在Python中使用'in'关键字在集合中查找项目时遇到麻烦

Question

I have this code here.

import spacy
nlp = spacy.load('en')
a = set(nlp('This is a test'))
b = nlp('is')
if b in a:
  print("Success")
else:
  print("Failed")

for some reason this output printed out "Failed". I expected it to succeed. I am new in using the spacy framework so I'm not quite sure how to do this right. How do I do this right?

Answer 1

The type(b) is a <class 'spacy.tokens.doc.Doc'> and you are comparing with a variable that is a set <class 'set'> . So try converting both the variables to set and then try the in method. And each item in the nlp tokens is a <class 'spacy.tokens.token.Token'> class rather than a string. So you have to convert them to compatible types before trying to use the in operator.

a = set(nlp('This is a test'))
a = {str(token) for token in a} # convert all token type to str

b = nlp('is')
b = str(set(b).pop()) # convert token to str, effectively same as b = 'is'
if b in a:
  print("Success")
else:
  print("Failed")

Answer 2

I don't think you can rely on the hash of the token for the set operation You can dig in and look at the .text attribute

import spacy
nlp = spacy.load('en')
a = set(x.text for x in nlp('This is a test'))
b = nlp('is').text
if b in a:
  print("Success")
else:
  print("Failed")

proof...

>>> import spacy
>>> nlp = spacy.load('en')
>>> a = set(x.text for x in nlp('This is a test'))
>>> b = nlp('is').text
>>> if b in a:
...   print("Success")
... else:
...   print("Failed")
... 
Success

Answer 3

@bboyjacks : Thanks for high-lightening this interesting question .

I just want to let you know that it doesn't specifically related to spaCy framework, it's more related to python concepts.

Answer provided by @John La Rooy above is correct but I put my version as you have asked the same in spaCy community as well (this may add some clarity to the solution).

Please check my answer below:

print(a) # prints -> {This, test, is}
print(b) # prints -> is

So it seems like 'in' operator should work but the catch is below:

print(type(a))          # prints -> <class 'set'>
print(type(a.pop()))    # prints -> <class 'spacy.tokens.token.Token'>
print(type(b))          # prints -> <class 'spacy.tokens.doc.Doc'>

Object with type [spacy.tokens.doc.Doc] == Object with type [spacy.tokens.token.Token] will always returns ' False '

We need to somehow convert them into same type and since we are not sure about the equal method defined in Token or Doc classes provided by spaCy, simply convert both into str class.

This conversion can be done as above shown by @John La Rooy or you can also try below complete running code.

import spacy
nlp = spacy.load('en')
a = set(nlp('This is a test'))
b = nlp('is')

if b.text in map(lambda token: token.text, a):
  print("Success")
else:
  print("Failed")

Feel free to comment for any further clarification, my responses might be have some delay but I will try my best to reply.