On Tue, Aug 21, 2001 at 03:12:57AM +0200, Paul Johnson wrote:
> On Mon, Aug 20, 2001 at 04:48:30PM -0800, Michael Fowler wrote:
> > Your "(value1 + values2 .. 11)" syntax can easily be duplicated using
> > slices, e.g. @values[1..11].
>
> Or, since you only need 10 elements, only use 10, and use
On Mon, Aug 20, 2001 at 04:48:30PM -0800, Michael Fowler wrote:
> On Mon, Aug 20, 2001 at 12:21:29PM -0400, Yacketta, Ronald wrote:
> > avg1 = (value1 + values2..11) / 10
> > avg2 = (value2 + values3..12) / 10
> > avg3 = (value3 + values4..13) / 10
> >
> > the math is the easy part, how would I g
On Mon, Aug 20, 2001 at 12:21:29PM -0400, Yacketta, Ronald wrote:
> avg1 = (value1 + values2..11) / 10
> avg2 = (value2 + values3..12) / 10
> avg3 = (value3 + values4..13) / 10
>
> the math is the easy part, how would I go about cleanly and efficiently
> running through a has and doing the above?
mail)" <[EMAIL PROTECTED]>
Sent: Monday, August 20, 2001 12:21 PM
Subject: assistance needed with data gathering/manipulation
> Folks,
>
> I have a series of ~ 350 data points, I have been asked to get a running
> average of each data
> point.
>
> IE:
>
> da
Folks,
I have a series of ~ 350 data points, I have been asked to get a running
average of each data
point.
IE:
data points 1 - 20
1: 30
2: 23
3: 1
4: 23
5: 34
6: 56
7: 85
9: 32
10: 89
11: 23
12: 34
13: 19
14: 94
15: 11
16: 19
17: 54
18: 23
19: 87
20 49
avg1 = (value1 + values2..11) / 10
av
