4 ms·

How To Read an Unlabeled Sales Chart

89 points by bslatkin 14y ago

willvarfar 14y ago
The article could have computed the numbers on various charts that others have published that do have the scale visible, to illustrate how well this works.
ars 14y ago
This assumes that all sales are equal, but if sales is a sum of dollars earned I assume this won't work (since each sale can be for a different amount).
- glogla 14y ago
  Sure, but then it gives you estimate of dollars earned.
  - gus_massa 14y ago
    It will give you a estimate of cents earned, but they are usually too small to appear in the graphic. In this case the first method doesn't work.
bbrizzi 14y ago
The first technique won't work if the sales are rounded off to the nearest pixel, which is pretty much almost always the case. The second method won't work if the sales do not follow a Poisson distribution, for instance a product will have better sales at launch and during advertising campaigns. The sales figures don't always float around a constant average. Also, I hadn't checked the math on the second method but the units don't add up. On the LHS, you have [sales / pixel]x[pixels] = [sales] On the RHS, you have ([sales / pixel]x[pixels])² = [sales]² IIRC, the left hand side should be squared.
- jules 14y ago
  Both sides are unitless: sales is not in dollars but the number of sales.
  - deleted 14y ago
    [deleted]
  - sesqu 14y ago
    The units are actually rates: sales per day.
- gus_massa 14y ago
  These methods also assume that the bottom level is 0. In this graph the bottom whirl is smaller, so probably it's a 0.
- sesqu 14y ago
  No, the left hand should not be squared. It's a special property of the Poisson process that applies to the numbers, not the units.
jychang 14y ago
Surprised no one on HN calculated this already. S = P / Var(P) Numbers of pixels: 13,6,16,29,6,16,13,6,19,13,19,9,26; followed by the large one. Mean ~ 14.692; Using Sample Variance ~ 53.397; Using Population Variance ~ 49.289; SalesPerPixel(Sample) ~ .275 This doesn't seem right, where does the math screw up? That said, this isn't very useful. The only time someone will show sales graphs is for special events... aka when it DOESN'T follow a possion distribution.
- jstanley 14y ago
  But, just like you did, the special event can be excluded from the poisson calculation. Then you know the scale and can figure out the values for the special event.
- sesqu 14y ago
  I think you underestimated the pixel counts. It's hard to be precise because of the bleeding, likely due to scaling, but adding one to the pixel counts results in a much more reasonable 0.294 sales per pixel. This would indicate the special event precipitated 25 sales.
lunchladydoris 14y ago
"Here's a table with more values (the Riemann zeta, unfortunately, is not implemented in Excel)" This made me laugh.
PeterisP 14y ago
The article fails to mention the very huge dependency on scale for such a methods. These methods will work on a sales graph built from numbers like [27, 41, 55, 73]. Part of these methods (not all) may work on a sales graph built from numbers like [27 000, 41 000, 55 000, 73 000] ... and give the exact same result as for [27, 41, 55, 73]. And none of them will give any reasonable results for numbers like [27 123, 41 432, 55 132, 73 433] - there's not enough information for that.

89 points by bslatkin 14y ago

The article could have computed the numbers on various charts that others have published that do have the scale visible, to illustrate how well this works.

ars 14y ago

This assumes that all sales are equal, but if sales is a sum of dollars earned I assume this won't work (since each sale can be for a different amount).

glogla 14y ago
Sure, but then it gives you estimate of dollars earned.
- gus_massa 14y ago
  It will give you a estimate of cents earned, but they are usually too small to appear in the graphic. In this case the first method doesn't work.

bbrizzi 14y ago

The first technique won't work if the sales are rounded off to the nearest pixel, which is pretty much almost always the case. The second method won't work if the sales do not follow a Poisson distribution, for instance a product will have better sales at launch and during advertising campaigns. The sales figures don't always float around a constant average. Also, I hadn't checked the math on the second method but the units don't add up. On the LHS, you have [sales / pixel]x[pixels] = [sales] On the RHS, you have ([sales / pixel]x[pixels])² = [sales]² IIRC, the left hand side should be squared.

jules 14y ago
Both sides are unitless: sales is not in dollars but the number of sales.
- deleted 14y ago
  [deleted]
- sesqu 14y ago
  The units are actually rates: sales per day.
gus_massa 14y ago
These methods also assume that the bottom level is 0. In this graph the bottom whirl is smaller, so probably it's a 0.
sesqu 14y ago
No, the left hand should not be squared. It's a special property of the Poisson process that applies to the numbers, not the units.

jychang 14y ago

Surprised no one on HN calculated this already. S = P / Var(P) Numbers of pixels: 13,6,16,29,6,16,13,6,19,13,19,9,26; followed by the large one. Mean ~ 14.692; Using Sample Variance ~ 53.397; Using Population Variance ~ 49.289; SalesPerPixel(Sample) ~ .275 This doesn't seem right, where does the math screw up? That said, this isn't very useful. The only time someone will show sales graphs is for special events... aka when it DOESN'T follow a possion distribution.

jstanley 14y ago
But, just like you did, the special event can be excluded from the poisson calculation. Then you know the scale and can figure out the values for the special event.
sesqu 14y ago
I think you underestimated the pixel counts. It's hard to be precise because of the bleeding, likely due to scaling, but adding one to the pixel counts results in a much more reasonable 0.294 sales per pixel. This would indicate the special event precipitated 25 sales.

lunchladydoris 14y ago

"Here's a table with more values (the Riemann zeta, unfortunately, is not implemented in Excel)" This made me laugh.

PeterisP 14y ago

The article fails to mention the very huge dependency on scale for such a methods. These methods will work on a sales graph built from numbers like [27, 41, 55, 73]. Part of these methods (not all) may work on a sales graph built from numbers like [27 000, 41 000, 55 000, 73 000] ... and give the exact same result as for [27, 41, 55, 73]. And none of them will give any reasonable results for numbers like [27 123, 41 432, 55 132, 73 433] - there's not enough information for that.