python operator.itemgetter

python四个魔法方法__len__,__getitem__,__setitem__,__delitem_

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class DictDemo:

def __init__(self,key,value):

self.dict = {}

self.dict[key] = value

def __getitem__(self,key):

return self.dict[key]

def __setitem__(self,key,value):

self.dict[key] = value

dictDemo = DictDemo('key0','value0')

print(dictDemo['key0']) #value0

dictDemo['key1'] = 'value1'

print(dictDemo['key1']) #value1

import weakref
实现weakref.ValueDict() descriptor 重写 set方法

import weakref  #实现通过id获取object

_id2obj_dict = weakref.WeakValueDictionary()

def remember(obj):
oid = id(obj)
_id2obj_dict[oid] = obj
return oid

def id2obj(oid):
return _id2obj_dict[oid]

----

operator

is
a built-in module providing a set of convenient operators. In two words

operator.itemgetter(n)

constructs
a callable that assumes iterable object (list, tuple, set) as input an fetches n-th element out of it.

So, you can't use

key=a[x][1]

there,
because python have no idea what

x

is.
Instead, you could use a

lambda

function
(

elem

is
just a variable name, no magic here):

a.sort(key=lambda elem: elem[1])

Or just ordinary function:

def get_second_elem(iterable):
return iterable[1]

a.sort(key=get_second_elem)

--------------------

>>> students = ['dave', 'john', 'jane']
>>> newgrades = {'john': 'F', 'jane':'A', 'dave': 'C'}
>>> sorted(students, key=newgrades.__getitem__)
['jane', 'dave', 'john']

>>> sorted(student_tuples, key=itemgetter(2), reverse=True)
[('john', 'A', 15), ('jane', 'B', 12), ('dave', 'B', 10)]

>>> sorted(student_objects, key=attrgetter('age'), reverse=True)
[('john', 'A', 15), ('jane', 'B', 12), ('dave', 'B', 10)]