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Dear Forum, [edit: excel attachment added] I have been successfully applying the Local Moran's Index (and Gi*) as part of my PhD, but I want to know exactly how the calculation is achieved. I am an archaeologist and I will be writing this in my thesis, so I must present it in a way other non-statisticians can understand it. I decided to produce a worked example, for Gi* it was simple but I'm having difficulty calculating a single instance of the Local Moran's i (Li). Thank you in advance for any help you can give me, I really want to crack this as I am just starting my writing up phase (and it's bugging me)! The ESRI implemented equation is here: http://resources.arcgis.com/en/help/main/10.1/index.html#//005p00000012000000 Here is the local neighbourhood (of a larger dataset 11x11 cells), the target cell is 100. 73|96|82 96|100|92 87|93|82 Neighbourhood average = 87.63 Here is my workings, there is an error which I point out as you go through it. To get Si^2 sum of al the neighbours minus the mean then squared this is divided by the number of neighbouring cells minus 1 (the target cell) this calculation is then subtracted from the mean squared so: �??(73-87.63)�??^2+ �??(96-87.63)�??^2+ �??(87-87.63)�??^2+ �??(96-87.63)�??^2+ �??(93-87.63)�??^2+ �??(82-87.63)�??^2+ �??(92-87.63)�??^2+ �??(82-87.63)�??^2 =465.88 then... Si^2 = 465.88/(8-1)-7678.14= -7611.59 We have to take the value of a neighbour and subtract this from the average of neighbours and multiplied by the search radius (wij) so: ((73-87.63)*1.75)=-25.60 ((96-87.63)*1.75)=14.6475 ((87-87.63)*1.75)=-1.1025 ((96-87.63)*1.75)=14.6475 ((93-87.63)*1.75)=9.3975 ((82-87.63)*1.75)=-9.8525 ((92-87.63)*1.75)=7.6475 ((82-87.63)*1.75)=-9.8525 results summed = -0.07 It should equal -11240.16 This would create the calculation: I-index= (100-87.63)/(-7611.)* -11240.16=18.27 Which is what ArcGIS produces, where have I gone wrong? Thanks, Gary
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02-09-2014
12:25 PM
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Dear forum, I am trying to uderstand exactly how the Local Moran equation works, the equation can be found here: http://resources.esri.com/help/9.3/arcgisdesktop/com/gp_toolref/spatial_statistics_tools/how_cluster_and_outlier_analysis_colon_anselin_local_moran_s_i_spatial_statistics_works.htm For my example I have chosen a single location with a 8 cell neighbourhood, this is within a grid of 11x11 cells (121 squares), the total of all these squares is 5598 and the value of the target cell is 100. For this cell ArcGIS gives the following results: I-index:18,274 Z-score:6.70446 P-value: 0 I am trying to work through he equation but I am having trouble understanding all the algebra. xi= value of the target cell = 100 X(bar) = the Average of all the values in the 121 cell grid = 46 below the division line is Si2 [ for this you take the sum of all the values but not the target cell: 5498 this is then multiplied by the result of xj minus the mean (X(bar)) xj= the neighbours? this is then squared and divided over the number of cells (121) minus 1 = 120 the mean squared is then subtracted from this. ] So this is Si^2 so: 100 - 46 ------------- Si^2 This is multiplied by the Sum of the neighbours in the neighbourhood (701) j not equal to i - so not including the target cell then (xj - mean) This gives you the I-index: Ii this can then be used to create the Z-score as Esri indicate. The values in the local neighbourhood for a single calculation are: 73|94|82 96|100|92 87|93|82 Can anyone help me with this? Many thanks Gary
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10-23-2013
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