Python String Exercise with Solutions

• Use string indexing to get the character present at the given index.
• Use str1[0] to get the first character of a string and add it to the result variable
• Next, get the middle character’s index by dividing string length by 2. x = len(str1) /2. Use str1[x] to get the middle character and add it to the result variable
• Use str1[len(str1)-1] to get the last character of a string and add it to the result variable
• print result variable to display new string

str1 = 'James'
print("Original String is", str1)

# Get first character
res = str1[0]

# Get string size
l = len(str1)
# Get middle index number
mi = int(l / 2)
# Get middle character and add it to result
res = res + str1[mi]

# Get last character and add it to result
res = res + str1[l - 1]

print("New String:", res)

• Get the middle character’s index using x = len(str1) /2.
• Use string slicing to get the middle three characters starting from the middle index to the next two character str1[middle_index-1:middle_index+2]

def get_middle_three_chars(str1):
    print("Original String is", str1)

    # first get middle index number
    mi = int(len(str1) / 2)

    # use string slicing to get result characters
    res = str1[mi - 1:mi + 2]
    print("Middle three chars are:", res)

get_middle_three_chars("JhonDipPeta")
get_middle_three_chars("JaSonAy")

• First, get the middle index number of s1 by dividing s1’s length by 2
• Use string slicing to get the character from s1 starting from 0 to the middle index number and store it in x
• concatenate x and s2. x = x + s2
• concatenate x and remeaning character from s1
• print x

def append_middle(s1, s2):
    print("Original Strings are", s1, s2)

    # middle index number of s1
    mi = int(len(s1) / 2)

    # get character from 0 to the middle index number from s1
    x = s1[:mi:]
    # concatenate s2 to it
    x = x + s2
    # append remaining character from s1
    x = x + s1[mi:]
    print("After appending new string in middle:", x)

append_middle("Ault", "Kelly")

• Get the first character from both strings, concatenate them, and store them in variable x
• Get the middle character from both strings, concatenate them, and store them in variable y
• Get the last character from both strings, concatenate them, and store them in variable x
• In the end, join x, y, and z and save it in the result variable
• print the result

def mix_string(s1, s2):
    # get first character from both string
    first_char = s1[0] + s2[0]

    # get middle character from both string
    middle_char = s1[int(len(s1) / 2):int(len(s1) / 2) + 1] + s2[int(len(s2) / 2):int(len(s2) / 2) + 1]

    # get last character from both string
    last_char = s1[len(s1) - 1] + s2[len(s2) - 1]

    # add all
    res = first_char + middle_char + last_char
    print("Mix String is ", res)

s1 = "America"
s2 = "Japan"
mix_string(s1, s2)

• Create two lists lower and upper
• Iterate a string using a for loop
• In each loop iteration, check if the current character is the lower or upper case using the islower() string function.
• If a character is the lower case, add it to the lower list, else add it to the upper list
• to join the lower and upper list using a join() function.
• convert list to string
• print the final string

str1 = "PYnAtivE"
print('Original String:', str1)
lower = []
upper = []
for char in str1:
    if char.islower():
        # add lowercase characters to lower list
        lower.append(char)
    else:
        # add uppercase characters to lower list
        upper.append(char)

# Join both list
sorted_str = ''.join(lower + upper)
print('Result:', sorted_str)

• Iterate each character from a string using a for loop
• In each loop iteration, check if the current character is the alphabet using an isalpha() function. If yes, increase the character counter. Check if it is a digit using the isdigit() function and increase the digit counter; otherwise, increase the symbol counter.
• Print the value of each counter

def find_digits_chars_symbols(sample_str):
    char_count = 0
    digit_count = 0
    symbol_count = 0
    for char in sample_str:
        if char.isalpha():
            char_count += 1
        elif char.isdigit():
            digit_count += 1
        # if it is not letter or digit then it is special symbol
        else:
            symbol_count += 1

    print("Chars =", char_count, "Digits =", digit_count, "Symbol =", symbol_count)

sample_str = "P@yn2at&#i5ve"
print("total counts of chars, Digits, and symbols \n")
find_digits_chars_symbols(sample_str)

s1 = "Abc"
s2 = "Xyz"

# get string length
s1_length = len(s1)
s2_length = len(s2)

# get length of a bigger string
length = s1_length if s1_length > s2_length else s2_length
result = ""

# reverse s2
s2 = s2[::-1]

# iterate string 
# s1 ascending and s2 descending
for i in range(length):
    if i < s1_length:
        result = result + s1[i]
    if i < s2_length:
        result = result + s2[i]

print(result)

def string_balance_test(s1, s2):
    flag = True
    for char in s1:
        if char in s2:
            continue
        else:
            flag = False
    return flag


s1 = "Yn"
s2 = "PYnative"
flag = string_balance_test(s1, s2)
print("s1 and s2 are balanced:", flag)

s1 = "Ynf"
s2 = "PYnative"
flag = string_balance_test(s1, s2)
print("s1 and s2 are balanced:", flag)

str1 = "Welcome to USA. usa awesome, isn't it?"
sub_string = "USA"

# convert string to lowercase
temp_str = str1.lower()

# use count function
count = temp_str.count(sub_string.lower())
print("The USA count is:", count)

Solution 1: Use string functions

• Iterate each character from a string s1 using a loop
• In the body of a loop, check if the current character is digit using the isdigit() function
• If it is a digit, then add it to the sum variable
• In the end, calculate the average by dividing the total by the count of digits

input_str = "PYnative29@#8496"
total = 0
cnt = 0
for char in input_str:
    if char.isdigit():
        total += int(char)
        cnt += 1

# average = sum / count of digits
avg = total / cnt
print("Sum is:", total, "Average is ", avg)

Solution 2: Use regular expression

import re

input_str = "PYnative29@#8496"
digit_list = [int(num) for num in re.findall(r'\d', input_str)]
print('Digits:', digit_list)

# use the built-in function sum
total = sum(digit_list)

# average = sum / count of digits
avg = total / len(digit_list)
print("Sum is:", total, "Average is ", avg)

• create an empty dictionary to store the result. character is the key, and the count is the value
• Iterate each character from a string s1 using a loop
• In the body of a loop, use the count() function to find how many times a current character appeared in a string
• Add key-value pair in a dictionary

str1 = "Apple"

# create a result dictionary
char_dict = dict()

for char in str1:
    count = str1.count(char)
    # add / update the count of a character
    char_dict[char] = count
print('Result:', char_dict)

Solution 1: Negative String slicing

str1 = "PYnative"
print("Original String is:", str1)

str1 = str1[::-1]
print("Reversed String is:", str1)

Solution 2: Using the reversed() function

str1 = "PYnative"
print("Original String is:", str1)

str1 = ''.join(reversed(str1))
print("Reversed String is:", str1)

str1 = "Emma is a data scientist who knows Python. Emma works at google."
print("Original String is:", str1)

index = str1.rfind("Emma")
print("Last occurrence of Emma starts at index:", index)

str1 = "Emma-is-a-data-scientist"
print("Original String is:", str1)

# split string
sub_strings = str1.split("-")

print("Displaying each substring")
for sub in sub_strings:
    print(sub)

Solution 1: Using the loop and if condition

str_list = ["Emma", "Jon", "", "Kelly", None, "Eric", ""]
res_list = []
for s in str_list:
    # check for non empty string
    if s:
        res_list.append(s)
print(res_list)

Solution 2: Using the built-in function filter()

str_list = ["Emma", "Jon", "", "Kelly", None, "Eric", ""]

# use built-in function filter to filter empty value
new_str_list = list(filter(None, str_list))

print("After removing empty strings")
print(new_str_list)

Solution 1: Use string functions translate() and maketrans().

The string.punctuation constant contain all special symbols.

import string

str1 = "/*Jon is @developer & musician"
print("Original string is ", str1)

new_str = str1.translate(str.maketrans('', '', string.punctuation))

print("New string is ", new_str)

Solution 2: Using regex replace pattern in a string

import re

str1 = "/*Jon is @developer & musician"
print("Original string is ", str1)

# replace special symbols with ''
res = re.sub(r'[^\w\s]', '', str1)
print("New string is ", res)

str1 = 'I am 25 years and 10 months old'
print("Original string is", str1)

# Retain Numbers in String
# Using list comprehension + join() + isdigit()
res = "".join([item for item in str1 if item.isdigit()])

print(res)

str1 = "Emma25 is Data scientist50 and AI Expert"
print("The original string is : " + str1)

res = []
# split string on whitespace
temp = str1.split()

# Words with both alphabets and numbers
# isdigit() for numbers + isalpha() for alphabets
# use any() to check each character

for item in temp:
    if any(char.isalpha() for char in item) and any(char.isdigit() for char in item):
        res.append(item)

print("Displaying words with alphabets and numbers")
for i in res:
    print(i)

• Use the string.punctuation constant to get the list of all punctuations
• Iterate each symbol from a punctuations
• Use string function replace() to replace the current special symbol in a string with #

import string

str1 = '/*Jon is @developer & musician!!'
print("The original string is : ", str1)

# Replace punctuations with #
replace_char = '#'

# string.punctuation to get the list of all special symbols
for char in string.punctuation:
    str1 = str1.replace(char, replace_char)

print("The strings after replacement : ", str1)