Get the number of letters in a number

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.'
    )

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.'