Counting how many times "I before E except after C" is correct

ibeforeeexceptafterc.py

"""
Inspired by this bogus TIL: https://www.reddit.com/r/todayilearned/comments/54c05w/til_there_are_923_words_in_the_english_language/

Downloaded this corpus: https://sourceforge.net/projects/wordlist/files/SCOWL/2016.06.26/scowl-2016.06.26.zip/download?use_mirror=pilotfiber

From http://wordlist.aspell.net/
SCOWL (Spell Checker Oriented Word Lists) 

The results of running this script are:

Total words: 643702
I before E odds: 3.1 to 1
I before E after C odds: 3.9 to 1
{'ie': 25800, 'cei': 351, 'ei': 8284, 'cie': 1382, '[not_c]ei': 7933}
"""

import glob
from collections import Counter

tracking_strings = 'cei', 'cie', 'ie', 'ei'
counts = {}

word_set = set()
ei_not_after_c = '[not_c]ei'

for fn in glob.glob('final/*'):
    for line in open(fn):
        norm_word = line.strip().lower()
        if norm_word in word_set:
            continue
        word_set.add(norm_word)
        for str_to_track in tracking_strings:
            if str_to_track in norm_word:
                counts[str_to_track] = counts.get(str_to_track, 0) + 1
        
        
        if norm_word.find('ei') >= 0 and 'cei' not in norm_word:
            counts[ei_not_after_c] = counts.get(ei_not_after_c, 0) + 1

print('Total words: %d' % len(word_set))
print('I before E odds: %.1f to 1' % (counts['ie'] * 1.0 / counts['ei']))
print('I before E after C odds: %.1f to 1' % (counts['cie'] * 1.0 / counts['cei']))
print(counts)