Python 字符串中短语前的先行词数

假设我有一个短语列表：

list = ['new york', 'school', 'new']

还有一根绳子

text = 'i am going to a school in new york and therefore i have to buy a new uniform to go to new york'

我想找出每个短语前面的单词数（仅用于第一次出现），ie输出应为：

new york = 7
school = 5
new = 7

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你知道我如何才能有效地实现这一点吗？

天真的方法，没有任何性能或NLP考虑：

lst = ['new york', 'school', 'new']  # do not use 'list' as a name
text = 'i am going to a school in new york and therefore i have to buy a new uniform to go to new york'

{p: len(text[:text.find(p)].strip().split()) for p in lst}
# {'new york': 7, 'school': 5, 'new': 7}

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使用

计数

和

索引

：

lst = ['new york', 'school', 'new']
text = 'i am going to a school in new york and therefore i have to buy a new uniform to go to new york'

for x in lst:
    print(f"{x} = {text.count(' ', 0, text.index(x))}")

# new york = 7
# school = 5                                                   
# new = 7

count

统计

text

中的空白，从开始到第一次出现与该短语前面的单词数相同的短语

lst = ['new york', 'school', 'new']
text = 'i am going to a school in new york and therefore i have to buy a new uniform to go to new york'

这将为您提供正在搜索其计数的字符串和字符串的计数

for x in lst:
    print(x +": "+str(len(text[0:text.index(x)].split(' ')) -1))

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首先，您需要一些东西来标记您的

文本

和

列表的元素

。您是否使用了一些函数/模块作为标记器？new不应该也是7吗？不，我只是在spacesYes上拆分文本，应该是7我修复了这个问题