本章主要介绍了Python中tuple和list的一些知识。
列表解析和生成器表达式
使用()可以得到一个生成器，可以看出，生成器的内存使用更少。
>>> a = [i for i in range(10)]
>>> g = (i for i in range(10))
>>> a
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> g
at 0x107deca50>
>>> a.__sizeof__()
168
>>> g.__sizeof__()
48
Python3中的列表推导没有了变量泄漏问题
python2中：
>>> x = 'my var'
>>> dummy = [x for x in 'ABC']
>>> x
'C'
python3中：
>>> x = 'my var'
>>> dummy = [x for x in 'ABC']
>>> x
'my var'
>>> dummy
['A', 'B', 'C']
使用*处理剩下的元素
>>> a, b , *part, c = range(10)
>>> part
[2, 3, 4, 5, 6, 7, 8]
关于namedtuple
namedtuple Returns a new tuple subclass named typename.
>>> from collections import namedtuple
>>> Point = namedtuple('Point', ['x', 'y'], verbose=True)
from builtins import property as _property, tuple as _tuple
from operator import itemgetter as _itemgetter
from collections import OrderedDict
class Point(tuple):
'Point(x, y)'
__slots__ = ()
_fields = ('x', 'y')
def __new__(_cls, x, y):
'Create new instance of Point(x, y)'
return _tuple.__new__(_cls, (x, y))
@classmethod
def _make(cls, iterable, new=tuple.__new__, len=len):
'Make a new Point object from a sequence or iterable'
result = new(cls, iterable)
if len(result) != 2:
raise TypeError('Expected 2 arguments, got %d' % len(result))
return result
def _replace(_self, **kwds):
'Return a new Point object replacing specified fields with new values'
result = _self._make(map(kwds.pop, ('x', 'y'), _self))
if kwds:
raise ValueError('Got unexpected field names: %r' % list(kwds))
return result
def __repr__(self):
'Return a nicely formatted representation string'
return self.__class__.__name__ + '(x=%r, y=%r)' % self
def _asdict(self):
'Return a new OrderedDict which maps field names to their values.'
return OrderedDict(zip(self._fields, self))
def __getnewargs__(self):
'Return self as a plain tuple. Used by copy and pickle.'
return tuple(self)
x = _property(_itemgetter(0), doc='Alias for field number 0')
y = _property(_itemgetter(1), doc='Alias for field number 1')
可以看出，Point类继承自tuple类，通过__slots__把__dict__设置为空，只有两个可读属性x和y。
使用*操作列表
使用 * 复制列表时候，可能是同一个元素的引用。
比如：
>>> a = [[]]*3
>>> a
[[], [], []]
>>> a[0].append(1)
>>> a
[[1], [1], [1]]
一个关于+=的谜题
tuple是不可变的，但是如果tuple里面有可变的元素，比如：
>>> atuple = (1,2,3,[1,2,3,4])
>>> atuple[3] += [5,6]
Traceback (most recent call last):
File "", line 1, in
TypeError: 'tuple' object does not support item assignment
>>> atuple
(1, 2, 3, [1, 2, 3, 4, 5, 6])
这里会出现异常，但是对tuple的修改也变了。
说明在tuple中最好不要放可以改变的元素。