Geographic Information Systems Asked by DarkCygnus on September 22, 2020
I have some single-band rasters (NDVI) on QGIS, for which I can obtain their distribution histogram by going to Right-click > Properties > Histogram Tab > Compute Histogram
. This is an example of what I get:
This is the same procedure (and also works) for multi-band images (RGB etc). However, I wonder what is the default bin size that QGIS used to calculate and render this histogram? I obtained this histogram without modifying any setting on the Prefs/Actions tab.
In the case of RGB images surely the bin size should be >=1
, as pixels can only be in the discrete interval of [0,255]
and it wouldn’t make sense to have a lower bin size (say, 0.5 pixel value).
However, in the case of NDVI and similar indexes non-integer bin sizes are viable, even more given that the NDVI lies between [-1,1]
. So, I wonder how QGIS determines this bin size, as I need to know what value was used to properly compare with other histograms I produce manually.
I am using QGIS 2.14.12-Essen
Doing some browsing and with some input from Jakob on comments, we have these two links that give some insight on the way QGis decides the bin count.
On this link it's stated in a comment, emphasis mine:
param binCount Number of bins (intervals,buckets). If 0, the number of bins is decided automatically according to data type, raster size etc.
Then on this other link we can see the actual code where binCount
is calculated, if unspecified by the histogram call. Summing up what's specified there:
min((hist.width * hist.height) , ceiling(hist.maximum - hist.minimum +1))
hist.width * hsit.height
.We also have these comments worth mentioning here:
// There is no best default value, to display something reasonable in histogram chart,binCount should be small, OTOH, to get precise data for cumulative cut, the number should be big.
// Because it is easier to define fixed lower value for the chart, we calc optimum binCount for higher resolution (to avoid calculating that where histogram() is used. In any case, it does not make sense to use more than width*height;
Finally, the histogram height and width are calculating depending on the sample size and the resolution of the image to obtain the histogram from. The code for that can be found on these lines.
Correct answer by DarkCygnus on September 22, 2020
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