Given a String list, extract frequency of specific characters in the whole strings list.
Input : test_list = [“neveropen is best for Lazyroar”], chr_list = [‘e’, ‘b’, ‘g’, ‘f’]
Output : {‘g’: 3, ‘e’: 7, ‘b’: 1, ‘f’ : 2}
Explanation : Frequency of certain characters extracted.Input : test_list = [“neveropen”], chr_list = [‘e’, ‘g’]
Output : {‘g’: 2, ‘e’: 4}
Explanation : Frequency of certain characters extracted.
Method #1 : Using join() + Counter()
In this, we concatenate all the strings, and then Counter() performs task of getting all frequencies. Last step is to get only specific characters from List in dictionary using dictionary comprehension.
Python3
# Python3 code to demonstrate working of # Specific Characters Frequency in String List# Using join() + Counter()from collections import Counter# initializing liststest_list = ["neveropen is best for Lazyroar"]# printing original listprint("The original list : " + str(test_list))# char list chr_list = ['e', 'b', 'g']# dict comprehension to retrieve on certain Frequenciesres = {key:val for key, val in dict(Counter("".join(test_list))).items() if key in chr_list} # printing result print("Specific Characters Frequencies : " + str(res)) |
The original list : ['neveropen is best for Lazyroar']
Specific Characters Frequencies : {'g': 3, 'e': 7, 'b': 1}
Time complexity: O(n), where n is the length of the input string.
Auxiliary space: O(k), where k is the number of characters in the character list (chr_list).
Method #2 : Using chain.from_iterable() + Counter() + dictionary comprehension
In this, task of concatenation is done using chain.from_iterable() rather than join(). Rest all tasks are done as above method.
Python3
# Python3 code to demonstrate working of # Specific Characters Frequency in String List# Using chain.from_iterable() + Counter() + dictionary comprehensionfrom collections import Counterfrom itertools import chain# initializing liststest_list = ["neveropen is best for Lazyroar"]# printing original listprint("The original list : " + str(test_list))# char list chr_list = ['e', 'b', 'g']# dict comprehension to retrieve on certain Frequencies# from_iterable to flatten / joinres = {key:val for key, val in dict(Counter(chain.from_iterable(test_list))).items() if key in chr_list} # printing result print("Specific Characters Frequencies : " + str(res)) |
The original list : ['neveropen is best for Lazyroar']
Specific Characters Frequencies : {'g': 3, 'e': 7, 'b': 1}
Time complexity: O(n), where n is the length of the input string
Auxiliary space: O(k), where k is the length of the character list.
Method #3: Using count() method.count() method returns the number of times a particular element occurs in a sequence.
Python3
# Python3 code to demonstrate working of# Specific Characters Frequency in String List# initializing liststest_list = ["neveropen is best for Lazyroar"]# printing original listprint("The original list : " + str(test_list))# char listchr_list = ['e', 'b', 'g']d=dict()for i in chr_list: d[i]=test_list[0].count(i)res=d # printing resultprint("Specific Characters Frequencies : " + str(res)) |
The original list : ['neveropen is best for Lazyroar']
Specific Characters Frequencies : {'e': 7, 'b': 1, 'g': 3}
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #4: Using operator.countOf() method
Python3
# Python3 code to demonstrate working of# Specific Characters Frequency in String Listimport operator as op# initializing liststest_list = ["neveropen is best for Lazyroar"]# printing original listprint("The original list : " + str(test_list))# char listchr_list = ['e', 'b', 'g']d=dict()for i in chr_list: d[i]=op.countOf(test_list[0],i)res=d # printing resultprint("Specific Characters Frequencies : " + str(res)) |
The original list : ['neveropen is best for Lazyroar']
Specific Characters Frequencies : {'e': 7, 'b': 1, 'g': 3}
Time Complexity: O(N), where n is the length of the given string
Auxiliary Space: O(N)
Method #5: Using loop and conditional statement
- Initialize the list of strings test_list with a single string.
- Print the original list test_list.
- Initialize the list of characters chr_list whose frequency we want to count in the string.
- Initialize an empty dictionary res to store the result.
- Join all the strings in test_list into a single string using the join() method and store it in a variable joined_str.
- Loop through each character in joined_str and count its frequency if it is present in the chr_list.
- If a character is present in the chr_list and also present in the res dictionary, increment its count by 1. If a character is present in the chr_list but not in the res dictionary, add it to the dictionary with a count of 1.
- Print the result dictionary res which will contain the frequencies of the specific characters in the original string.
Python3
# initializing liststest_list = ["neveropen is best for Lazyroar"]# printing original listprint("The original list : " + str(test_list))# char listchr_list = ['e', 'b', 'g']# initializing dictionary for resultres = {}# loop through each character in the test_list and count their frequencyfor char in "".join(test_list): if char in chr_list: if char in res: res[char] += 1 else: res[char] = 1 # printing resultprint("Specific Characters Frequencies : " + str(res)) |
The original list : ['neveropen is best for Lazyroar']
Specific Characters Frequencies : {'g': 3, 'e': 7, 'b': 1}
Time complexity: O(n), where n is the length of the string in the test_list.
Auxiliary space: O(n), since the result dictionary will have at most n/3 key-value pairs if all the characters in the string are from the chr_list.
