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Performing Raster Calculation?
Iterate through all fields in attribute table and multiply by fieldPerforming unweighted overlay on multiple raster layersscript to draw a square where the raster pixels contained within are a certain type and greater than XConditional Combine Rasters and Track which Raster pixel usedReclassify raster based on Z-values in Arcmap 10.4.1Combining two reclassified raster images in ArcGIS Desktop?How to extend Raster layer without pixel value data losingExtracting multi values to raster?calculate 'predicted class' raster in QGISMax value for each cell
This is my formal:
("Standort_ger_fre_mosaic" ^ 0.4) *
("Habitattradition_raster_rec.tif" ^ 0.3) *
("Vollständigkeit_rec.tif" ^ 0.1) *
("Vorkommen_ungestörter_Lebensräume_rec.tif" ^ 0.1) *
("Vorkommen_von_Arten_dynamischer_Lebensräume_rec.tif"^ 0.1)
I would like to charge the pixels of the above-mentioned layers together.
As a final product, I would like to have a file that has different values for each pixel.
When I do the computation, my final product has only one value in each pixel. This is equal to 0
I am working with ArcMap 10.6
arcgis-desktop spatial-analyst raster-calculator
New contributor
add a comment |
This is my formal:
("Standort_ger_fre_mosaic" ^ 0.4) *
("Habitattradition_raster_rec.tif" ^ 0.3) *
("Vollständigkeit_rec.tif" ^ 0.1) *
("Vorkommen_ungestörter_Lebensräume_rec.tif" ^ 0.1) *
("Vorkommen_von_Arten_dynamischer_Lebensräume_rec.tif"^ 0.1)
I would like to charge the pixels of the above-mentioned layers together.
As a final product, I would like to have a file that has different values for each pixel.
When I do the computation, my final product has only one value in each pixel. This is equal to 0
I am working with ArcMap 10.6
arcgis-desktop spatial-analyst raster-calculator
New contributor
add a comment |
This is my formal:
("Standort_ger_fre_mosaic" ^ 0.4) *
("Habitattradition_raster_rec.tif" ^ 0.3) *
("Vollständigkeit_rec.tif" ^ 0.1) *
("Vorkommen_ungestörter_Lebensräume_rec.tif" ^ 0.1) *
("Vorkommen_von_Arten_dynamischer_Lebensräume_rec.tif"^ 0.1)
I would like to charge the pixels of the above-mentioned layers together.
As a final product, I would like to have a file that has different values for each pixel.
When I do the computation, my final product has only one value in each pixel. This is equal to 0
I am working with ArcMap 10.6
arcgis-desktop spatial-analyst raster-calculator
New contributor
This is my formal:
("Standort_ger_fre_mosaic" ^ 0.4) *
("Habitattradition_raster_rec.tif" ^ 0.3) *
("Vollständigkeit_rec.tif" ^ 0.1) *
("Vorkommen_ungestörter_Lebensräume_rec.tif" ^ 0.1) *
("Vorkommen_von_Arten_dynamischer_Lebensräume_rec.tif"^ 0.1)
I would like to charge the pixels of the above-mentioned layers together.
As a final product, I would like to have a file that has different values for each pixel.
When I do the computation, my final product has only one value in each pixel. This is equal to 0
I am working with ArcMap 10.6
arcgis-desktop spatial-analyst raster-calculator
arcgis-desktop spatial-analyst raster-calculator
New contributor
New contributor
edited 15 hours ago
PolyGeo♦
53.8k1781244
53.8k1781244
New contributor
asked 16 hours ago
Tim HolstTim Holst
1
1
New contributor
New contributor
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Instead of using "^", use the "Power" function as follows:
(Power("Standort_ger_fre_mosaic",0.4)) *
(Power("Habitattradition_raster_rec.tif",0.3)) *
...
add a comment |
This is probably a rounding issue. Please try converting your image values to float before the exponent. Also I suggest to use Power.
Power(Float("Standort_ger_fre_mosaic"),0.4) ) *
Power(Float("Habitattradition_raster_rec.tif"),0.3) ...
I think the Power function alone does the trick. I tried Power("image",0.4) with an uint8 image and it worked just fine
– kowalski
16 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Instead of using "^", use the "Power" function as follows:
(Power("Standort_ger_fre_mosaic",0.4)) *
(Power("Habitattradition_raster_rec.tif",0.3)) *
...
add a comment |
Instead of using "^", use the "Power" function as follows:
(Power("Standort_ger_fre_mosaic",0.4)) *
(Power("Habitattradition_raster_rec.tif",0.3)) *
...
add a comment |
Instead of using "^", use the "Power" function as follows:
(Power("Standort_ger_fre_mosaic",0.4)) *
(Power("Habitattradition_raster_rec.tif",0.3)) *
...
Instead of using "^", use the "Power" function as follows:
(Power("Standort_ger_fre_mosaic",0.4)) *
(Power("Habitattradition_raster_rec.tif",0.3)) *
...
answered 16 hours ago
kowalskikowalski
135210
135210
add a comment |
add a comment |
This is probably a rounding issue. Please try converting your image values to float before the exponent. Also I suggest to use Power.
Power(Float("Standort_ger_fre_mosaic"),0.4) ) *
Power(Float("Habitattradition_raster_rec.tif"),0.3) ...
I think the Power function alone does the trick. I tried Power("image",0.4) with an uint8 image and it worked just fine
– kowalski
16 hours ago
add a comment |
This is probably a rounding issue. Please try converting your image values to float before the exponent. Also I suggest to use Power.
Power(Float("Standort_ger_fre_mosaic"),0.4) ) *
Power(Float("Habitattradition_raster_rec.tif"),0.3) ...
I think the Power function alone does the trick. I tried Power("image",0.4) with an uint8 image and it worked just fine
– kowalski
16 hours ago
add a comment |
This is probably a rounding issue. Please try converting your image values to float before the exponent. Also I suggest to use Power.
Power(Float("Standort_ger_fre_mosaic"),0.4) ) *
Power(Float("Habitattradition_raster_rec.tif"),0.3) ...
This is probably a rounding issue. Please try converting your image values to float before the exponent. Also I suggest to use Power.
Power(Float("Standort_ger_fre_mosaic"),0.4) ) *
Power(Float("Habitattradition_raster_rec.tif"),0.3) ...
answered 16 hours ago
radouxjuradouxju
41.2k144121
41.2k144121
I think the Power function alone does the trick. I tried Power("image",0.4) with an uint8 image and it worked just fine
– kowalski
16 hours ago
add a comment |
I think the Power function alone does the trick. I tried Power("image",0.4) with an uint8 image and it worked just fine
– kowalski
16 hours ago
I think the Power function alone does the trick. I tried Power("image",0.4) with an uint8 image and it worked just fine
– kowalski
16 hours ago
I think the Power function alone does the trick. I tried Power("image",0.4) with an uint8 image and it worked just fine
– kowalski
16 hours ago
add a comment |
Tim Holst is a new contributor. Be nice, and check out our Code of Conduct.
Tim Holst is a new contributor. Be nice, and check out our Code of Conduct.
Tim Holst is a new contributor. Be nice, and check out our Code of Conduct.
Tim Holst is a new contributor. Be nice, and check out our Code of Conduct.
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