Python – Bigrams Frequency in String
Last Updated :
12 Apr, 2023
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
from collections import Counter
test_str = 'geeksforgeeks'
print("The original string is : " + str(test_str))
res = Counter(test_str[idx : idx + 2] for idx in range(len(test_str) - 1))
print("The Bigrams Frequency is : " + str(dict(res)))
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Output :
The original string is : geeksforgeeks 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
from collections import Counter
test_str = 'geeksforgeeks'
print("The original string is : " + str(test_str))
res = Counter(map(''.join, zip(test_str, test_str[1:])))
print("The Bigrams Frequency is : " + str(dict(res)))
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Output :
The original string is : geeksforgeeks 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
test_str = 'geeksforgeeks'
print("The original string is : " + str(test_str))
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
print("The Bigrams Frequency is : " + str(freq_dict))
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Output
The original string is : geeksforgeeks
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
test_str = 'geeksforgeeks'
print("The original string is : " + str(test_str))
freq_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)
print("The Bigrams Frequency is : " + str(freq_dict))
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Output
The original string is : geeksforgeeks
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
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