基于空间规则匹配器选择Pandas DataFrame的行

Question

我需要根据基于spacy规则的配对结果分割一只熊猫DataFrame。以下是我尝试过的。

import pandas as pd
import numpy as np
import spacy
from spacy.matcher import Matcher

df = pd.DataFrame([['Eight people believed injured in serious SH1 crash involving truck and three cars at Hunterville',
 'Fire and emergency responding to incident at Mataura, Southland ouvea premix site',
 'Civil Defence Minister Peeni Henare heartbroken over Northland flooding',
 'Far North flooding: New photos reveal damage to roads']]).T
df.columns = ['col1']

nlp = spacy.load("en_core_web_sm")

flood_pattern = [{'LOWER': 'flooding'}]

matcher = Matcher(nlp.vocab, validate=True)
matcher.add("FLOOD_DIS", None, flood_pattern)
titles = (_ for _ in df['col1'])
g = (d for d in nlp.pipe(titles) if matcher(d))
x = list(g)

df2 = df[df['col1'].isin(x)]
df2

这会产生一个空的DataFrame。但是，它应该从df中提取以下两行。

1. 民防部长皮尼·赫纳雷因北部洪水
2. 远北洪水而心碎:新照片显示道路遭到破坏

Answer 1

你可以做下面的事。

titles = (_ for _ in df['col1'])
g = (d for d in nlp.pipe(titles) if matcher(d))


A = []
for i in range(len(df)):
    doc = nlp(next(titles))
    if len(matcher(doc)) == 1:
        A.append(str(doc))
df2 = df[df['col1'].isin(A)]

Answer 2

试试这个：

matcher.add("FLOOD_DIS", None, flood_pattern)
matches = [True if matcher(doc) else False for doc in nlp.pipe(df['col1'])]
df2 = df[matches][['col1']]