Replace missing white spaces in a string with the least frequent character using Pandas

Let’s create a program in python which will replace the white spaces 
in a string with the character that occurs in the string very least 
using the Pandas library. 

Example 1: 

String S = "akash loves gfg" 
here:
'g' comes: 2 times
's' comes: 2 times
'a' comes: 2 times
'h' comes: 1 time 
'o' comes: 1 time 
'k' comes: 1 time 
'v' comes: 1 time 
'e' comes: 1 time 
'f' comes: 1 time 
'l' comes: 1 time 

In this example, there are 7 characters with least frequency 1 so, there can be
7 valid outputs One of the possible output is given below: 
So, the Output String will be: "akashlloveslgfg".

Example 2: 

string ="goodd noon" 
here:
g comes: 1 time 
o comes: 4 times 
d comes: 2 times 
n comes: 2 times 
So the character with the least frequency 1 is g So here white spaces will be 
replaced by the character g and the output will be: 
"gooddgnoon" 

Now, Let’s see the implementation:

Python3

# importing pandas library
import pandas as pd

# taking string with white spaces
newstr1 = 'akash loves gfg'

# printing the original string
print("Original String given by user:",
      newstr1)

# converting string into
# list of characters
ser = pd.Series(list(newstr1))

# counting the frequency
# of characters
element_freq = ser.value_counts()

# printing character and their
# respective frequency
print(element_freq)

current_freq = element_freq.dropna().index[-1]

# function element_freq.dropna()
# will  Return a new Series with
# missing values removed
result = "".join(ser.replace(' ',
                             current_freq))

print(result)

Output:

The overall time complexity  is O(n log n).

The auxiliary space complexity  is O(n), where n is the length of the input string.

My Personal Notes arrow_drop_up

Save