re.findall（）查找包含负项的所有双字母组

Question

I am required to use the re.findall() function to find all bigrams that contain a negative term ("never" or "not") as the first word in the following text:

He jests at scars that never felt a wound. JULIET appears above at a window But, soft! what light through yonder window breaks? It is the east, and Juliet is the sun. Arise, fair sun, and kill the envious moon, Who is already sick and pale with grief, That thou her maid art far more fair than she: Be not her maid, since she is envious; Her vestal livery is but sick and green And none but fools do wear it; cast it off. It is my lady, O, it is my love! O, that she knew she were! She speaks yet she says nothing: what of that? Her eye discourses; I will answer it. I am too bold, 'tis not to me she speaks: Two of the fairest stars in all the heaven, Having some business, do entreat her eyes To twinkle in their spheres till they return. What if her eyes were there, they in her head? The brightness of her cheek would shame those stars, As daylight doth a lamp; her eyes in heaven Would through the airy region stream so bright That birds would sing and think it were not night. See, how she leans her cheek upon her hand! O, that I were a glove upon that hand, That I might touch that cheek!

I have no problem trying to find one word, but I am at a loss with finding bigrams.

import re
inp = input("please enter an expression: ")
print (re.findall(r'\b(?:never|not)\b', inp))

['never', 'not', 'not', 'not]

How do I get

['never felt', 'not her', 'not to', 'not right']

Answer 1

If you want to also wrap a word just after not or never you need to extend your regex to this,

\b(?:never|not)\s+[a-zA-Z]+

Here, \\s+ will match one or more whitespace and [a-zA-Z]+ will match one english word having one or more characters.

Regex Demo

Python code demo

import re

s = '''He jests at scars that never felt a wound. JULIET appears above at a window But, soft! what light through yonder window breaks? It is the east, and Juliet is the sun. Arise, fair sun, and kill the envious moon, Who is already sick and pale with grief, That thou her maid art far more fair than she: Be not her maid, since she is envious; Her vestal livery is but sick and green And none but fools do wear it; cast it off. It is my lady, O, it is my love! O, that she knew she were! She speaks yet she says nothing: what of that? Her eye discourses; I will answer it. I am too bold, 'tis not to me she speaks: Two of the fairest stars in all the heaven, Having some business, do entreat her eyes To twinkle in their spheres till they return. What if her eyes were there, they in her head? The brightness of her cheek would shame those stars, As daylight doth a lamp; her eyes in heaven Would through the airy region stream so bright That birds would sing and think it were not night. See, how she leans her cheek upon her hand! O, that I were a glove upon that hand, That I might touch that cheek!'''
print(re.findall(r'\b(?:never|not)\s+[a-zA-Z]+', s))

Prints,

['never felt', 'not her', 'not to', 'not night']

Edit: As you said, you want to discard matches that are followed by a space and a character, you can use a negative look ahead and extend current regex like this,

\b(?:never|not)\s+[a-zA-Z]+\b(?! a\b)

Here I have used \\b before the negative lookahead to avoid partial match of word and \\b after a in negative lookahead avoids matching words that are not just a but something more like add or and etc.

Regex Demo where matches are discard if followed by space and a char

Answer 2

x=input()
m = re.findall(r'\b(?:never|not)\b\s+[\w]+', x)
print(m)
# output
['never felt', 'not her', 'not to', 'not night']