[Libreoffice-bugs] [Bug 133746] Conditional Colour Scale Formatting incorrectly rendered: MAX & MEAN colours are reversed

bugzilla-daemon at bugs.documentfoundation.org bugzilla-daemon at bugs.documentfoundation.org
Sun Jun 7 09:22:04 UTC 2020


https://bugs.documentfoundation.org/show_bug.cgi?id=133746

--- Comment #6 from Colin <that.man.colin at gmail.com> ---
(In reply to Colin from comment #5)
> (In reply to Mike Kaganski from comment #4)
> > This is not a bug. Using Percentile for intermediate value, you ask it for
> > such a value that would represent exactly half of the population see [1]
> > that refers to [2]. Having total of 46 values in the data set used in the
> > conditional format, and 26 of them being 1, there's no value that would
> > represent half of the population (23 values). So it necessarily ends up with
> > 1 as the percentile value, thus being equal to max, and so effectively being
> > equal to 2-color scale, as the third color is never used.
> > 
> > Possibly what was wanted is using *Percent*, not *Percentile*.
> > 
> > [1] https://help.libreoffice.org/6.4/en-US/text/scalc/01/05120000.html
> > [2] https://wiki.documentfoundation.org/Faq/Calc/142
> 
> I think I have to disagree. I have other spreadsheets that also have the
> same scale of "fixed" percentage results 20,40,60,80,100 and they don't fail
> because there is no 50%. It's not actually possible to divide a week by a
> whole number of days to get 50% and I am only targeting the five weekdays
> which also imposes the same impossibility. Also, the remedy as mentioned in
> my comment #3, whilst setting the mean to "real" 0.5 never encounters a
> calculated 0.5 but gives a "bug free" result.
> Why does deleting any five random rows that also don't contain any 50%
> values obviate the bug?
> I'm not convinced the colour scale is a standard deviation function.

My apologies, it appears I do have one cell containing 0.5 but the exercise of
randomly removing 5 rows is unaffected by the inclusion or exclusion of the
source data and therefore the value of this cell.

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