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Importance of the "Number to word converter - A cool python project"
Number to word converter is needed in the cheque, especially in the production cheque. In the production sector and bank sector, both numeric digit and written word numbers are equally important to maintain the formality.
"standard dictionary numbers" for "Number to word converter - A python project for practice"
Digits in value with its name (one to billion)
Digits/Value In Word Name
1 one
10 ten
100 hundred
1,000 thousand
10,000 ten thousand
100,000 hundred thousand
1,000,000 million
10,000,000 ten million
100,000,000 hundred million
1,000,000,000 billion
Digits in number with places (one to billion)
1 0 0 0 0 0 0 0 0 0
| | | | | | | | | |____ Ones'
| | | | | | | | |__ ____ Tens'
| | | | | | | |__ __ ____ Hundreds'
| | | | | | |__ __ __ ____ Thousands'
| | | | | |__ __ __ __ ____ Tens Thousands'
| | | | |__ __ __ __ __ ____ Hundred Thousands'
| | | |__ __ __ __ __ __ ____ One Millions'
| | |__ __ __ __ __ __ __ ____ Ten Millions'
| |__ __ __ __ __ __ __ __ ____ Hundred Millions'
|__ __ __ __ __ __ __ __ __ ____ One Billions'
Algorithm for "Number to word converter - A python project for practice"
Large number name rule
-> initialize the dictionary for large number names. For this post, the large number of names is limited to zero to a nonillion.
Number rules
->Zero Rule
If the input value is zero, then the output will be zero.
NumberIntoWord+='zero'
->3-Digit Rule
3-Digit words are: "thousand", "million", "billion", "trillion", "quadrillion", "quintillion", "sextillion", "septillion", "octillion", "nonillion". (From Right to the Left )
The 3-digit words will be appeared in the text according to the hierarchy. (Right to left method)
NumberIntoWord+=3-DigitWords[i]
->Hundred Rule
If the input three-digit number is divisible by 1 hundred, then the text 'hundred' will be appended.
NumberIntoWord+='hundred'
Else, 'hundred and' will be appended to the word.
NumberIntoWord+='hundred and'
->Tens Rule
If a three-digit group's tens section is two or higher, the corresponding '-ty' word (twenty, thirty, forty, fifty, sixty, etc.) is appended to the output final text (NumberIntoWord), followed by the name of the third digit (unless the third digit is a zero).
NumberIntoWord+='ty'
There is no text added if the tens and units are both zero.
NumberIntoWord+=''
->Recombination or Comma Rule
For the recombining of the final word as an output, the three-digit group is followed by a comma and a large number name.
If the three-digit group (million, billion, etc) does not have any hundred, then the final comma in the NumberIntoWord is replaced by a 'and'. (ex: two million and two)
NumberIntoWord+=','
NumberIntoWord.replace([:]word, ',', 'and')
->Negative Rule
Check the first of the input number and if the sign'-' (negative), then the text 'negative' will be appended.
NumberIntoWord+='negative'
Python code for "Number to word converter - A python project for practice"
OneDigitWords = {
'0': ["zero"],
'1': ["one"],
'2': ["two", "twen"],
'3': ["three", "thir"],
'4': ["four", "for"],
'5': ["five", "fif"],
'6': ["six"],
'7': ["seven"],
'8': ["eight"],
'9': ["nine"],
}
TwoDigitWords = ["ten", "eleven", "twelve"]
Hundred = "hundred"
LargeMathWords = ["thousand", "million", "billion", "trillion", "quadrillion", "quintillion", "sextillion", "septillion", "octillion", "nonillion"]
def IsNumberNegative(NumberInToWord, InputNumber):
if InputNumber.startswith('-'):
NumberInToWord.append("(Negative)")
InputNumber = InputNumber[1:]
return NumberInToWord, InputNumber
def NumberInToWordConverter(InputNumber):
NumberInToWord = []
NumberInToWord, InputNumber=IsNumberNegative(NumberInToWord, InputNumber)
if len(InputNumber) % 3 != 0 and len(InputNumber) > 3:
InputNumber = InputNumber.zfill(3 * (((len(InputNumber) - 1) // 3) + 1))
SumOfList = [InputNumber[i:i + 3] for i in range(0, len(InputNumber), 3)]
skip = False
for i, number in enumerate(SumOfList):
if number != '000': skip = False
for _ in range(len(number)):
number = number.lstrip('0')
if len(number) == 1:
if (len(SumOfList) > 1 or (len(SumOfList) == 1 and len(SumOfList[0]) == 3)) and i == len(
SumOfList) - 1 and (NumberInToWord[-1] in LargeMathWords or Hundred in NumberInToWord[-1]):
NumberInToWord.append("and")
NumberInToWord.append(OneDigitWords[number][0])
number = number[1:]
break
if len(number) == 2:
if number[0] != '0':
if (len(SumOfList) > 1 or (len(SumOfList) == 1 and len(SumOfList[0]) == 3)) and i == len(SumOfList) - 1:
NumberInToWord.append("and")
if number.startswith('1'):
if int(number[1]) in range(3):
NumberInToWord.append(TwoDigitWords[int(number[1])])
else:
number = OneDigitWords[number[1]][1 if int(number[1]) in range(3, 6, 2) else 0]
NumberInToWord.append(number + ("teen" if not number[-1] == 't' else "een"))
else:
NumberInToWord.append(OneDigitWords[number[0]][1 if int(number[0]) in range(2, 6) else 0] + (
"ty " if number[0] != '8' else 'y ') + (OneDigitWords[number[1]][0] if number[1] != '0' else ""))
break
else:
number = number[1:]
continue
if len(number) == 3:
if number[0] != '0':
NumberInToWord.append(OneDigitWords[number[0]][0] + " " + Hundred)
if number[1:] == '00': break
number = number[1:]
if len(SumOfList[i:]) > 1 and not skip:
NumberInToWord.append(LargeMathWords[len(SumOfList[i:]) - 2])
skip = True
NumberInToWord = " ".join(map(str.strip, NumberInToWord))
return NumberInToWord[0].lstrip().upper() + NumberInToWord[1:].rstrip().lower() if "negative" not in NumberInToWord else NumberInToWord[:11].lstrip() + \
NumberInToWord[11].upper() + NumberInToWord[
12:].rstrip().lower()
if __name__ == "__main__":
while True:
try:
InputNumber = input("\n Enter any number to convert it into words or 'x' to stop: ")
if InputNumber == "x" or InputNumber == "X":
break
int(InputNumber)
print(InputNumber, ": In Word :", NumberInToWordConverter(InputNumber))
except ValueError:
print("Error: You type an invalid number!")Sample Input:
Sample Output:
Understand the program: "Number to word converter - A python project for practice"
Large number name rule
LargeMathWords = ["thousand", "million", "billion", "trillion", "quadrillion", "quintillion", "sextillion", "septillion", "octillion", "nonillion"]
Number rules (dictionary initialization)
OneDigitWords = {
'0': ["zero"],
'1': ["one"],
'2': ["two", "twen"],
'3': ["three", "thir"],
'4': ["four", "for"],
'5': ["five", "fif"],
'6': ["six"],
'7': ["seven"],
'8': ["eight"],
'9': ["nine"],
}
TwoDigitWords = ["ten", "eleven", "twelve"]
Hundred = "hundred"Number rules (zero rules)
'0': ["zero"],Number rules (3-digit rule)
if (len(SumOfList) > 1 or (len(SumOfList) == 1 and len(SumOfList[0]) == 3)) and i == len(
SumOfList) - 1 and (NumberInToWord[-1] in LargeMathWords or Hundred in NumberInToWord[-1]):if len(SumOfList[i:]) > 1 and not skip:
NumberInToWord.append(LargeMathWords[len(SumOfList[i:]) - 2])
skip = TrueNumber rules (hundred rules)
if len(number) == 3:
if number[0] != '0':
NumberInToWord.append(OneDigitWords[number[0]][0] + " " + Hundred)
if number[1:] == '00': break
number = number[1:]Number rules (tens rules)
number = OneDigitWords[number[1]][1 if int(number[1]) in range(3, 6, 2) else 0]
NumberInToWord.append(number + ("teen" if not number[-1] == 't' else "een"))Number rules (Recombination or Comma rule)
(OneDigitWords[number[1]][0] if number[1] != '0' else ""))if (len(SumOfList) > 1 or (len(SumOfList) == 1 and len(SumOfList[0]) == 3)) and i == len(
SumOfList) - 1 and (NumberInToWord[-1] in LargeMathWords or Hundred in NumberInToWord[-1]):
NumberInToWord.append("and")Number rules (Negative rule)
def IsNumberNegative(NumberInToWord, InputNumber):
if InputNumber.startswith('-'):
NumberInToWord.append("(Negative)")
InputNumber = InputNumber[1:]
return NumberInToWord, InputNumberA recent python cool project has been discussed in this link.
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