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I have two shapefiles-- one is labeled islands, the other is unlabeled parcels of land on some of the islands. My issue is that I need to label the islands that have parcels of land. My instinct is to do a "Select by Attribute" for "ISLAND"has"PARCEL" but a) I don't know the actual code for this and b) I can't figure out how to make my program select by attribute using multiple layers or even if this is the right way to go about it.

I'm using ArcMap. This seems like a relatively simple task but I'm having great difficulty with it.

Any suggestions?

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  • Select islands by location using parcels
    – FelixIP
    Commented Sep 12, 2015 at 1:01
  • Are you doing this manually or aer you creating a program using python and arcpy? You can only use arcpy.SelectLayerByAttribute_management to select Islands with names and parcels with numbers separately. Then you would use arcpy.SelectLayerByLocation_management to find the SUBSET_SELECTION of named Islands that intersect with the layer containing selected numbered Parcels. You could alternatively use arcpy.SpatialJoin_analysis fof these selections with Islands as the target and the One_To_One option and the KEEP_COMMON option to create a new shapefile of Islands that meet your criteria. Commented Sep 12, 2015 at 1:02
  • @RichardFairhurst Your comment looks like it would make a valid brief answer.
    – PolyGeo
    Commented Jan 10, 2017 at 7:04

1 Answer 1

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As commented by @RichardFairhurst:

Are you doing this manually or are you creating a program using python and arcpy? You can only use arcpy.SelectLayerByAttribute_management to select Islands with names and parcels with numbers separately. Then you would use arcpy.SelectLayerByLocation_management to find the SUBSET_SELECTION of named Islands that intersect with the layer containing selected numbered Parcels. You could alternatively use arcpy.SpatialJoin_analysis fof these selections with Islands as the target and the One_To_One option and the KEEP_COMMON option to create a new shapefile of Islands that meet your criteria.

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