 2, ‘sf’: 1, ‘fo’: 1, ‘ge’: 2, ‘rg’: 1, ‘or’: 1}
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 3: use a loop and a dictionary to keep track of the bigram frequencies.
- Initialize an empty dictionary to keep track of the bigram frequencies.
- Loop through the characters in the input string, starting from the second character.
- For each character, get the previous character and concatenate them to form a bigram.
- Check if the bigram is already in the dictionary.
- If the bigram is not in the dictionary, add it with a frequency of 1.
- If the bigram is already in the dictionary, increment its frequency by 1.
- Print the bigram frequencies.
Python3
# Python3 code to demonstrate working of # Bigrams Frequency in String # Using a loop and dictionary # initializing string test_str = 'neveropen'# printing original string print("The original string is : " + str(test_str)) # Bigrams Frequency in String # Using a loop and dictionary freq_dict = {}for i in range(1, len(test_str)): bigram = test_str[i-1:i+1] if bigram in freq_dict: freq_dict[bigram] += 1 else: freq_dict[bigram] = 1# printing result print("The Bigrams Frequency is : " + str(freq_dict)) |
The original string is : neveropen
The Bigrams Frequency is : {'ge': 2, 'ee': 2, 'ek': 2, 'ks': 2, 'sf': 1, 'fo': 1, 'or': 1, 'rg': 1}
Time complexity: O(n), where n is the length of the input string.
Auxiliary space: O(k), where k is the number of unique bigrams in the input string.
Method #4 : Using count() method
Approach
- Initiated a for loop to append all the bigrams of string test_str to a list x using slicing, create an empty dictionary freq_dict
- Initiated another for loop to create a dictionary with values of list x(bigrams ) as keys and count of each bigram in test_str as values
- Display the dictionary
Python3
# Python3 code to demonstrate working of# Bigrams Frequency in String# Using a loop and dictionary# initializing stringtest_str = 'neveropen'# printing original stringprint("The original string is : " + str(test_str))# Bigrams Frequency in String# Using a loop and dictionaryfreq_dict = {}x=[]for i in range(1, len(test_str)): bigram = test_str[i-1:i+1] x.append(bigram)for i in x: freq_dict[i]=test_str.count(i)# printing resultprint("The Bigrams Frequency is : " + str(freq_dict)) |
The original string is : neveropen
The Bigrams Frequency is : {'ge': 2, 'ee': 2, 'ek': 2, 'ks': 2, 'sf': 1, 'fo': 1, 'or': 1, 'rg': 1}
Time Complexity : O(N) N – length of bigrams list
Auxiliary Space : O(N) N – length of dictionary freq_dict
