python dict pop key value

星期二, 2015-01-27 20:34 — shiping1

self.data.pop(key, *args) ←------ why is there a second argument? ... If key is in the dictionary,
 remove it and return its value, else return default. 

class a(object):
    data={'a':'aaa','b':'bbb','c':'ccc'}
    def pop(self, key, *args):
            return self.data.pop(key, *args)#what is this mean.

b=a()
print b.pop('a',{'b':'bbb'})
print b.data

self.data.pop(key, *args) ←------ why is there a second argument?

up vote34down voteaccepted

The pop method of dicts (like self.data, i.e. {'a':'aaa','b':'bbb','c':'ccc'}, here) takes two arguments -- see the docs

The second argument, default, is what pop returns if the first argument, key, is absent. (If you callpop with just one argument, key, it raises an exception if that key's absent).

In your example, print b.pop('a',{'b':'bbb'}), this is irrelevant because 'a' is a key in b.data. But if you repeat that line...:

b=a()
print b.pop('a',{'b':'bbb'})
print b.pop('a',{'b':'bbb'})
print b.data

you'll see it makes a difference: the first pop removes the 'a' key, so in the second pop the default argument is actually returned (since 'a' is now absent from b.data).

来自 http://stackoverflow.com/questions/1990802/in-python-what-does-dict-popa-b-mean

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