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Any recommended ratio of zone area / raster cell size when running small input zones with ArcGIS Tabulate Area Tools?

Esri mentions:

If the zone input is a feature dataset with relatively small features, keep in mind that the resolution of the information needs to be appropriate relative to the resolution of the value raster. If the areas of single features are similar to or smaller than the area of single cells in the value raster, in the feature-to-raster conversion some of these zones may not be represented.

To demonstrate this, try converting the feature dataset to a raster with the appropriate feature-to-raster conversion tool and specify the resolution to be that of the Value raster. The result from this conversion will give an indication about what the default output of the zonal operation will be.

If you have fewer results in the output than you may have expected, you need to determine an appropriate raster resolution that will represent the detail of your feature input, and use this resolution as the Cell Size of the Raster Analysis Settings of the Environment.

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The answer depends on the precision you need.

To obtain some quantitative guidance, we may conceptualize this process in two steps: rasterization of the zones followed by a summary operation. We can't say much in general about the latter, but let's pretend it's a simple statistical summary, such as a count (from which you would obtain the area), average, or sum, that is roughly proportional to the values in all the zone's cells. (A range, by contrast, depends only on two of the values in the cells; because of that it would be harder to analyze the effects on a range.)

In the first step, a cell is assigned to any zone containing its central point. To assess the possible error of this discretization step, imagine repeating it many times but changing just the origin of the grid randomly. If the area of a zone is A and the cellsize is c, the cell areas are all c^2, whence the expected number of cells assigned to the zone is A/c^2. As a rough approximation we could analyze this situation as if the cell assignments were statistically independent of each other. (They're not--they depend on the zone's shape--but that's another thing that is difficult to make any generalizations about.) That makes the distribution of cell counts approximately Poisson, implying the standard deviation is sqrt(A)/c. The relative error, then, is sqrt(A)/c divided by A/c^2, equal to c/sqrt(A). If you are uncomfortable with relative errors of this size, you need a grid with finer resolution.

An example may help fix these ideas. Suppose you can tolerate a relative error of 10% in your zonal statistics and that the zone of smallest area is about 0.9 acre (40,000 square feet). With a cellsize of 100 feet we would have A = 40,000, c = 100, and c/sqrt(A) = 100 / 200 = 50%. That much of an error in the cell count would likely translate to at least that much error in the statistics, so it's too big. To reduce it to 10% or less, you can't change A, so all you can do is decrease c by a factor of 5 or more. Indeed, with c = 20 feet, c/sqrt(A) = 20 / 200 = 10%.

You probably would want even a smaller cellsize in this example, because as a rule of thumb there's about a 1/3 chance that the actual number of cells used to represent the smallest zone would differ by more than 10% from the proper (expected) value and even a 1/20 chance that the number would differ by more than twice that. (This is an application of the well-known 68-95-99.7 rule; 1/3 approximates 100% - 68% and 1/20 equals 100% - 95%.)

The calculations can be further simplified by considering the numbers of cells expected for each zone. In the example, a cellsize of c = 100 would give an expected number of cells of only four (10,000 square feet each). Its square root, sqrt(4) = 2, is 50% of 4: that was too high. To meet a 10% threshold you want the square root of the number of cells to be just 10% = 1/10 of that number: the solution obviously is 10^2 = 100 cells. To meet a 10% threshold with a greater margin of safety--say, by halving the cellsize--you would want the square root of the number to be just 10%/2 = 5% = 1/20, implying 20^2 = 400 cells.

In short, a good rule of thumb is that to meet x% relative precision requirements, you would aim to have at least n = 100/x^2 or (for more safety) even (200/x)^2 cells within all zones, including the smallest. From this you can work out the cellsize c from the zone areas A and n: c = sqrt(A/n).


Note that the answer does not depend on the zone area:cell size ratio; it depends on the ratio of the square root of the area to the cell size. This is the Pizza Principle in action.

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