如果在Python中我将一个列表放在元组中,我可以安全地更改该列表的内容吗？

Question

元组内部的值只是对列表的引用,如果我更改列表中的值,一切都仍然有序,对吧？我想确保如果我这样做,我将不会开始遇到混乱的错误.

Answer 1

元组是不可变的,你可能不会改变它们的内容.

有了清单

>>> x = [1,2,3]
>>> x[0] = 5
>>> x
[5, 2, 3]

有一个元组

>>> y = tuple([1,2,3])
>>> y
(1, 2, 3)
>>> y[0] = 5   # Not allowed!

Traceback (most recent call last):
  File "<pyshell#20>", line 1, in <module>
    y[0] = 5
TypeError: 'tuple' object does not support item assignment

但如果我理解你的问题,请说你有

>>> a = [1,2,3]
>>> b = [4,5,6]
>>> t = (a,b)
>>> t
([1, 2, 3], [4, 5, 6])

您可以将内部列表修改为

>>> t[0][0] = 5
>>> t
([5, 2, 3], [4, 5, 6])

Answer 2

元组是不可变的 - 你不能改变它们的结构

>>> a = []
>>> tup = (a,)
>>> tup[0] is a # tup stores the reference to a
True
>>> tup[0] = a # ... but you can't re-assign it later
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'tuple' object does not support item assignment
>>> tup[0] = 'string' # ... same for all other objects
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'tuple' object does not support item assignment

或大小

>>> del tup[0] # Nuh uh
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'tuple' object does not support item deletion
>>> id(tup)
139763805156632
>>> tup += ('something',) # works, because it creates a new tuple object:
>>> id(tup) # ... the id is different
139763805150344

在你创建它们之后.

另一方面,存储在元组中的可变对象不会失去可变性,例如,您仍然可以使用list方法修改内部列表:

>>> a = []
>>> b, c = (a,), (a,) # references to a, not copies of a
>>> b[0].append(1)
>>> b
([1],)
>>> c
([1],)

元组可以存储任何类型的对象,尽管包含列表(或任何其他可变对象)的元组不可清除:

>>> hash(b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: 'list'

上面演示的行为确实会导致混淆错误.