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nltk - How to split an NLP parse tree to clauses (independent and subordinate)?

Given an NLP parse tree like

(ROOT (S (NP (PRP You)) (VP (MD could) (VP (VB say) (SBAR (IN that) (S (NP (PRP they)) (ADVP (RB regularly)) (VP (VB catch) (NP (NP (DT a) (NN shower)) (, ,) (SBAR (WHNP (WDT which)) (S (VP (VBZ adds) (PP (TO to) (NP (NP (PRP$ their) (NN exhilaration)) (CC and) (NP (FW joie) (FW de) (FW vivre))))))))))))) (. .)))

Original sentence is "You could say that they regularly catch a shower, which adds to their exhilaration and joie de vivre."

How could the clauses be extracted and reverse engineered? We would be splitting at S and SBAR (to preserve the type of clause, eg subordinated)

 - (S (NP (PRP You)) (VP (MD could) (VP (VB say) 
 - (SBAR (IN that) (S (NP (PRP they)) (ADVP (RB regularly)) (VP (VB catch) (NP (NP (DT a) (NN shower))
 - (, ,) (SBAR (WHNP (WDT which)) (S (VP (VBZ adds) (PP (TO to)
   (NP (NP (PRP$ their) (NN exhilaration)) (CC and) (NP (FW joie) (FW
   de) (FW vivre))))))))))))) (. .)))

to arrive at

 - You could say
 - that they regularly catch a shower 
 - , which adds to their exhilaration and joie de vivre.

Splitting at S and SBAR seems very easy. The problem seems to be stripping away all the POS tags and chunks from the fragments.

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1 Answer

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You can use Tree.subtrees(). For more information check NLTK Tree Class.

Code:

from nltk import Tree

parse_str = "(ROOT (S (NP (PRP You)) (VP (MD could) (VP (VB say) (SBAR (IN that) (S (NP (PRP they)) (ADVP (RB regularly)) (VP (VB catch) (NP (NP (DT a) (NN shower)) (, ,) (SBAR (WHNP (WDT which)) (S (VP (VBZ adds) (PP (TO to) (NP (NP (PRP$ their) (NN exhilaration)) (CC and) (NP (FW joie) (FW de) (FW vivre))))))))))))) (. .)))"
#parse_str = "(ROOT (S (SBAR (IN Though) (S (NP (PRP he)) (VP (VBD was) (ADJP (RB very) (JJ rich))))) (, ,) (NP (PRP he)) (VP (VBD was) (ADVP (RB still)) (ADJP (RB very) (JJ unhappy))) (. .)))"

t = Tree.fromstring(parse_str)
#print t

subtexts = []
for subtree in t.subtrees():
    if subtree.label()=="S" or subtree.label()=="SBAR":
        #print subtree.leaves()
        subtexts.append(' '.join(subtree.leaves()))
#print subtexts

presubtexts = subtexts[:]       # ADDED IN EDIT for leftover check

for i in reversed(range(len(subtexts)-1)):
    subtexts[i] = subtexts[i][0:subtexts[i].index(subtexts[i+1])]

for text in subtexts:
    print text

# ADDED IN EDIT - Not sure for generalized cases
leftover = presubtexts[0][presubtexts[0].index(presubtexts[1])+len(presubtexts[1]):]
print leftover

Output:

You could say 
that 
they regularly catch a shower , 
which 
adds to their exhilaration and joie de vivre
 .

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