How to get all noun phrases in Spacy

Question

I am new to Spacy and I would like to extract "all" the noun phrases from a sentence. I'm wondering how I can do it. I have the following code:

import spacy

nlp = spacy.load("en")

file = open("E:/test.txt", "r")
doc = nlp(file.read())
for np in doc.noun_chunks:
    print(np.text)

But it returns only the base noun phrases, that is, phrases which don't have any other NP in them. That is, for the following phrase, I get the result below:

Phrase: We try to explicitly describe the geometry of the edges of the images.

Result: We, the geometry, the edges, the images.

Expected result: We, the geometry, the edges, the images, the geometry of the edges of the images, the edges of the images.

How can I get all the noun phrases, including nested phrases?

Answer 1

Please see commented code below to recursively combine the nouns. Code inspired by the Spacy Docs here

import spacy

nlp = spacy.load("en")

doc = nlp("We try to explicitly describe the geometry of the edges of the images.")

for np in doc.noun_chunks: # use np instead of np.text
    print(np)

print()

# code to recursively combine nouns
# 'We' is actually a pronoun but included in your question
# hence the token.pos_ == "PRON" part in the last if statement
# suggest you extract PRON separately like the noun-chunks above

index = 0
nounIndices = []
for token in doc:
    # print(token.text, token.pos_, token.dep_, token.head.text)
    if token.pos_ == 'NOUN':
        nounIndices.append(index)
    index = index + 1


print(nounIndices)
for idxValue in nounIndices:
    doc = nlp("We try to explicitly describe the geometry of the edges of the images.")
    span = doc[doc[idxValue].left_edge.i : doc[idxValue].right_edge.i+1]
    span.merge()

    for token in doc:
        if token.dep_ == 'dobj' or token.dep_ == 'pobj' or token.pos_ == "PRON":
            print(token.text)