Python | Find frequency of given character at every position in list of lists

Given a list of lists, the task is to find the frequency of a character at every position of sub-list in list of lists.

Input : lst = [['X', 'Y', 'X'], ['Z', 'Y', 'X'],
               ['Y', 'Y', 'Y'], ['Z', 'Z', 'X'],
               ['Y', 'Z', 'X']], character = 'X'

Output: [0.2, 0.0, 0.8]

Explanation: We have 3 elements in each sublist, we have to find position of ‘X’ at position 0, 1 and 2. For Position 0 in all sublist we have – ‘x’ in first sub list at zero position, ‘z’ in second sub list at zero position, ‘y’ in third sub list at zero position, ‘z’ in fourth sub list at zero position and ‘y’ in fifth sub list at zero position. So, we have 1 occurrence of ‘x’ at position 1 in all sub list so, Occurrence = 1/5 = .2 For Position 1 we don’t have any occurrence of ‘x’ in sub list so, Occurrence = 0/5 = 0. For Position 2 we have 4 occurrence of ‘x’ in sub list so, Occurrence = 4/5 = 0.8   Let’s discuss certain ways in which this can be performed.
Method #1 : Using Iteration

Python3

# Python code to find frequency of a character
# at every position of list in list of lists.

# Input list initialization
Input = [['X', 'Y', 'X'], ['Z', 'Y', 'X'],
        ['Y', 'Y', 'Y'], ['Z', 'Z', 'X'],
        ['Y', 'Z', 'X']]
Output = []

# Character Initialization
character = 'X'

# Output list initialization
for elem in range(len(Input[0])):
    Output.append(0)

# Using iteration
for elem in Input:
    for x, y in enumerate(elem):
        if y == character:
            Output[x]+= 1
for x, y in enumerate(Output):
    Output[x] = y / len(Input)

# Printing
print("Initial list of list is :", Input)
print("Occurrence of 'X' in list is", Output)

Output

Initial list of list is : [['X', 'Y', 'X'], ['Z', 'Y', 'X'], ['Y', 'Y', 'Y'], ['Z', 'Z', 'X'], ['Y', 'Z', 'X']]
Occurrence of 'X' in list is [0.2, 0.0, 0.8]

Time complexity: O(N*M), where N is the number of lists in the Input list and M is the maximum length of any list in the Input list.

Auxiliary space: O(M), where M is the maximum length of any list in the Input list.

Method #2 : Using zip

Python3

# Python code to find frequency of a character
# at every position of list in list of lists.

# Input list initialization
Input = [['X', 'Y', 'X'], ['Z', 'Y', 'X'],
        ['Y', 'Y', 'Y'], ['Z', 'Z', 'X'],
        ['Y', 'Z', 'X']]

Output = []

# Character initialization
character = 'X'

# Using zip
Output = [elem.count(character)/len(elem)
                for elem in zip(*Input)]

# Printing
print("Initial list of list is :", Input)
print("Occurrence of 'X' in list is", Output)

Output

Initial list of list is : [['X', 'Y', 'X'], ['Z', 'Y', 'X'], ['Y', 'Y', 'Y'], ['Z', 'Z', 'X'], ['Y', 'Z', 'X']]
Occurrence of 'X' in list is [0.2, 0.0, 0.8]

The time complexity of this code is O(m*n), where m is the number of sublists in the input list and n is the length of the longest sublist. 

The space complexity is O(n), where n is the length of the longest sublist.

Method #3: Using Pandas

Python3

# Python code to find frequency of a character
# at every position of list in list of lists.

import pandas as pd

# Input list initialization
Input = [['X', 'Y', 'X'],
    ['Z', 'Y', 'X'],
    ['Y', 'Y', 'Y'],
    ['Z', 'Z', 'X'],
    ['Y', 'Z', 'X']]

# Defining character
character = 'X'

# using pandas
Output = pd.DataFrame(Input)
Output = Output.where(Output == character, 0).where(Output != character, 1)

# Printing
print("Initial list of list is :", Input)
print("Occurrence of 'X' in list is\n", Output.mean())

Output:

Initial list of list is : [[‘X’, ‘Y’, ‘X’], [‘Z’, ‘Y’, ‘X’], [‘Y’, ‘Y’, ‘Y’], [‘Z’, ‘Z’, ‘X’], [‘Y’, ‘Z’, ‘X’]] Occurrence of ‘X’ in list is 0 0.2 1 0.0 2 0.8 dtype: float64

 Method #4 : Using operator.countOf()

Python3

# Python code to find frequency of a character
# at every position of list in list of lists.
import operator as op
# Input list initialization
Input = [['X', 'Y', 'X'], ['Z', 'Y', 'X'],
         ['Y', 'Y', 'Y'], ['Z', 'Z', 'X'],
         ['Y', 'Z', 'X']]

Output = []

# Character initialization
character = 'X'

# Using zip
Output = [op.countOf(elem,character)/len(elem) for elem in zip(*Input)]

# Printing
print("Initial list of list is :", Input)
print("Occurrence of 'X' in list is", Output)

Output

Initial list of list is : [['X', 'Y', 'X'], ['Z', 'Y', 'X'], ['Y', 'Y', 'Y'], ['Z', 'Z', 'X'], ['Y', 'Z', 'X']]
Occurrence of 'X' in list is [0.2, 0.0, 0.8]

Time Complexity: O(N*N)
Auxiliary Space:  O(N*N)

Method #5: Using List Comprehension

By using list comprehension we can iterate through the rows of the input list and count the occurrences of the character ‘X’ in each row. Then, we can divide this count by the length of the row to get the frequency of the character at each position.

Python3

# Python code to find frequency of a character
# at every position of list in list of lists.

# Input list initialization
Input = [['X', 'Y', 'X'], ['Z', 'Y', 'X'],
         ['Y', 'Y', 'Y'], ['Z', 'Z', 'X'],
         ['Y', 'Z', 'X']]

# Character initialization
character = 'X'

# Using List Comprehension
Output = [sum(row[i] == character for row in Input)/len(Input) for i in range(len(Input[0]))]

# Printing
print("Initial list of list is :", Input)
print("Occurrence of 'X' in list is", Output)

Output

Initial list of list is : [['X', 'Y', 'X'], ['Z', 'Y', 'X'], ['Y', 'Y', 'Y'], ['Z', 'Z', 'X'], ['Y', 'Z', 'X']]
Occurrence of 'X' in list is [0.2, 0.0, 0.8]

The time complexity of this code is O(nm), where n is the number of rows in the input list and m is the length of each row.

The space complexity of this code is O(m), where m is the length of each row in the input list.