[gdal-dev] Raster statistics

Tue Aug 8 13:22:01 PDT 2017

Paul,

You could call gdal_calc.py and pass it the numpy formulas on the command line…Otherwise it might be best to bring the raster in to OpenCV.

--Eric

From: gdal-dev [mailto:gdal-dev-bounces at lists.osgeo.org] On Behalf Of Paul Meems
Sent: Thursday, August 03, 2017 2:51 PM
To: Chris Waigl
Cc: gdal-dev at lists.osgeo.org
Subject: Re: [gdal-dev] Raster statistics

Thanks Chris for your reply.

I forgot to mention I'm not using GDAL with Python.
I use it with C++ and/or C#.

Paul

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2017-08-03 20:05 GMT+02:00 Chris Waigl <cwaigl at alaska.edu<mailto:cwaigl at alaska.edu>>:
I would not use gdal for this particular task. I presume you have the band data in a 2D numpy array. Then I’d get the 80th percentile for example with np.percentile() and use a boolean expression to generate a mask for the array (droneraster > perc80value ).

Chris

--
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On Aug 3, 2017, at 5:43 AM, Paul Meems <bontepaarden at gmail.com<mailto:bontepaarden at gmail.com>> wrote:

I have a drone raster file which I want to use for some calculation.
Before the calculation, I need to loose some extreme values.
I want to do something like a percentile calculation where you get all values, order them and loose the top 10%.
For this, I need to get all values first which can be slow when using a large file.

I looked at the statistics (band.GetStatistics) but that doesn't work well.
I thought I could use 2 times the standard deviation added to the mean to get roughly 97%.
But with these statistics:
    STATISTICS_MAXIMUM=33.186080932617
    STATISTICS_MEAN=24.840205979603
    STATISTICS_MINIMUM=1.5951598882675
    STATISTICS_STDDEV=4.7285348016053
Mean + 2*std is larger than the max.

So I moved to the histogram. It is also very fast, but I'm not sure how to use it.
I have this:
  256 buckets from 1.53322 to 33.248:
  410 77 66 66 65 58 56 45 42 87 57 72 61 65 68 70 73 82 93 ...
Does this mean, bucket 1 = 410 that I have 410 pixels of value 1.53322 and the second bucket means I have 77 pixels between 1.53322 and 1.657? 1.657 = 1.53322 + ((33.248 - 1.53322)/256)

Is this a good approach? Or can/should I use a different one.

Thanks,

Paul
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