Eliminate the use of global variables; makes the example cleaner.
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@@ -7,7 +7,7 @@ RECURSION_LIMIT = 9500
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# this doesn't just rule recursion: it rules the depth of the call stack
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sys.setrecursionlimit(RECURSION_LIMIT+10)
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def count(word_list, word_freqs):
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def count(word_list, stopwords, wordfreqs):
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# What to do with an empty list
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if word_list == []:
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return
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@@ -16,33 +16,33 @@ def count(word_list, word_freqs):
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word = word_list[0]
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if word not in stopwords:
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if word in word_freqs:
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word_freqs[word] += 1
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wordfreqs[word] += 1
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else:
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word_freqs[word] = 1
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wordfreqs[word] = 1
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# The inductive case, what to do with a list of words
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else:
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# Process the head word
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count([word_list[0]], word_freqs)
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count([word_list[0]], stopwords, wordfreqs)
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# Process the tail
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count(word_list[1:], word_freqs)
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count(word_list[1:], stopwords, wordfreqs)
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def wf_print(word_freq):
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if word_freq == []:
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def wf_print(wordfreq):
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if wordfreq == []:
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return
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if len(word_freq) == 1:
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(w, c) = word_freq[0]
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if len(wordfreq) == 1:
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(w, c) = wordfreq[0]
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print w, '-', c
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else:
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wf_print([word_freq[0]])
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wf_print(word_freq[1:])
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wf_print([wordfreq[0]])
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wf_print(wordfreq[1:])
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stopwords = set(open('../stop_words.txt').read().split(','))
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stop_words = set(open('../stop_words.txt').read().split(','))
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words = re.findall('[a-z]{2,}', open(sys.argv[1]).read().lower())
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word_freqs = {}
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# Theoretically, we would just call count(words, word_freqs)
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# Try doing that and see what happens.
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for i in range(0, len(words), RECURSION_LIMIT):
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count(words[i:i+RECURSION_LIMIT], word_freqs)
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count(words[i:i+RECURSION_LIMIT], stop_words, word_freqs)
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wf_print(sorted(word_freqs.iteritems(), key=operator.itemgetter(1), reverse=True)[:25])
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