在Python中使用交集查找多于1个单词

Question

我目前正在使用以下方法查找一个或多个单词是否在字符串中：

stopwords=set(["Good to Soft","Slow","Standard to Slow","Standard","Standard to Fast","Fast","Heavy","Soft to Heavy","Soft","Yeliding to Soft","Yeilding","Good to Yeilding","Good","Good to Firm","Firm","Hard"])
Going = stopwords.intersection(set(testtext[0].split()))

我的问题是，如果字符串有Standard To Slow，那么交集似乎只拾取Standard，Slow-我想要Standard To Slow。你知道吗

还有什么我可以改进的吗？你知道吗

为了更清楚地解释我需要什么结果，我将使用以下示例：

我有一份要去的地方的清单

["Good to Soft","Slow","Standard to Slow","Standard","Standard to Fast","Fast","Heavy","Soft to Heavy","Soft","Yeliding to Soft","Yeilding","Good to Yeilding","Good","Good to Firm","Firm","Hard"]

我想找出哪些在我的字符串中，例如：

2:20 - H Brown & Son Recycling Maiden Hurdle (Class4) �4,094 Good 2m3f207y

这将返回“好”，因为这是匹配

2:20 - H Brown & Son Recycling Maiden Hurdle (Class4) �4,094 Good to Soft 2m3f207y

这将返回“好到软”，因为这是匹配，而不是“软”

Answer 1

交集只能找到匹配的元素。以下是您要比较的列表：

["Good to Soft","Slow","Standard to Slow","Standard","Standard to Fast","Fast","Heavy","Soft to Heavy","Soft","Yeliding to Soft","Yeilding","Good to Yeilding","Good","Good to Firm","Firm","Hard"]

testtext[0].split() == ["Standard", "to", "Slow"]

如果只希望它与testtext[0]中的内容匹配，则不应使用split()来创建列表：

Going = stopwords.intersection({testtext[0]})