Re: How to create a tuple quickly with list comprehension?

[Hatem Nassrat]

on Wed Jun 13 10:17:24 CEST 2007, Diez B. Roggisch deets at nospam.web.de wrote:

>markacy wrote:

>On 13 Cze, 09:45, "fdu.xia... at gmail.com" <fdu.xia... at
gmail.comwrote :

>>Hi all,

I can use list comprehension to create list quickly. So I
expected that I
can created tuple quickly with the same syntax. But I
found that the
same syntax will get a generator, not a tuple. Here is my
example:

In [147]: a = (i for i in range(10))

In [148]: b = [i for i in range(10)]

In [149]: type(a)
Out[149]: <type 'generator'>

In [150]: type(b)
Out[150]: <type 'list'>

[...]

>You should do it like this:

>>>>a = tuple([i for i in range(10)])
type(a)

<type 'tuple'>

[...]

>No need to create the intermediate list, a generator
expression works just
fine:

a = tuple(i for i in range(10))

Well I have looked into this and it seems that using the list
comprehension is faster, which is reasonable since generators require
iteration and stop iteration and what not.

# If you really really want a tuple, use [24] style
# if you need a generator use [27] style (without the tuple keyword
# off course)

In [24]: %timeit tuple([x for x in range(1000)])
10000 loops, best of 3: 185 µs per loop

In [25]: %timeit tuple([x for x in range(1000)])
1000 loops, best of 3: 195 µs per loop

In [26]: %timeit tuple([x for x in range(1000)])
10000 loops, best of 3: 194 µs per loop

############### ############### ############### ####

In [27]: %timeit tuple((x for x in range(1000)))
1000 loops, best of 3: 271 µs per loop

In [28]: %timeit tuple((x for x in range(1000)))
1000 loops, best of 3: 253 µs per loop

In [29]: %timeit tuple((x for x in range(1000)))
1000 loops, best of 3: 276 µs per loop

Thanks

--
Hatem Nassrat