如何将 2 列表中的单词与 Python 中没有子字符串匹配的另一个单词字符串进行匹配？

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]

我还有一本名为 overallDict 的字典，其中包含推文。键值对是 {ID: Tweet text) 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 中分离出至少有一个关键字的推文。因此，推文必须包含来自 slangNames 的任何关键字和来自 riskNames 的任何关键字。所以从上面的例子来看，我的代码应该 return 键 1 和 3，即

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

但我的代码正在提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' 的东西都会被拾取。我如何将这些作为整个单词而不是子字符串进行匹配？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，因为我们将在第一场比赛中短路。如果两个集合的交集是一个空集合，则这两个集合是不相交的，因此如果 if not slangNames.isdisjoint(spl) 为真，则至少出现一个常用词。

如果 MS Contin 实际上是一个词，您还需要注意：

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)

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