Get the number of letters in a number

jcao219 0 193 Views

This is not my code! Taken from this SO post.

You can use this to get the number of letters in the English representation of an integer, from 0 to 99.
So for example 5 (five) is len("five") which is 4.

len_of_num = lambda x: 3+int('1yrof7i9b1lsi207bozyzg2m7\
sclycst0zsczde5oks6zt8pedm\
nup5omwfx56b29',36)/10**x%10

for x in range(100):
    print "The length of the word for {0} is {1}.".format(x,
                                                         len_of_num(x))

Dani AI

Generated

Short summary: the compact, code‑golf trick pasted from StackOverflow is a neat way to get letter counts for 0..99 without writing a full parser. As observed, those packed-string solutions are brittle: they ignore hyphens/spaces, they assume a particular counting convention, and they don't scale past two digits.

Decide the counting rules first (these change the result): count only alphabetic characters (common), or include hyphens/spaces; use British "and" in numbers like "one hundred and one" or omit it (US style). Once the rule is fixed, a clear, maintainable approach is to build a small number-to-words generator (units, tens, hundreds, then group thousands/millions), then strip non-letters and take the length. That makes behavior explicit and easily extendable.

units = {0:'zero',1:'one',2:'two',3:'three',4:'four',5:'five',6:'six',7:'seven',8:'eight',9:'nine',
         10:'ten',11:'eleven',12:'twelve',13:'thirteen',14:'fourteen',15:'fifteen',16:'sixteen',
         17:'seventeen',18:'eighteen',19:'nineteen'}
tens = {20:'twenty',30:'thirty',40:'forty',50:'fifty',60:'sixty',70:'seventy',80:'eighty',90:'ninety'}

def _under_100(n):
    if n < 20: return units[n]
    t = (n//10)*10
    u = n%10
    return tens[t] + ('-' + units[u] if u else '')

def _under_1000(n, use_and=False):
    h, r = divmod(n, 100)
    if h:
        if r:
            sep = ' and ' if use_and else ' '
            return units[h] + ' hundred' + sep + _under_100(r)
        return units[h] + ' hundred'
    return _under_100(r)

def num_to_words(n, use_and=False):
    if n == 0: return 'zero'
    scales = ['', 'thousand', 'million', 'billion']
    parts, i = [], 0
    while n:
        chunk = n % 1000
        if chunk:
            w = _under_1000(chunk, use_and)
            parts.append(w + (' ' + scales[i] if scales[i] else ''))
        n //= 1000; i += 1
    return ' '.join(reversed(parts))

def letter_count(n, use_and=False):
    w = num_to_words(n, use_and)
    return sum(1 for c in w if c.isalpha())

Notes and tips: for bulk work precompute counts for 0..999 and compose larger numbers arithmetically to avoid repeated string building. If localization or very large numbers are needed, consider a tested library such as num2words. Clarify the hyphen/space/and rules upfront so results are reproducible.

Member Avatar for TrustyTony
TrustyTony 888 ex-Moderator

Seems to work until 20, then does not count dash in numbers like twenty-one, over 99 does not work.

Here my test and suggestion if you need number as string using more conventinal base than 36 (16)

from __future__ import print_function   ## let's practice Python 3 printing
import sys
sys.path.append('D:\test')  ## put here the location of the snippet
from intoeng import inttoeng ## replace with name of the file you saved integer to English snippet

def numberlength(x):
    try:
        return int('433544355436688779886aacbbaccb6aacbbaccb599baa9bba599baa9bba599baa9bba7bbdccbddc6aacbbaccb6aacbbaccb'[x],16)
    except IndexError:
        raise ValueError,'Number word length out of range 0..99'

##  string copied from run from:
##  res=''
##  for i in range(100):
##      res+=hex(len(inttoeng(i)))[-1]
##  print res


### from original post
len_of_num = lambda x: 3+int('1yrof7i9b1lsi207bozyzg2m7\
sclycst0zsczde5oks6zt8pedm\
nup5omwfx56b29',36)/10**x%10

for x in range(101): ## extended to see behaviour with wrong input
    # this is test of string, will of course match as the
    # inttoeng is source of information in string,
    # 100 will raise error from numberlength if next line uncommented
    len_of_num=numberlength 
    diff = len(inttoeng(x))-len_of_num(x)
    print("The length of the word for {0} is {1}".
          format(x,
                 len_of_num(x),
                 ),
          ", {0}, difference {1}".
          format(len(inttoeng(x)), diff) if diff
          else ' matching.'
    )
Member Avatar for TrustyTony
TrustyTony 888 ex-Moderator

Posted my version to SO link with original '-' does not count counts:

n=input();r='%i is '%n
while n-4:n=int('43354435543668877988699baa9bba699baa9bba588a998aa9588a998aa9588a998aa97aacbbaccb699baa9bba699baa9bba'[n],16);r+='%i.\n%i is '%(n,n)
print r+'magic.'
Edited by TrustyTony because: n/a
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