python - 如何在Python中将2列表中的单词与另一个单词字符串进行匹配而无需子字符串匹配？

Question

我有 2 个包含关键字的列表:

slangNames = [Vikes, Demmies, D, MS Contin]
riskNames = [enough, pop, final, stress, trade]

我还有一个名为 overallDict 的字典，其中包含推文。键值对是{ID:推文文本)例如:

{1:"Vikes is not enough for me", 2:"Demmies is okay", 3:"pop a D"}

我试图仅从 slangNames 和riskNames 中分离出那些至少具有一个关键字的推文。因此，推文必须包含 slangNames 中的任何关键字以及riskNames 中的任何关键字。因此，从上面的示例来看，我的代码应该返回键 1 和 3，即

{1:"Vikes is not enough for me", 3:"pop a D"}. 

但是我的代码正在拾取子字符串而不是完整的单词。所以基本上，任何带有字母“D”的东西都会被选中。如何将它们匹配为整个单词而不是子字符串？请帮忙。谢谢!

到目前为止我的代码如下:

for key in overallDict:
    if any(x in overallDict[key] for x in strippedRisks) and (any(x in overallDict[key] for x in strippedSlangs)):
        output.append(key)

Answer 1

将 slangNames 和riskNames存储为集合，拆分字符串并检查是否有任何单词出现在两个集合中

slangNames = set(["Vikes", "Demmies", "D", "MS", "Contin"])
riskNames = set(["enough", "pop", "final", "stress", "trade"])
d =  {1: "Vikes is not enough for me", 2:"Demmies is okay", 3:"pop a D"}

for k,v in d.items():
    spl = v.split() # split once
    if any(word in slangNames for word in spl) and any(word  in riskNames for word in spl):
        print(k,v)

输出:

1 Vikes is not enough for me
3 pop a D

或者不使用set.isdisjoint:

slangNames = set(["Vikes", "Demmies", "D", "MS", "Contin"])
riskNames = set(["enough", "pop", "final", "stress", "trade"])
d =  {1: "Vikes is not enough for me", 2:"Demmies is okay", 3:"pop a D"}

for k,v in d.items():
    spl = v.split()
    if not slangNames.isdisjoint(spl) and not riskNames.isdisjoint(spl):
        print(k, v)

使用any应该是最有效的，因为我们会在第一场比赛中短路。如果两个集合的交集是空集，则它们是不相交的，因此如果 if not slangNames.isdisjoint(spl) 为 True，则至少会出现一个常见单词。

如果 MS Contin 实际上是一个单词，您还需要捕获这一点:

import re
slangNames = set(["Vikes", "Demmies", "D"])
r = re.compile(r"\bMS Contin\b")
riskNames = set(["enough", "pop", "final", "stress", "trade"])
d =  {1: "Vikes is not enough for me", 2:"Demmies is okay", 3:"pop a D"}

for k,v in d.items():
    spl = v.split()
    if (not slangNames.isdisjoint(spl) or r.search(v)) and not riskNames.isdisjoint(spl):
        print(k,v)