how to make a iterable object like this?

Question

Question 1

how to make an iterable object like this:

0 1 2 3 4 5 4 3 2 1 0 1 2 3 4 5 4 3 2 1 0 1 2 3 4 5 4 3 2 1 ....

Question 2

if use list(obj) on the object above would this eat up machine's memory? how to prevent it?

Please don't use python2

Also, if this is homework, we appreciate it if you use the `homework` tag so we can help you learn instead of just handing over the answers. — Martijn Pieters, Sep 04 '12 at 11:08

Martijn Pieters · Accepted Answer · 2012-09-04T11:19:07.353

3

You can make an infinite generator that counts up and down:

def updown(n):
    while True:
        for i in range(n):
            yield i
        for i in range(n - 2, 0, -1):
            yield i

uptofive = updown(6)
for i in range(20):
    print uptofive.next(),

would output:

0 1 2 3 4 5 4 3 2 1 0 1 2 3 4 5 4 3 2 1

You cannot prevent list(updown(6)) from trying to consume all memory, no. As the doctor would say: "Then don't do that!".

Use .next() calls instead, or use your generator with another statement that limits the number of times you iterate over the generator. The itertools.islice() function would do just that:

import itertools
list(itertools.islice(updown(6), 20))

edited Sep 04 '12 at 11:19

answered Sep 04 '12 at 11:13

Martijn Pieters

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You have a duplicate "5" in there. – mhawke Sep 04 '12 at 11:15
In Python 2, `xrange` might be a bit better in this context. – cdarke Sep 04 '12 at 11:15
@cdarke: for small ranges, don't bother, and it'll keep it portable with minimum fuss. – Martijn Pieters Sep 04 '12 at 11:16
1

@mhawke: What dupe 5? *innocent whistle* – Martijn Pieters Sep 04 '12 at 11:18
@MartijnPieters: I'm afraid that later comers will not see the joke :) – mhawke Sep 04 '12 at 11:20
Spoiler alert: `(n - 1)` changed to `(n - 2)` – cdarke Sep 04 '12 at 11:28
@MartijnPieters: I don't violently disagree, but the op specifically asked about iterable objects, and might extend your code to a larger range if this was just a sample. Anyway the `print` breaks portability to Py3. – cdarke Sep 04 '12 at 11:31
@Max: Only the print line won't work in 3; use `print(next(uptofive), end='')` in 3. – Martijn Pieters Sep 05 '12 at 06:46

mhawke · Answer 2 · 2012-09-04T11:51:59.577

1

An alternative is to use itertools.cycle():

from itertools import cycle

def oscillator(start, stop):
    # assumes stop >= start
    return cycle(range(start, stop+1) + range(stop-1, start, -1))

o = oscillator(0, 5)
for i in range(30):
    print o.next(),

edited Sep 04 '12 at 11:51

answered Sep 04 '12 at 11:46

mhawke

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how to make a iterable object like this?

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