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3

The way I would handle this problem would be to generate a new empty field and populate it using the field calculator. This way you will not lose any data and can keep your original data without any modifications to the fields. "New Field" = reclass(!sens!, !S2_adjusted!, !S1_adjusted!) Pre-Logic Script Code: def reclass(a, b, c): if (a) == 'S1': ...


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According to ISO 19115 every geographic information and services has its own metadata. It provides information about the identification, the extent, the quality, the spatial and temporal schema, spatial reference, and distribution of digital geographic data. Date and scale are part of the metadata. For the projection/coordinate system it's a part of your ...


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Though not a true answer in the frame of the original question that I had, I was able to alter the text of my legend by turning on Text Wrapping (under the layout tab) and playing around with the label width. I did discover also that in order to un-gray-out the preserve aspect ratio radio button, the Fixed Frame radio button must be checked to allow you to ...


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You have two options I can think of off hand. 1 You could use the select by location tool, and select all polygons that intersect the points. Then reverse the selection in the attribute table, thereby switching the selection from those that have points within them, to those that do not interact with the points at all. 2 You could run the spatial join ...


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It's possible to do without duplicating the layer. In the Classification dialog click on exclusion: Setup the query to exclude all values equal to zero. Then open the Legend tab and select 'Show symbol for excluded data'. Setup symbology for the zero value. Return to Classification and set to Manual with 2 classes ensure 0 as the mid value. Edit the ...


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If you have an 'Advanced' license of ArcMap you can use the 'Generate Near Table tool' and select the Angle option, from the output table you can then use the 'Select by Attribute tool' to select the angle representing the direction. N = 0 NE = 0.1-89.9 E = 90 SE = 90.1 - 179.9 S = 180 SW = 180.1 - 269.9 W = 270 NW = 270.1 - 359.9


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You have 11 points and a very flat variogram. This means there appears to be no spatial structure and your data is random noise. Any best prediction at a non-sample location is going to be the sample average. You could modify the variogram binning or variogram parameters and get something with a bit of spatial structure, but that might be more luck than ...


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