An aggressive little script I used to trim out duplicate fortunes.

This commit is contained in:
Ruslan Ermilov 2005-02-07 21:15:16 +00:00
parent 6325a5d282
commit 0b4978b4ea
Notes: svn2git 2020-12-20 02:59:44 +00:00
svn path=/head/; revision=141477

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#!/usr/local/bin/python
#
# $FreeBSD$
#
# an aggressive little script for trimming duplicate cookies
import re, sys
wordlist = [
'hadnot',
'donot', 'hadnt',
'dont', 'have', 'more', 'will', 'your',
'and', 'are', 'had', 'the', 'you',
'am', 'an', 'is', 'll', 've', 'we',
'a', 'd', 'i', 'm', 's',
]
def hash(fortune):
f = fortune
f = f.lower()
f = re.sub('[\W_]', '', f)
for word in wordlist:
f = re.sub(word, '', f)
# f = re.sub('[aeiouy]', '', f)
# f = re.sub('[^aeiouy]', '', f)
f = f[:30]
# f = f[-30:]
return f
def edit(datfile):
dups = {}
fortunes = []
fortune = ""
for line in file(datfile):
if line == "%\n":
key = hash(fortune)
if not dups.has_key(key):
dups[key] = []
dups[key].append(fortune)
fortunes.append(fortune)
fortune = ""
else:
fortune += line
for key in dups.keys():
if len(dups[key]) == 1:
del dups[key]
o = file(datfile + '~', "w")
for fortune in fortunes:
key = hash(fortune)
if key in dups:
print '\n' * 50
for f in dups[key]:
if f != fortune:
print f, '%'
print fortune, '%'
if raw_input("Remove last fortune? ") == 'y':
del dups[key]
continue
o.write(fortune + "%\n")
o.close()
assert len(sys.argv) == 2
edit(sys.argv[1])