Number to Words Conversion in Python Without Any Library

This python program converts given integer number to its equivalent words without using any external library.

This number to word conversion python program supports up to 12 digits number. For large number, we encourage you to modify this program :)

Python Source Code: Number to Words Conversion

# Number to Words

# Main Logic
ones = ('Zero', 'One', 'Two', 'Three', 'Four', 'Five', 'Six', 'Seven', 'Eight', 'Nine')

twos = ('Ten', 'Eleven', 'Twelve', 'Thirteen', 'Fourteen', 'Fifteen', 'Sixteen', 'Seventeen', 'Eighteen', 'Nineteen')

tens = ('Twenty', 'Thirty', 'Forty', 'Fifty', 'Sixty', 'Seventy', 'Eighty', 'Ninety', 'Hundred')

suffixes = ('', 'Thousand', 'Million', 'Billion')

def process(number, index):
    
    if number=='0':
        return 'Zero'
    
    length = len(number)
    
    if(length > 3):
        return False
    
    number = number.zfill(3)
    words = ''
 
    hdigit = int(number[0])
    tdigit = int(number[1])
    odigit = int(number[2])
    
    words += '' if number[0] == '0' else ones[hdigit]
    words += ' Hundred ' if not words == '' else ''
    
    if(tdigit > 1):
        words += tens[tdigit - 2]
        words += ' '
        words += ones[odigit]
    
    elif(tdigit == 1):
        words += twos[(int(tdigit + odigit) % 10) - 1]
        
    elif(tdigit == 0):
        words += ones[odigit]

    if(words.endswith('Zero')):
        words = words[:-len('Zero')]
    else:
        words += ' '
     
    if(not len(words) == 0):    
        words += suffixes[index]
        
    return words;
    
def getWords(number):
    length = len(str(number))
    
    if length>12:
        return 'This program supports upto 12 digit numbers.'
    
    count = length // 3 if length % 3 == 0 else length // 3 + 1
    copy = count
    words = []
 
    for i in range(length - 1, -1, -3):
        words.append(process(str(number)[0 if i - 2 < 0 else i - 2 : i + 1], copy - count))
        count -= 1;

    final_words = ''
    for s in reversed(words):
        temp = s + ' '
        final_words += temp
    
    return final_words
# End Main Logic

# Reading number from user
number = int(input('Enter any number: '))
print('%d in words is: %s' %(number, getWords(number)))

Output

Run 1:
----------
Enter any number: 0
0 in words is: Zero 

Run 2:
----------
Enter any number: 17
17 in words is: Seventeen  

Run 3:
----------
Enter any number: 100
100 in words is: One Hundred  

Run 4:
----------
Enter any number: 100001
100001 in words is: One Hundred Thousand One  

Run 5:
----------
Enter any number: 1234567890
1234567890 in words is: One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety  

For 'and' Requirement

In case if you need to append 'and' if number is over one hundred and either of the last two digits are not a zero then you can modify above program as:

# Number to Words

# Main Logic
ones = ('Zero', 'One', 'Two', 'Three', 'Four', 'Five', 'Six', 'Seven', 'Eight', 'Nine')

twos = ('Ten', 'Eleven', 'Twelve', 'Thirteen', 'Fourteen', 'Fifteen', 'Sixteen', 'Seventeen', 'Eighteen', 'Nineteen')

tens = ('Twenty', 'Thirty', 'Forty', 'Fifty', 'Sixty', 'Seventy', 'Eighty', 'Ninety', 'Hundred')

suffixes = ('', 'Thousand', 'Million', 'Billion')

def process(number, index, ln):
    print(">>", ln)
    if number=='0':
        return 'Zero'
    
    length = len(number)
    
    if(length > 3):
        return False
    
    number = number.zfill(3)
    words = ''
 
    hdigit = int(number[0])
    tdigit = int(number[1])
    odigit = int(number[2])
    
    words += '' if number[0] == '0' else ones[hdigit]
    words += ' Hundred ' if not words == '' else ''
    
    if index==0 and ln>3:
        words+=' and '
    elif words=='':
        words+=''
    elif index==0 and tdigit==0 and odigit==0:
        words+=''
    elif index==0:
        words+= ' and '
    else:
        words+=''
    
    if(tdigit > 1):
        words += tens[tdigit - 2]
        words += ' '
        words += ones[odigit]
    
    elif(tdigit == 1):
        words += twos[(int(tdigit + odigit) % 10) - 1]
    elif(tdigit == 0):
        words += ones[odigit]
    if(words.endswith('Zero')):
        words = words[:-len('Zero')]
    else:
        words += ' '
     
    if(not len(words) == 0):    
        words += suffixes[index]
        
    return words;
    
def getWords(number):
    length = len(str(number))
    
    if length>12:
        return 'This program supports upto 12 digit numbers.'
    
    count = length // 3 if length % 3 == 0 else length // 3 + 1
    copy = count
    words = []
 
    for i in range(length - 1, -1, -3):
        words.append(process(str(number)[0 if i - 2 < 0 else i - 2 : i + 1], copy - count, length))
        count -= 1;

    final_words = ''
    for s in reversed(words):
        temp = s + ' '
        final_words += temp
    
    return final_words
# End Main Logic

# Reading number from user
number = int(input('Enter any number: '))
print('%d in words is: %s' %(number, getWords(number)))

Output

Run 1:
----------
Enter any number: 0
0 in words is: Zero 

Run 2:
----------
Enter any number: 17
17 in words is: Seventeen  

Run 3:
----------
Enter any number: 100
100 in words is: One Hundred  

Run 4:
----------
Enter any number: 100001
100001 in words is: One Hundred Thousand and One  

Run 5:
----------
Enter any number: 1234567890
1234567890 in words is: One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred and Ninety