Python NLP库，用于判断笑话是否令人反感

Question

是否有一个python库包含一个方法，通过给出一个笑话来决定这个笑话是否令人反感?

我发现了存储库SemEval，但我认为有人已经做了一个用户友好的库来做这件事。

Answer 1

在HuggingFace中有一些预训练模型，你可以简单地使用它们来检测句子是否具有攻击性。例如，你可以使用twitter-roberta-base-offensive模型:

from transformers import AutoModelForSequenceClassification
from transformers import TFAutoModelForSequenceClassification
from transformers import AutoTokenizer
import numpy as np
from scipy.special import softmax
import csv
import urllib.request
# Preprocess text (username and link placeholders)
def preprocess(text):
new_text = []
for t in text.split(" "):
t = '@user' if t.startswith('@') and len(t) > 1 else t
t = 'http' if t.startswith('http') else t
new_text.append(t)
return " ".join(new_text)
task='offensive'
MODEL = f"cardiffnlp/twitter-roberta-base-{task}"
tokenizer = AutoTokenizer.from_pretrained(MODEL)
# download label mapping
labels=[]
mapping_link = f"https://raw.githubusercontent.com/cardiffnlp/tweeteval/main/datasets/{task}/mapping.txt"
with urllib.request.urlopen(mapping_link) as f:
html = f.read().decode('utf-8').split("n")
csvreader = csv.reader(html, delimiter='t')
labels = [row[1] for row in csvreader if len(row) > 1]
# PT
model = AutoModelForSequenceClassification.from_pretrained(MODEL)
model.save_pretrained(MODEL)
text = "Good night 😊"
text = preprocess(text)
encoded_input = tokenizer(text, return_tensors='pt')
output = model(**encoded_input)
scores = output[0][0].detach().numpy()
scores = softmax(scores)
ranking = np.argsort(scores)
ranking = ranking[::-1]
for i in range(scores.shape[0]):
l = labels[ranking[i]]
s = scores[ranking[i]]
print(f"{i+1}) {l} {np.round(float(s), 4)}")