QGIS – Removing Extreme Values from a Calculated Raster in QGIS

landsatqgisraster-calculator

I have a Landsat GeoTIFF with a multiplicative scaling factor of 0.01.

When I load the tif into QGIS and multiply the raster by 0.01 extreme values are created (see attached picture).

Looking at the calculated and source layer, it looks kike the extreme values are due to differing CRS between the Landsat and calculated layer:

I've tried to address this at the calculation stage, but without success. Equally, I've tried setting values below 0 to 'no data', and a variety of other suggestions found online but I can't get it to behave – none of the raster calc solutions works (normally QGIS says there is formula error and won't proceed).

The Landsat tif used can be downloaded from here.

Can you help?

Best Answer

I don't know why you say that "I have a Landsat GeoTIFF with a multiplicative scaling factor of 0.01" but, in your picture, you have a landsat 9 product with ID equals to:

LC09_L2SP_039035_20220703_20220705_02_T1

and, exhibited image (Las Vegas, Nevada), it is for ST_B10 band. In this case, for converting values to kelvin, the multiplicative scaling factor is 0.00341802 and the adding scaling factor is 149. So, conversion formula in raster calculator is as follows:

"LC09_L2SP_039035_20220703_20220705_02_T1_ST_B10@1" * 0.00341802 + 149

After applying above formula, I got result of following picture. There is not "extreme values" and surface temperature values (Kelvin degrees) are as expected. It can also be observed that there are not colored areas outside image (in your case red areas) reflecting correctly no data values.

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I wouldn't clamp the values, as suggested by the pseudocode, because the curvature artifacts do not necessarily correspond to high-curvature areas. Better cures for the problem include:

Consider smoothing the DEM before computing the curvature. This would slightly reduce most curvatures but might give reasonable curvature values at all locations. Smoothing the DEM can be done with local medians or means.
Find a robust way to estimate curvature. (You won't find this in any Arc* software, but with the availability of general-purpose tools like R, this is a feasible recommendation.) In general, curvature is estimated by fitting a quadratic (or some specialized low-order polynomial) in each neighborhood. The usual fitting method is least-squares, which can be sensitive to slight local variations. By expanding the neighborhood and using a fitting method that downweights outlying residuals (such as IWLS), you might be able to get representative and accurate curvature estimates everywhere.
Replace the extreme curvature values with NoData. This is still a draconian measure, but would be appropriate if the extreme values are really artifacts of the interpolation (and don't, therefore, reflect truly high curvatures).
Choose an interpolation method that produces smoother surfaces!

The 'map algebra' syntax for replacing values with NoData uses the SetNull command, as in

SetNull(Abs([Curvature grid]) > 10*Abs(m), [Curvature grid])

By the way, m should not be the median curvature: the median often will be close to zero and won't reflect typical large curvatures (positive or negative) A large multiple of the median absolute curvature might be a reasonable choice. Better yet, examine the histogram of curvatures to find where to cut off the values.

If the SetNull approach pokes too many NoData holes in the grid, think about whether the curvature grid should be smoothly varying. If so, and if those holes tend to be isolated (or in narrow stripes--which would be a strong indicator that a poor interpolator was used), you can fill them by computing a focal mean and pasting the mean values over the holes. The pasting is done with the conditional operator, as in

Con(IsNull([Modified curvature grid]), [Focal mean grid], [Modified curvature grid])

Note the use of IsNull to detect the NoData cells.

[GIS] Calculating EVI from Sentinel 2 imagery gives difference in results between ArcMap & QGIS

Please stick to one question per post in the future. It makes answering significantly easier.

Question 1 - How to calculate EVI:
The formula that you are using is "correct", in the sense that you are using the constants calculated for MODIS.

Question 2 - Why does QGis and ArcGIS provide different results:
Minuscule differences in results most likely originate from different resampling methods used when combining the spectral bands. It is a difficult discover the reason for the differences, without having the data available.

Question 3 - How to remove extreme values from a raster calculation:
Removing those values is best done by adjusting your raster calculation. One approach would be to simply add a parenthesis around the entire equation and multiply it by a pair of 'greater than' and 'less than' calculations.
Example:

(EVI) * (EVI > -1) * (EVI < 1) + (EVI > 1)*1 + (EVI < -1)*(-1)

The above outlined calculation would set all the values below -1 and above 1 to -1 and 1, respectively.

Question 4 - When to apply the scale factors:
The scale factors should be applied when you are using the bands. If you applied them afterwards, to the EVI, then the effect of the constants in the equation would be changed, and your calculations would become incorrect. As such, you use is currently the right one.

Best Answer

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