lookbehind in a for loop

How are you doing?

Am kinda stuck with this problem, I need to use a for loop to find a word that ends with 'ing' and is preceded by a tag that is IN, I come from a background of C and j开发者_C百科ava and there its easy to do, but I cant yet grasp how to do it in python!!

I searched around and here is what I think i need to do:

for word, tag in list:
    if word.endswith('ing'):
       //use regular expression here which should look like this '(?<=\bIN\b)ing'

now ofcourse there are some problems there, first I the I need to look at the previous tag not word, the regular expression is probably is wrong and more importantly this just sounds too complicated, am I missing something here, is there a way to just use the index of the word ending with 'ing' to look at the tag behind it like I would have done using java for example??

Thank you in advance and sorry if its a silly question, its like my second time trying to write python and am still rusty with it =)

EDIT: more explanation on what I need to do, and an example here is what am trying to solve, sometimes pos_tag mistakes a VBG for a noun, so i need to write a method that given a tagged list (for example [('Cultivate', 'NNP'), ('peace', 'NN'), ('by', 'IN'), ('observing', 'NN'), ('justice', 'NN')] corrects this problem and returns [('Cultivate', 'NNP'), ('peace', 'NN'), ('by', 'IN'), ('observing', 'VBG'), ('justice', 'NN')] ) notice how observing has changed

EDIT2: problem solved now, here is the solution def transform(li): for i in xrange(len(li)): if li[i][0].endswith('ing') and i > 0 and li[i-1][1]: li[i] = (li[i], 'VBG')

thank you guys all for your help =D appreciated it

Based on your comment, sounds like you want this:

def transform(li):
    new_li = []
    prev_tag = None
    for word, tag in li:
        if word.endswith('ing') and prev_tag == 'NN':
            tag = 'VBG'
        new_li += [(word, tag)]
        prev_tag = tag
    return new_li

You can also do this in-place:

def transform(li):
    for i in xrange(len(li)):
        if li[i][0].endswith('ing') and i > 0 and li[i-1][1]:
            li[i] = (li[i], 'VBG')

Note that I renamed list to li. list is the type-name for a Python list and overriding it is a bad idea.

This does the change in place

for index,(word, _tag) in enumerate(li):
    if word.endswith('ing') and i > 0 and li[index-1][1] == 'IN':
        li[index] = word, 'VBG'

enumerate allows you to iterate over a list in a foreach fashion, but also get access to the current index. I quite like it, but I sometimes worry if I overuse it and should use something like for i in xrange(10): ... instead.

previousWord = ""
previousTag = ""

for word, tag in list:
    if word.endswith('ing'):
       //use regular expression here which should look like this '(?<=\bIN\b)ing'
       //use previousWord and previousTag here
    previousWord = word
    previousTag = tag

Your solution is somewhat driven by having immutable tuples as the data pairs in your list. The easiest way then is to create the new list you want in total:

li=[('Cultivate', 'NNP'), 
    ('peace', 'NN'), 
    ('by', 'IN'), 
    ('observing', 'NN'), 
    ('justice', 'NN')]

lnew=[]    

for word, tag in li:
    if word.endswith('ing') and tag == 'NN':
        tag='VBG'
    lnew.append((word,tag))

for word, tag in lnew:
    print word, tag

Somewhat wasteful if you have thousands or millions...

If this is your data and your format that you control, you may wish to consider using a dictionary instead of a list of tuples. Then you can loop the dictionary through more naturally and modify in place:

ld={'justice': 'NN', 'Cultivate': 'NNP', 'peace': 'NN', 
    'observing': 'NN', 'by': 'IN'}

for word, tag in ld.items():
    if word.endswith('ing') and tag == 'NN':
       ld[word]='VBG'

In large data sets, the dictionary approach is faster and more memory efficient. Consider that.