I reworked my numbers_to_name program so I can actually follow the logic.
Since I think the best way to learn is to follow something through to
completion, I'll put it online so I learn how to do that. But naturally,
lest I look dumb, I'd appreciate criticism, improvements of bad form, or if
anyone can break it, let me know ;') . Seems to work with a bit of random
testing, but ya never know. If you did, Msoft wouldn't be putting out those
updates all the time. Works with Py 27 or 33, but that brings up a
question: I think most of my programs work with both if I don't do anything
exotic, except for input(), which is the fly in the ointment. Instead of
using the routine below, could I redefine input() to use the routine, or it
that a bad idea?
I have a feeling the leftpad routine could be simplified, but it isn't
coming to me.
# Using C:\Python33\python.exe on Win 7 in c:\python33\jimprogs - not
Windows-specific
# Also works with python 2.7
import sys
# Data
ones = {'1': 'one', '2': 'two', '3': 'three', '4': 'four', '5': 'five',
'6': 'six',
'7': 'seven', '8': 'eight', '9': 'nine'}
tens = {'2': 'twenty', '3': 'thirty', '4': 'forty', '5': 'fifty', '6':
'sixty',
'7': 'seventy', '8': 'eighty', '9': 'ninety'}
doubles = {'0': 'ten', '1': 'eleven', '2': 'twelve', '3': 'thirteen', '4':
'fourteen',
'5': 'fifteen', '6': 'sixteen', '7': 'seventeen', '8': 'eighteen', '9':
'nineteen'}
powers_of_1000 = (' thousand', ' million', ' billion', ' trillion',
' quadrillion', ' quintillion', ' sextillion', ' septillion', ' octillion',
' nonillion', ' decillion')
'''add these later, and also option for dollars-and-cents ending.
'vigintillion', 'novemdecillion', 'octodecillion', 'septendecillion',
'sexdecillion', 'quindecillion', 'quattuordecillion', 'tredecillion',
'duodecillion', 'undecillion',
'decillion', 'nonillion'
'''
# Functions
def make_triplets_from_input():
'''Enter a positive integer. A list of triplets in the same order will
be returned.
Raise ValueError to loop on non-integer input, or return 'zero'
trigger-value of
'xxx' if all zeroes are entered. If input is okay, strip leading
zeroes, then
zero-pad leftmost triplet to three characters, so all slices are
triplets. Adjust
input for different Python versions.'''
while True:
try:
if sys.version[0:3] == '2.7':
numbers_str = original = raw_input('Enter a positive
integer, space \
separated if desired.')
elif sys.version[0:3] == '3.3':
numbers_str = original = input('Enter a positive integer,
space \
separated if desired.')
else:
print('Python versions 2.7 and 3.3 only supported')
sys.exit()
numbers_str = ''.join(numbers_str.split())
if numbers_str == '' or numbers_str == ' ': raise ValueError
numbers_int = int(numbers_str)
if numbers_int == 0:
print('Zero')
sys.exit()
numbers_str = str(numbers_int)
if len(numbers_str) > 36: raise ArithmeticError
break
except KeyboardInterrupt:
print('Program cancelled by user')
sys.exit()
except ValueError as err:
print(original, "is not an integer.")
continue
except ArithmeticError as err:
print(original, "is too big.\n999...decillion is max: 10**37-1
or 36 chars \
or 12 groups of 3 chars")
leftpad = len(numbers_str) % 3 # zero-pad left, so we get all
3-character triplets
if leftpad == 0: leftpad = ''
elif leftpad == 2: leftpad = '0'
else: leftpad = '00'
numbers_str = leftpad + numbers_str
triplets = [numbers_str[x:x+3] for x in range(0,len(numbers_str),3)]
return (triplets, len(triplets))
def numbers_to_name(triplets, triplen):
'''Create a name from each triplet and append the power of ten'''
triplen -= 2
number_name = ''
for triplet in triplets:
triplet_name = ''
first, second, third, last_two = (triplet[0], triplet[1],
triplet[2], triplet[1:])
if triplet == '000':
triplen -= 1
continue
if first > '0':
if last_two == '00': # special case - snip extra space separator
triplet_name += ones.get(first) + ' hundred'
else:
triplet_name += ones.get(first) + ' hundred '
if second == '0':
if third > '0':
triplet_name += ones.get(third)
elif second == '1':
triplet_name += doubles.get(third)
elif second > '1':
triplet_name += tens.get(second)
if third > '0':
triplet_name += '-' + ones.get(third)
number_name += triplet_name
if triplen > -1:
number_name += powers_of_1000[triplen] + ', '
triplen -= 1
if number_name[-2] == ',':
number_name = number_name[0:-2] # special case - snip extraneous
ending comma
return number_name
# Main Program - turn numeric input into number-names
triplets, triplen = make_triplets_from_input()
print(numbers_to_name(triplets, triplen))
--
Jim
"I asked God for a bike, but I know God doesn't work that way. So I stole a
bike and asked for forgiveness." ~ Anonymous
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