如何将一系列条件映射为字典中的键？

Question

我知道你可以使用字典作为switch语句的替代,如下所示:

def printMessage(mystring):
    # Switch statement without a dictionary
    if mystring == "helloworld":
        print "say hello"
    elif mystring == "byeworld":
        print "say bye"
    elif mystring == "goodafternoonworld":
        print "good afternoon"


def printMessage(mystring):
    # Dictionary equivalent of a switch statement
    myDictionary = {"helloworld": "say hello",
                    "byeworld": "say bye",
                    "goodafternoonworld": "good afternoon"}
    print myDictionary[mystring]

但是,如果使用条件,除了返回true为false的等于(==)之外,这些不能轻易映射,即:

if i > 0.5:
    print "greater than 0.5"
elif i == 5:
    print "it is equal to 5"
elif i > 5 and i < 6:
    print "somewhere between 5 and 6"

以上内容不能直接转换为字典键值对:

# this does not work
mydictionary  = { i > 0.5: "greater than 0.5" }

可以使用lambda,因为它是可散列的,但是从地图中获取结果字符串的唯一方法是将相同的lambda对象传递给字典,而不是在lambda的求值为真时:

x = lambda i: i > 0.5
mydictionary[x] = "greater than 0.5"
# you can get the string by doing this:
mydictionary[x]
# which doesnt result in the evaluation of x

# however a lambda is a hashable item in a dictionary
mydictionary = {lambda i: i > 0.5: "greater than 0.5"}

有没有人知道在lambda评估和返回值之间创建映射的技术或方法？(这可能类似于函数式语言中的模式匹配)

Answer 1

你的条件是连续的; 你想一个接一个地测试,而不是在这里将少量键映射到一个值.改变条件的顺序可能会改变结果; 样本中的5结果值"greater than 0.5",而不是"it is equal to 5".

使用元组列表:

myconditions  = [
    (lambda i: i > 0.5, "greater than 0.5"),
    (lambda i: i == 5, "it is equal to 5"),
    (lambda i: i > 5 and i < 6, "somewhere between 5 and 6"),
]

之后你可以依次访问每一个,直到匹配:

for test, message in myconditions:
    if test(i):
        return message

重新排序测试将改变结果.

字典适用于您的第一个示例,因为存在针对由字典优化的多个静态值的简单相等性测试,但此处没有这样的简单等式.