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Python中dict详解

#字典的添加、删除、修改操作 dict = {"a" : "apple", "b" : "banana", "g" : "grape", "o" : "orange"} dict["w"] = "watermelon" del(dict["a"]) dict["g"] = "grapefruit" print dict.pop("b") print dict dict.clear() print dict #字典的遍历 dict = {"a" : "apple", "b" : "banana", "g" : "grape", "o" : "orange"} for k in dict:     print "dict[%s] =" % k,dict[k] #字典items()的使用 dict = {"a" : "apple", "b" : "banana", "c" : "grape", "d" : "orange"} #每个元素是一个key和value组成的元组,以列表的方式输出 print dict.items() #调用items()实现字典的遍历 dict = {"a" : "apple", "b" : "banana", "g" : "grape", "o" : "orange"} for (k, v) in dict.items():     print "dict[%s] =" % k, v #调用iteritems()实现字典的遍历 dict = {"a" : "apple", "b" : "banana", "c" : "grape", "d" : "orange"} print dict.iteritems() for k, v in dict.iteritems():     print "dict[%s] =" % k, v for (k, v) in zip(dict.iterkeys(), dict.itervalues()):     print "dict[%s] =" % k, v #使用列表、字典作为字典的值 dict = {"a" : ("apple",), "bo" : {"b" : "banana", "o" : "orange"}, "g" : ["grape","grapefruit"]} print dict["a"] print dict["a"][0] print dict["bo"] print dict["bo"]["o"] print dict["g"] print dict["g"][1] dict = {"a" : "apple", "b" : "banana", "c" : "grape", "d" : "orange"} #输出key的列表 print dict.keys() #输出value的列表 print dict.values() #每个元素是一个key和value组成的元组,以列表的方式输出 print dict.items() dict = {"a" : "apple", "b" : "banana", "c" : "grape", "d" : "orange"} it = dict.iteritems() print it #字典中元素的获取方法 dict = {"a" : "apple", "b" : "banana", "c" : "grape", "d" : "orange"} print dict print dict.get("c", "apple")          print dict.get("e", "apple") #get()的等价语句 D = {"key1" : "value1", "key2" : "value2"} if "key1" in D:     print D["key1"] else:     print "None" #字典的更新 dict = {"a" : "apple", "b" : "banana"} print dict dict2 = {"c" : "grape", "d" : "orange"} dict.update(dict2) print dict #udpate()的等价语句 D = {"key1" : "value1", "key2" : "value2"} E = {"key3" : "value3", "key4" : "value4"} for k in E:     D[k] = E[k] print D #字典E中含有字典D中的key D = {"key1" : "value1", "key2" : "value2"} E = {"key2" : "value3", "key4" : "value4"} for k in E:     D[k] = E[k]

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