Re: [Numpy-discussion] reverse cumsum?
On 7/6/10 10:42 , numpy-discussion-request@scipy.org wrote:
Date: Tue, 06 Jul 2010 10:02:57 -0400 From: Alan G Isaac<aisaac@american.edu> Subject: Re: [Numpy-discussion] reverse cumsum? To: Discussion of Numerical Python<numpy-discussion@scipy.org> Message-ID:<4C333791.5010505@american.edu> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
On 7/6/2010 9:56 AM, Ken Basye wrote:
Is there a simple way to get a cumsum in reverse order?
> x = np.arange(10) > x[::-1].cumsum()[::-1] array([45, 45, 44, 42, 39, 35, 30, 24, 17, 9])
Is that what you want?
Alan Isaac
Or, you can do: In [1]: a = np.arange(10) In [5]: np.sum(a) - np.cumsum(a) Out[5]: array([45, 44, 42, 39, 35, 30, 24, 17, 9, 0]) Jonathan
On Tue, Jul 6, 2010 at 11:25 AM, Jonathan Stickel <jjstickel@vcn.com> wrote:
On 7/6/10 10:42 , numpy-discussion-request@scipy.org wrote:
Date: Tue, 06 Jul 2010 10:02:57 -0400 From: Alan G Isaac<aisaac@american.edu> Subject: Re: [Numpy-discussion] reverse cumsum? To: Discussion of Numerical Python<numpy-discussion@scipy.org> Message-ID:<4C333791.5010505@american.edu> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
On 7/6/2010 9:56 AM, Ken Basye wrote:
Is there a simple way to get a cumsum in reverse order?
>> x = np.arange(10) >> x[::-1].cumsum()[::-1] array([45, 45, 44, 42, 39, 35, 30, 24, 17, 9])
Is that what you want?
Alan Isaac
Or, you can do:
In [1]: a = np.arange(10) In [5]: np.sum(a) - np.cumsum(a) Out[5]: array([45, 44, 42, 39, 35, 30, 24, 17, 9, 0])
Jonathan _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Alternately: In [11]: reversed(np.arange(10).cumsum()) Out[11]: <reversed object at 0x8b9d14c> In [12]: [i for i in _] Out[12]: [45, 36, 28, 21, 15, 10, 6, 3, 1, 0] reversed(x) is, as you can see, not an array but an iterator which will return the cumulative sums in reverse. If you need an array specifically, you can convert it fairly easily, though it does need two type conversions: In [10]: np.array(list(reversed(np.arange(10).cumsum()))) Out[10]: array([45, 36, 28, 21, 15, 10, 6, 3, 1, 0]) or, if you like list comps: In [14]: np.array([i for i in reversed(np.arange(10).cumsum())]) Out[14]: array([45, 36, 28, 21, 15, 10, 6, 3, 1, 0]) I think Ken's suggestion may be the best so far, but as perl users say, timtowtdi. --Josh
On 7/6/2010 3:37 PM, Joshua Holbrook wrote:
In [10]: np.array(list(reversed(np.arange(10).cumsum()))) Out[10]: array([45, 36, 28, 21, 15, 10, 6, 3, 1, 0])
That might appear to match the subject line but does not match the OP's example output, which was [45, 45, 44, 42, 39, 35, 30, 24, 17, 9]. You are giving the equivalent of x.cumsum()[::-1], while the OP asked for the equivalent of x[::-1].cumsum()[::-1]. fwiw, Alan Isaac
On Tue, Jul 6, 2010 at 2:23 PM, Alan G Isaac <aisaac@american.edu> wrote:
On 7/6/2010 3:37 PM, Joshua Holbrook wrote:
In [10]: np.array(list(reversed(np.arange(10).cumsum()))) Out[10]: array([45, 36, 28, 21, 15, 10, 6, 3, 1, 0])
That might appear to match the subject line but does not match the OP's example output, which was [45, 45, 44, 42, 39, 35, 30, 24, 17, 9].
You are giving the equivalent of x.cumsum()[::-1], while the OP asked for the equivalent of x[::-1].cumsum()[::-1].
fwiw, Alan Isaac _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Oh snap. Good call--idk what I was thinking. Tired, I guess. :) In that case, if you were going to use reversed() things would get a bit nastier: In [13]: np.array(list(reversed(np.array([9-i for i in xrange(10)]).cumsum()))) Out[13]: array([45, 45, 44, 42, 39, 35, 30, 24, 17, 9]) ...which is gross enough that this approach is probably worth abandoning.
I think Ken's suggestion may be the best so far...
I meant to say Alan's suggestion, i.e. x[::-1].cumsum()[::-1].
participants (3)
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Alan G Isaac -
Jonathan Stickel -
Joshua Holbrook