topic Re: finding the midpoint of observations in a three year period in SAS Programming

finding the midpoint of observations in a three year period

89974114 — Mon, 19 Mar 2018 17:30:18 GMT

I have a dataset which represents the volume of sales over three years:

data test;
input one two three average;
datalines;
10 20 30 .
20 30 40 .
10 30 50 .
10 10 10 .
;
run;

I'm looking for a way to find the middle point of the three years, the average sale point

the updated dataset would read

data test;
input one two three average;
datalines;
10 20 30 2
20 30 40 1.5
10 30 50 2.1
10 10 10 1.5
;
run;

So essentially looking for what part of the three years the halfway point of the sales occurred.

Appreciate.

EDIT: what I've been trying with the weight and proc means

I've been trying to use proc means and weight function but it doesn't give me the average point of the three years

proc means data=test noprint;
var one two three;
var one+two+three=total;
var (one+two+three)/3=Average; 
var Average/weight=Average_Year;

output out=testa2
    sum(Total) = 
    mean(Total) = ;
run;

Re: finding the midpoint of observations in a three year period

PaigeMiller — Mon, 19 Mar 2018 17:34:10 GMT

@89974114 wrote:

I'm looking for a way to find the middle point of the three years, the average sale point

the updated dataset would read
data test;
input one two three average;
datalines;
10 20 30 2
20 30 40 1.5
10 30 50 2.1
10 10 10 1.5
;
run;

I have to admit I am not following this, I don't see how you have computed the average value.

Re: finding the midpoint of observations in a three year period

89974114 — Mon, 19 Mar 2018 17:37:38 GMT

so the first value would be 10+20+30=60, then midway is 30 which is 10+20 so 2 years

I'm looking for the average point throughout the three years based on the volume of sales

the second would be 20+30+40=90 / 2 = 45 ,

45-20 = 25 then 25/30 = 5/6th of a year so my mistake the second line should be 1 + 5/6th years

Re: finding the midpoint of observations in a three year period

89974114 — Mon, 19 Mar 2018 17:38:34 GMT

So i'm thinking if each volume is given a weight based on total volume, how far through the three years is the middle point

Re: finding the midpoint of observations in a three year period

PaigeMiller — Mon, 19 Mar 2018 19:49:15 GMT

so the first value would be 10+20+30=60, then midway is 30 which is 10+20 so 2 years

Still not following this at all.

the second would be 20+30+40=90 / 2 = 45 ,

45-20 = 25 then 25/30 = 5/6th

So in the first example, there is no subtraction happening, but in this example there is a subtraction in the math?

Re: finding the midpoint of observations in a three year period

PGStats — Tue, 20 Mar 2018 18:07:59 GMT

So, I guess you want something like

data test;
input one two three;
midpoint = sum(one, two, three) / 2;
if midpoint < one then halfSalesPoint = midpoint / one;
else if midpoint < one + two then halfSalesPoint = 1 + (midpoint - one) / two;
else halfSalesPoint = 2 + (midpoint - one - two) / three;
drop midpoint;
datalines;
10 20 30
20 30 40
10 30 50
10 10 10
;

proc print data=test; run;

Re: finding the midpoint of observations in a three year period

89974114 — Tue, 20 Mar 2018 06:28:25 GMT

Maybe think of it like a cumulative frequency curve, as you add all three observations together you move from 0 to 100% of the value, think of the x axis as 1-3 years and the y-axis as 0 to 100%, you are looking for the point that aligns 50% on the y-axis and the middle point of the x-axis.

Re: finding the midpoint of observations in a three year period

89974114 — Tue, 20 Mar 2018 06:38:00 GMT

I can't fault that it gives the right answers, appreciate. I am wondering if there is a more efficient solution though as I'll be performing this datastep with millions of entries

Re: finding the midpoint of observations in a three year period

ballardw — Tue, 20 Mar 2018 15:39:45 GMT

@89974114 wrote:

I can't fault that it gives the right answers, appreciate. I am wondering if there is a more efficient solution though as I'll be performing this datastep with millions of entries

What is inefficient about the accepted solution?

Re: finding the midpoint of observations in a three year period

89974114 — Tue, 20 Mar 2018 15:42:14 GMT

There must be a function in the proc means or a proc command which allows you to find the weighted average with respect to the observation length

Re: finding the midpoint of observations in a three year period

PGStats — Tue, 20 Mar 2018 16:31:56 GMT

I just don't see how this pr/oblem can be expressed as a weighted average. To me, the solution to your problem amounts to finding the inverse of a piecewise linear function.

This little datastep should run very fast. I doubt that any proc can do much better.

Re: finding the midpoint of observations in a three year period

ballardw — Tue, 20 Mar 2018 17:03:39 GMT

@89974114 wrote:

There must be a function in the proc means or a proc command which allows you to find the weighted average with respect to the observation length

If the actual concern is that your actual problem involves more than 3 variables whose names are actually ordinal data values then you might be looking at an array to hold the listed variables and compare the mean to an iteratively created cumulative total.

Re: finding the midpoint of observations in a three year period

89974114 — Wed, 21 Mar 2018 07:01:25 GMT

I was being unreasonably pedantic yesterday, my apologies.