Sometimes, while working with huge amount of data, we can have a problem in which we need to know count of just specific sized strings which are greater than a specific length. This kind of problem can occur during validation cases across many domains. Let’s discuss certain ways to handle this in Python strings list.
Method #1 : Using list comprehension + len()
The combination of above functionalities can be used to perform this task. In this, we iterate for all the strings and return only strings which have length greater than K checked using len(). The count is extracted using len().
Python3
# Python3 code to demonstrate working of# Threshold Size Greater Strings Frequency# using list comprehension + len()# initialize list test_list = ['gfg', 'is', 'best', 'for', 'neveropen']# printing original list print("The original list : " + str(test_list))# initialize K K = 3# Threshold Size Greater Strings Frequency# using list comprehension + len()res = len([ele for ele in test_list if len(ele) >= K])# printing resultprint("The frequency of threshold K sized strings are : " + str(res)) |
The original list : ['gfg', 'is', 'best', 'for', 'neveropen'] The frequency of threshold K sized strings are : 4
Method #2 : Using filter() + len() + lambda
The combination of above functionalities can be used to perform this task. In this, we extract the elements using filter() and logic is compiled in a lambda function. The count is extracted using len().
Python3
# Python3 code to demonstrate working of# Threshold Size Greater Strings Frequency# using filter() + lambda + len()# initialize list test_list = ['gfg', 'is', 'best', 'for', 'neveropen']# printing original list print("The original list : " + str(test_list))# initialize K K = 3# Threshold Size Greater Strings Frequency# using filter() + lambda + len()res = len(list(filter(lambda ele: len(ele) >= K, test_list)))# printing resultprint("The frequency of threshold K sized strings are : " + str(res)) |
The original list : ['gfg', 'is', 'best', 'for', 'neveropen'] The frequency of threshold K sized strings are : 4
Method #3: Using a for loop
- Initialize a counter variable to 0.
- Loop through each element in the list using a for loop.
- For each element, check if its length is greater than or equal to K.
- If it is, increment the counter variable.
- After the loop finishes, the value of the counter variable is the frequency of threshold K sized strings.
- Print the frequency.
Python3
test_list = ['gfg', 'is', 'best', 'for', 'neveropen']print("The original list : " + str(test_list))K = 3count = 0for ele in test_list: if len(ele) >= K: count += 1print("The frequency of threshold K sized strings are : " + str(count)) |
The original list : ['gfg', 'is', 'best', 'for', 'neveropen'] The frequency of threshold K sized strings are : 4
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(1), since we only use a counter variable.
Method 4: Using numpy
Python3
# Python3 code to demonstrate working of# Threshold Size Greater Strings Frequency# using numpyimport numpy as np# initialize list test_list = ['gfg', 'is', 'best', 'for', 'neveropen']# printing original list print("The original list : " + str(test_list))# initialize K K = 3# Threshold Size Greater Strings Frequency# using numpyres = np.sum(np.array([len(ele) >= K for ele in test_list]))# printing resultprint("The frequency of threshold K sized strings are : " + str(res)) |
Output:
The original list : ['gfg', 'is', 'best', 'for', 'neveropen'] The frequency of threshold K sized strings are : 4
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(n) auxiliary space to create the intermediate list of Boolean values.
