I'm trying to write a small program to loop through a list of words and find the numeric order of the letters in the string, e.g. HELLO will return 21345 (I don't want to double up numbers) and PROSPER will be 3527416.
I'm having trouble thinking of a way I can do this. Any ideas? I imagine I'll have to use the ord() function, but that's about as far as I've gotten.

3 Years
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Last Post by matmunn14

It wasn't for homework, it's part of a utility I was writing to solve the hat for headline puzzles. I managed to solve the problem, though it is a little bit unwieldy. I'll post code when I can get back to my desktop


Solution I came up with was:

import string, re

word = raw_input("Word to test: ")
letters_list = list(string.ascii_uppercase)
next_val = 1
order = ["" for i in range(100)]

for i in letters_list:
    starts = [match.start() for match in re.finditer(re.escape(i), word)]
    for j in starts:
        order[j] = str(next_val)
        next_val += 1

for i in range(len(order)):
    if len(order[i]) == 0:
        chopped = order[:i]

print ''.join(chopped)

Here is my best solution

word = raw_input("Word to test: ")
uword = word.upper()

L = sorted((c, pos) for (pos, c) in enumerate(uword))
M = sorted((pos, n) for (n, (c, pos)) in enumerate(L, 1))
result = ''.join(str(n) for (pos, n) in M)

print result

I also have a solution using objects from module collections:

from collections import defaultdict, deque

word = raw_input("Word to test: ")
uword = word.upper()

D = defaultdict(deque)
for i, c in enumerate(sorted(uword), 1):

L = [D[c].popleft() for c in uword]
result = ''.join(L)

print result

What will you do if the word has more than 9 letters, such as internationalization ?

Edited by Gribouillis


The code works fine for 13 character strings containing multiple occurrences of the same letter, tested with METAMORPHOSIS and BIRTHSTONE

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