Question

民间，你会如何用字典重写'checkme'函数中的if / elif？

def dosomething(queue):
    ...

def checkme(queue):
  """ Consume Message """
  if queue == 'foo':
    username = 'foo'
    password = 'vlTTdhML'
  elif queue == 'bar':
    username = 'bar'
    password = 'xneoYb2c'
  elif queue == 'baz':
    username = 'baz'
    password = 'wnkyVsBI'
  ...
  dosomething(queue)

def main():
    checkme('foo')
    checkme('bar')
    checkme('baz')

Answer 1

你可以这样做：

CHECK_ME = {'foo': 'vlTTdhML', 'bar': 'xneoYb2c', 'baz': 'wnkyVsBI'}

def checkme(queue):
    username, password = queue, CHECK_ME.get(queue)
    #May be some more check here, like
    if not password:
        print 'password is none'
        #Or do something more relevant here

    #rest of the code.

Answer 2

看起来你依赖于副作用，尤其是dosomething(queue)部分，所以我认为在我的解决方案中处理完全正常，但我更愿意这样做这不依赖于副作用。

def checkme(queue):

    class to_do_dict(dict):
        def __missing__(self, itm):
            dosomething(itm)

    to_do = to_do_dict({
        "foo":("foo", "v1TTdhML"),
        "bar":("bar", "xneoYb2c")})

    username, password = to_do[queue]

Answer 3

试试这个：

passwords = {'foo':'vlTTdhML', 'bar':'xneoYb2c', 'baz':'wnkyVsBI'}
username, password = queue, passwords[queue]

以上假设每个用户都在字典中有一个密码。如果情况并非如此，那么最好安全地使用它：

username, password = queue, passwords.get(queue, None)

无论哪种方式，您都可以在最后调用dosomething(queue)。正如问题中目前所述，始终会调用dosomething。

Answer 4

你可以这样做：

{'foo'：{'username'：'foo'，'password'：'vlTTdhML'}}

只需按照您的意愿添加词典。

Answer 5

嵌套词典可以解决问题。

设置用户名和密码：

queue = {}

queue["foo"] = {"username": "foo", "password": "vlTTdhML" }
queue["bar"] = {"username": "bar", "password": "xneoYb2c" }

并检查用户名/密码是否存在：

if queue.get("foo"):
 username = queue["foo"]["username"]
 password = queue["foo"]["password"]

else:
 # username does not exist, so do something
 print "username does not exist"

python，将if / elif替换为字典

5 个答案: