Question

我有以下元组

text =[('Michael', 'PERSON'), ('Jordan', 'PERSON'), ("'s", 'O'), ('legacy', 'O'), ('in', 'O'), ('the', 'O'), ('90', 'O'), ("'s", 'O'), ('shows', 'O'), ('that', 'O'), ('he', 'O'), ('was', 'O'), ('the', 'O'), ('biggest', 'O'), ('player', 'O'), ('ever', 'O'), ('in', 'O'), ('the', 'O'), ('NBA', 'ORGANIZATION'), ('.', 'O')]

最初的句子是“迈克尔·乔丹（Michael Jordan）在90年代的表演中显示出他是NBA最大的球员。”

我需要删除归类为“ PERSON”的元素

我所做的

new_text = [x for x in text if x[1] != "PERSON"]
sentence= " ".join(x[0] for x in new_text)
print(sentence)

我得到的输出是

's legacy in the 90 's shows that he was the biggest player ever in the NBA .

请在开始时注意"'s"。

现在我被困住了，因为我需要在成为"PERSON"之前删除该元素上有条件的“'s”元素。在此示例中，有2个"'s"，但是我只想删除在"PERSON"之后的那个。有什么建议吗？

感谢您的任何输入。

Answer 1

一种方法是使用text遍历zip及其转换版本，并根据以下条件保留字符串：

out = []
for i,j in zip(text[:-1], text[1:]):
    if j[0] == "'s":
        if i[1] == 'PERSON':
            continue
        else:
            out.append(j[0])
    else:
        if i[1] != 'PERSON':
            out.append(j[0])

' '.join(out)
"legacy in the 90 's shows that he was the biggest player ever in the NBA ."

Answer 2

在这里使用简单的for循环会更容易。请注意，enumerate用于检索前一个元素（text[pos-1]），但是，只有在存在前一个元素（pos > 0）时才能这样做。

#!/usr/bin/env python3

text =[('Michael', 'PERSON'), ('Jordan', 'PERSON'), ("'s", 'O'), ('legacy', 'O'), ('in', 'O'), ('the', 'O'), ('90', 'O'), ("'s", 'O'), ('shows', 'O'), ('that', 'O'), ('he', 'O'), ('was', 'O'), ('the', 'O'), ('biggest', 'O'), ('player', 'O'), ('ever', 'O'), ('in', 'O'), ('the', 'O'), ('NBA', 'ORGANIZATION'), ('.', 'O')]


new_text = []
for pos, (word, type_) in enumerate(text):
    if type_ == "PERSON":
        # we ignore words of type PERSON
        continue
    if word == "'s" and pos > 0 and text[pos-1][1] == "PERSON":
        # ignore 's if the previous word was of type PERSON
        continue 
    new_text.append((word, type_))


sentence= " ".join(x[0] for x in new_text)
print(sentence)shows

执行此脚本将产生以下文本：
legacy in the 90 's shows that he was the biggest player ever in the NBA .

Answer 3

如果找到range，您可以只使用O并向后看：

text =[('Michael', 'PERSON'), ('Jordan', 'PERSON'), ("'s", 'O'), ('legacy', 'O'), ('in', 'O'), ('the', 'O'), ('90', 'O'), ("'s", 'O'), ('shows', 'O'), ('that', 'O'), ('he', 'O'), ('was', 'O'), ('the', 'O'), ('biggest', 'O'), ('player', 'O'), ('ever', 'O'), ('in', 'O'), ('the', 'O'), ('NBA', 'ORGANIZATION'), ('.', 'O')]

filtered_text = []

for idx in range(len(text)):
  if text[idx][1] == "PERSON":
    continue

  if text[idx][1] == 'O' and idx > 0 and text[idx-1][1] == 'PERSON':
    continue

  filtered_text.append(text[idx][0])

sentence= " ".join(filtered_text)
print(sentence)

Answer 4

text = [('', j[1]) if j[0] == "'s" and text[i-1][1]=='PERSON' else j for i, j in enumerate(text)]
print(' '.join([i for i, j in text if j !='PERSON']))

输出

legacy in the 90 's shows that he was the biggest player ever in the NBA .

Answer 5

我来晚了，但是如果这只是要解决的一个条件，那么这也可以。一个非常简单的补充，您已经拥有的。

text =[('Michael', 'PERSON'), ('Jordan', 'PERSON'), ("'s", 'O'), ('legacy', 'O'), ('in', 'O'), ('the', 'O'), ('90', 'O'), ("'s", 'O'), ('shows', 'O'), ('that', 'O'), ('he', 'O'), ('was', 'O'), ('the', 'O'), ('biggest', 'O'), ('player', 'O'), ('ever', 'O'), ('in', 'O'), ('the', 'O'), ('NBA', 'ORGANIZATION'), ('.', 'O')]

new_text = [x for idx, x in enumerate(text) if x[1] != "PERSON" and not (idx > 1 and text[idx - 1][1] == "PERSON" and x[0] == "'s")]
sentence= " ".join(x[0] for x in new_text)
print(sentence)

输出如下

"legacy in the 90 's shows that he was the biggest player ever in the NBA ."

从其他元素为条件的元组中删除元素

5 个答案: