A more Pythonic way to iterate over a list with duplicate items

Question

Is there a more elegant and Pythonic way to handle counters in this solution to an exercise from the NLTK book? The exercise asks to print out the context (one word forward and one word back) for every verb of a particular type (tagged VN).

import nltk
wsj = nltk.corpus.treebank.tagged_words(simplify_tags=True)
cdf = nltk.ConditionalFreqDist((tag, word) for (word, tag) in wsj)
wordlist = cdf['VN'].keys()
# Bug 1: strange exceptions in the for loop when iterating over wsj
# Solution: wsj is a custom NLTK type "ConcatenatedCorpusView"
# cast wsj into a native python type "list" for better iteration.
# I am guessing ConcatenatedCorpusView chokes on empty tuples
wsj_list = list(wsj)
# Bug 2: repeated words return index of the first word only
# Solution: to deal with repeated words
# we keep indexing from the last location. The index method
# takes a second parameter
starts_at = 0
for t in wsj_list:
 if t[0] in wordlist and t[1] == 'VN':
 ndx = wsj_list.index(t,starts_at)
 starts_at = ndx + 1
 print wsj_list[ndx-1:ndx+1], ndx

Answer 1

You could use enumerate to get the index. It makes the code both simpler and more efficient, as the linear search of index is avoided. I would also suggest unpacking t to (word, tag) to improve readability.

for ndx, (word, tag) in enumerate(wsj_list):
 if word in wordlist and tag == 'VN':
 print wsj[ndx-1:ndx+1], ndx