#!/usr/bin/python
# -*- coding: utf-8 -*-
"""Identify words that are the same on a QWERTY keyboard modulo
difference in hands.
The objective here is to analyze the feasibility of one-handed typing
using T9-style disambiguation. That is, if we consider the two hands
equivalent and just choose the most common 1-gram possibility for each
word, how often will we choose a possibility the user didn't intend?
And what are the most troublesome such possibilities?
Results:
> wess: will (253955) well (116604) oils (789)
> tww: two (156114) too (70169) tow (363) yow (43) tww (32) yoo (6)
> vade: make (79054) made (68515) vade (6) madi (6)
> wwrsd: would (255198) world (59031)
> gave: have (473693) gave (22946) game (14943) hame (8)
> set: set (38769) let (25918) sit (8703) lit (1466) ley (173) sey (21)
> sede: like (149479) side (33418) sidi (53) sede (12)
> saw: saw (26734) law (25346) sao (182) lao (33)
> sat: say (67862) sat (10950) lay (9690) lat (22)
> swve: some (171368) love (20146)
...
> 1538199 total errors in 90080933 words, 1.7%
This makes me think:
1. It's probably a feasible alternative for full-speed English text,
if you have a keybinding to correct a missed guess once or twice a
minute;
2. 2-gram disambiguation would provide much better results, but would
not make missed guesses insignificant and eliminate the need for a
correction keybinding — 'make' and 'made'; 'set', 'let', and 'sit';
'say' and 'lay'; and other similar cases; would still have significant
missed-guess rates with any conceivable N-gram model over words.
"""
from __future__ import division
import string
keyboard = [('qwert', 'yuiop'),
('asdfg', 'hjkl;'),
('zxcvb', 'nm,./')]
assert all(len(left_n) == len(right_n) for left_n, right_n in keyboard)
left_side = ''.join(left for left, right in keyboard)
right_side = ''.join(''.join(reversed(right)) for left, right in keyboard)
translation = string.maketrans(right_side, left_side)
classes = {}
total_1grams = 0
for line in open('wordlist'): # wget http://canonical.org/~kragen/sw/wordlist
freq, word = line.split()
freq = int(freq)
total_1grams += freq
word_class = word.translate(translation)
if word_class not in classes:
classes[word_class] = []
classes[word_class].append((freq, word))
for word_class in classes:
classes[word_class] = sorted(classes[word_class], reverse=True)
errors = 0
errors_for = lambda fws: sum(freq for freq, word in fws[1:])
for word_class in sorted(classes.keys(),
key=lambda word_class: errors_for(classes[word_class]),
reverse=True):
collisions = classes[word_class]
if len(collisions) == 1:
continue
errors += errors_for(collisions)
print "%s: %s" % (word_class, ' '.join('%s (%s)' % (word, freq)
for freq, word in collisions))
print "%d total errors in %d words, %.2g%%" % (errors, total_1grams,
errors / total_1grams * 100)
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