为什么Python中没有元组理解？

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.

You can use a generator expression:

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

but parentheses were already taken for … generator expressions.

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)

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

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

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.

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))

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 …

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 ] )

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) 

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.

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)]