Python字典方法总结

2023-01-31 02:01:49 python 方法 字典

1.清空字典中元素清空,dict变为{}

    L.clear()-> None.  Remove all items from L

>>> L ={'shaw':23,'sam':36,"eric":40}

>>> L.clear()

>>> print L

{}

2.返回一个字典的浅复制

    L.copy()-> a shallow copy of L

>>> L ={'shaw':23,'sam':36,"eric":40}

>>> L.copy()

{'shaw': 23, 'sam': 36,'eric': 40}

3.用于创建一个新字典,以序列seq中元素做字典的键,value为字典所有键对应的初始值(默认为“None”)

    Lict.fromkeys(S[,v])-> New Lict with keys from S and values equal to v. v defaults toNone.

>>> seq =('shaw','sam','stiven')

>>> name =dict.fromkeys(seq)

>>> print"listone :{}".fORMat(name)

listone :{'stiven': None,'shaw': None, 'sam': None}

>>> name =dict.fromkeys(seq,1000)

>>> print"listone :{}".format(name)

listone :{'stiven': 1000,'shaw': 1000, 'sam': 1000}

4.返回指定键的值,如果值不在字典中返回默认值(None)

    D.get(k[,d])-> D[k] if k in D, else d.  d defaultsto None.

>>> L ={'shaw':23,'sam':36,"eric":40}

>>> L.get('shaw')

23

>>> printL.get('stiven')

None

5.用于判断键是否存在于字典中,如果键在字典dict里返回true,否则返回false

    L.has_key(k) -> True if D has a key k,else False

>>> L ={'shaw':23,'sam':36,"eric":40}

>>> L.has_key('sam')

True

>>>L.has_key('linux')

False

6.以列表的方式返回可遍历的(键, 值) 元组(键值对)

    L.items()-> list of D's (key, value) pairs, as 2-tuples

>>> L ={'shaw':23,'sam':36,"eric":40}

>>> L.items()

[('shaw', 23), ('sam', 36),('eric', 40)]

 

7.以列表的方式返回一个字典所有的键

    L.keys()-> a set-like object providing a view on L's keys

>>> L ={'shaw':23,'sam':36,"eric":40}

>>> L.keys()

['shaw', 'sam', 'eric']

8.删除某个键值对

    D.pop(k[,d])-> v, remove specified key and return the corresponding value. If key is notfound, d is returned if given, otherwise KeyError is raised

>>> L ={'shaw':23,'sam':36,"eric":40}

>>> L.pop('sam')

36

>>> L

{'shaw': 23, 'eric': 40}

9.默认删除字典中第一个键值对

    D.popitem()-> (k, v), remove and return some (key, value) pair as a 2-tuple; but raise KeyError if D is empty.

>>> L ={'shaw':23,'sam':36,"eric":40}

>>> L.popitem()

('shaw', 23)

>>> L

{'sam': 36, 'eric': 40}

10. setdefault()方法和get()方法类似,如果键不已经存在于字典中,将会添加键并将值设为默认值(如果dict中已有a,则不会被覆盖)

    D.setdefault(k[,d]) ->D.get(k,d), also set D[k]=d if k not in D

>>> L ={'shaw':23,'sam':36,"eric":40}

>>>L.setdefault('stiven')

>>> L

{'stiven': None, 'shaw': 23,'sam': 36, 'eric': 40}

>>>L.setdefault('mira',65)

65

>>> L

{'stiven': None, 'mira': 65,'shaw': 23, 'sam': 36, 'eric': 40}

>>>L.setdefault('shaw',18)

23

>>> L

{'stiven': None, 'mira': 65,'shaw': 23, 'sam': 36, 'eric': 40}

11.把字典dict2的键/值对更新到dict里

    L.update()

>>> L ={'shaw':23,'sam':36,"eric":40}

>>> A ={'book':45,'apple':13}

>>> L.update(A)

>>> L

{'book': 45, 'apple': 13,'shaw': 23, 'sam': 36, 'eric': 40}

12.返回dic所有的值

    L.values(…)

>>> L = {'book':45,'apple':13}

>>> L.values()

[45, 13]

 


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