问题:为什么Python中没有元组理解?

众所周知,列表理解

[i for i in [1, 2, 3, 4]]

并且有字典理解,例如

{i:j for i, j in {1: 'a', 2: 'b'}.items()}

(i for i in (1, 2, 3))

最终将成为生成器,而不是tuple理解力。这是为什么?

我的猜测是a tuple是不可变的,但这似乎并不是答案。

As we all know, there’s list comprehension, like

[i for i in [1, 2, 3, 4]]

and there is dictionary comprehension, like

{i:j for i, j in {1: 'a', 2: 'b'}.items()}

but

(i for i in (1, 2, 3))

will end up in a generator, not a tuple comprehension. Why is that?

My guess is that a tuple is immutable, but this does not seem to be the answer.


回答 0

您可以使用生成器表达式:

tuple(i for i in (1, 2, 3))

但是对于…生成器表达式,已经使用了括号。

You can use a generator expression:

tuple(i for i in (1, 2, 3))

but parentheses were already taken for … generator expressions.


回答 1

Raymond Hettinger(Python核心开发人员之一)在最近的一条推文中曾这样说过元组:

#python提示:通常,列表用于循环;结构的元组。列表是同质的;元组异构。列出可变长度。

(对我来说)支持这样的想法:如果序列中的项目相关性足以由生成器生成,那么它应该是一个列表。尽管元组是可迭代的,并且看起来只是一个不可变的列表,但它实际上与C结构的Python等效:

struct {
    int a;
    char b;
    float c;
} foo;

struct foo x = { 3, 'g', 5.9 };

成为Python

x = (3, 'g', 5.9)

Raymond Hettinger (one of the Python core developers) had this to say about tuples in a recent tweet:

#python tip: Generally, lists are for looping; tuples for structs. Lists are homogeneous; tuples heterogeneous. Lists for variable length.

This (to me) supports the idea that if the items in a sequence are related enough to be generated by a, well, generator, then it should be a list. Although a tuple is iterable and seems like simply a immutable list, it’s really the Python equivalent of a C struct:

struct {
    int a;
    char b;
    float c;
} foo;

struct foo x = { 3, 'g', 5.9 };

becomes in Python

x = (3, 'g', 5.9)

回答 2

从Python 3.5开始,您还可以使用splat *解包语法来解压缩生成器表达式:

*(x for x in range(10)),

Since Python 3.5, you can also use splat * unpacking syntax to unpack a generator expresion:

*(x for x in range(10)),

回答 3

正如另一位发帖人macm所述,从生成器创建元组的最快方法是tuple([generator])


性能比较

  • 清单理解:

    $ python3 -m timeit "a = [i for i in range(1000)]"
    10000 loops, best of 3: 27.4 usec per loop
    
  • 来自列表理解的元组:

    $ python3 -m timeit "a = tuple([i for i in range(1000)])"
    10000 loops, best of 3: 30.2 usec per loop
    
  • 生成器中的元组:

    $ python3 -m timeit "a = tuple(i for i in range(1000))"
    10000 loops, best of 3: 50.4 usec per loop
    
  • 打开包装的元组:

    $ python3 -m timeit "a = *(i for i in range(1000)),"
    10000 loops, best of 3: 52.7 usec per loop
    

我的python版本

$ python3 --version
Python 3.6.3

因此,除非性能不是问题,否则应始终从列表理解中创建一个元组。

As another poster macm mentioned, the fastest way to create a tuple from a generator is tuple([generator]).


Performance Comparison

  • List comprehension:

    $ python3 -m timeit "a = [i for i in range(1000)]"
    10000 loops, best of 3: 27.4 usec per loop
    
  • Tuple from list comprehension:

    $ python3 -m timeit "a = tuple([i for i in range(1000)])"
    10000 loops, best of 3: 30.2 usec per loop
    
  • Tuple from generator:

    $ python3 -m timeit "a = tuple(i for i in range(1000))"
    10000 loops, best of 3: 50.4 usec per loop
    
  • Tuple from unpacking:

    $ python3 -m timeit "a = *(i for i in range(1000)),"
    10000 loops, best of 3: 52.7 usec per loop
    

My version of python:

$ python3 --version
Python 3.6.3

So you should always create a tuple from a list comprehension unless performance is not an issue.


回答 4

理解通过循环或迭代项并将它们分配到容器中来工作,元组无法接收分配。

创建元组后,将无法对其进行追加,扩展或分配。修改元组的唯一方法是是否可以将其对象之一本身分配给它(是非元组容器)。因为元组仅持有对此类对象的引用。

另外-元组有自己的构造函数tuple(),您可以提供任何迭代器。这意味着要创建一个元组,您可以执行以下操作:

tuple(i for i in (1,2,3))

Comprehension works by looping or iterating over items and assigning them into a container, a Tuple is unable to receive assignments.

Once a Tuple is created, it can not be appended to, extended, or assigned to. The only way to modify a Tuple is if one of its objects can itself be assigned to (is a non-tuple container). Because the Tuple is only holding a reference to that kind of object.

Also – a tuple has its own constructor tuple() which you can give any iterator. Which means that to create a tuple, you could do:

tuple(i for i in (1,2,3))

回答 5

我最好的猜测是,他们用完了括号,并认为这对警告添加“丑陋”语法没有足够的帮助…

My best guess is that they ran out of brackets and didn’t think it would be useful enough to warrent adding an “ugly” syntax …


回答 6

元组不能像列表一样有效地附加。

因此,元组理解将需要在内部使用列表,然后转换为元组。

那将与您现在所做的相同:tuple([comprehension])

Tuples cannot efficiently be appended like a list.

So a tuple comprehension would need to use a list internally and then convert to a tuple.

That would be the same as what you do now : tuple( [ comprehension ] )


回答 7

括号不会创建元组。又名一个=(两个)不是元组。唯一的解决方法是一个=(两个)或一个=元组(两个)。因此,解决方案是:

tuple(i for i in myothertupleorlistordict) 

Parentheses do not create a tuple. aka one = (two) is not a tuple. The only way around is either one = (two,) or one = tuple(two). So a solution is:

tuple(i for i in myothertupleorlistordict) 

回答 8

我相信这只是为了清楚起见,我们不想用太多不同的符号来使语言混乱。同样tuple也不需要理解,可以使用列表,而速度差异可以忽略不计,而不是像dict理解而不是list理解。

I believe it’s simply for the sake of clarity, we do not want to clutter the language with too many different symbols. Also a tuple comprehension is never necessary, a list can just be used instead with negligible speed differences, unlike a dict comprehension as opposed to a list comprehension.


回答 9

我们可以从列表理解中生成元组。下一个将两个数字顺序加到一个元组中,并给出一个从0-9的列表。

>>> print k
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]
>>> r= [tuple(k[i:i+2]) for i in xrange(10) if not i%2]
>>> print r
[(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)]

We can generate tuples from a list comprehension. The following one adds two numbers sequentially into a tuple and gives a list from numbers 0-9.

>>> print k
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]
>>> r= [tuple(k[i:i+2]) for i in xrange(10) if not i%2]
>>> print r
[(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)]

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