你看到的解决方案是搜索每个字母。我的解决方案通过查看整个字符串来解决这个问题，它返回一个数组而不是第一个值。

myDict = {"Key1": "The dog ran over the bridge",
    "Key2": "The cat sleeps under the rock",
    "Key3": "The house is dark at night and the dog waits"}

def search(values, searchFor):
    listOfKeys = []
    for k in values.items():
        if searchFor in k[1]:
            listOfKeys.append(k[0])
    return listOfKeys

print(search(myDict, "dog"))

它将输出：

['Key1', 'Key3']

这里有一个使用列表解析的版本。https://docs.python.org/3/tutorial/datastructures.html#list-comprehensions

d = {
    "Key1": "The dog ran over the bridge",
    "Key2": "The cat sleeps under the rock",
    "Key3": "The house is dark at night and the dog waits",
}

def find(values, key):
    key = key.lower()
    return [k for k, v in values.items() if key in v.lower()]

print(find(d, "dog"))

如果这将是经常做的事情，那么确保dic值都是小写的，并以这种方式存储它们是值得的。

d = {
    "Key1": "The dog ran over the bridge",
    "Key2": "The cat sleeps under the rock",
    "Key3": "The house is dark at night and the dog waits",
}

for k in d:
    d[k] = d[k].lower()

def find(values, key):
    key = key.lower()
    return [k for k, v in values.items() if key in v]

print(find(d, "dog"))

我发现使用Array.filter（）更容易一些：

const myDict = {"Key1": "The dog ran over the bridge", 
          "Key2": "The cat sleeps under the rock", 
          "Key3": "The house is dark at night and the dog waits"}

const searchString = "dog";
const res = Object.keys(myDict).filter((key) => myDict[key].includes(searchString));

console.log(res);

Iterating Through .items（）应该会给予结果：

myDict = {"Key1": "The dog ran over the bridge", 
          "Key2": "The cat sleeps under the rock", 
          "Key3": "The house is dark at night and the dog waits"}

for key, value in myDict.items():
    if "dog" in value.lower():
        print(key)