Scott.Daniels at Acm
Dec 31, 2008, 3:30 PM
Post #4 of 4
(4816 views)
Permalink
|
|
Re: How to initialize an array with a large number of members ?
[In reply to]
|
|
David Lemper wrote:
> ... Python. Using version 3.0
> # script23
> from array import array
> < see failed initialization attempts below >
> tally = array('H',for i in range(75) : [0])
> tally = array('H',[for i in range(75) : 0])
> tally = array('H',range(75) : [0])
> tally = array('H',range(75) [0])
> Above give either Syntax error or TypeError
>
> All examples of sequences in docs show only a few members
> being initialized. Array doc says an iterator can be used,
> but doesn't show how.
First, [1,2,3] is iterable. You could look up iterators or
generators or list comprehension, but since it's the holidays:
import array
tally = array.array('H', (0 for i in range(75)))# a generator
tally = array.array('H', [0 for i in range(75)])# list comprehension
> What would you do for a 2 dimensional array ?
You'll have to wait for numpy for the most general 2-D,
but for list-of-array:
square = [.array.array('H', (i * 100 + j for j in range(75)))
for i in range(75)]
square[4][8]
--
http://mail.python.org/mailman/listinfo/python-list
|
