Arcpy Check Rows for 2 values and Loop

Question

How can I check each row for 2 values [counties & township]? Loop and if they match the dictionaries below, continue the If True statement than do something. My script right now is super slow. Processed 2 counties in 3 hours. #Import system modules import arcpy from arcpy import env import os import time counties = ["RIC",\ "ASD", "ATB", "TUS", "HOL", "STA", "SUM", "CAR", "TRU", "WAS", "FAI", "COL", "WAY", "GEA", "LAK", "HOC", "MRW", "MAR", "MAH", "FRA", "GUE", "HAS", "POR", "CRA", "CUY", "COS"] townships = ["SANDUSKY",\ "MILTON", "WELLER", "MORGAN", "ORWELL", "LENOX", "MONTGOMERY", "MADISON", "MIFFLIN", "WAYNE", "WALNUT CREEK", "ROME", "PERRY", "PIKE", "SPRINGFIELD", "TWINSBURG", "MONROE", "SUGAR CREEK", "TUSCARAWAS", "AUGUSTA", "BROWN", "ORANGE", "ROSE", "MESOPOTAMIA", "WARREN", "FAIRFIELD", "MILL", "SANDY", "FEARING", "VIOLET", "SALEM", "BUTLER", "KNOX", "ADAMS", "MUSKINGUM", "SALT CREEK", "SUGAR CREEK", "WAYNE", "WATERFORD", "WATERTOWN", "MONTVILLE", "GREEN", "MILTON", "PAINT", "WOOSTER", "BAUGHMAN", "CHESTER", "CHIPPEWA", "MADISON", "PAINESVILLE", "PERRY", "CONCORD", "FALLS", "NORTH BLOOMFIELD", "BIG ISLAND", "GRAND PRAIRIE", "CANFIELD", "ELLSWORTH", "AUSTINTOWN", "FRANKLIN", "WESTLAND", "MADISON", "MONROE", "TULLY", "SCOTT", "NORTH", "RIPLEY", "MARION", "PLEASANT", "VALLEY", "SALT CREEK", "GREENFIELD", "AUBURN", "LIBERTY", "CHARDON", "EAST UNION", "BERLIN", "NEWBURY", "PARKMAN", "PAINT", "BLOOM", "HARDY", "DEERFIELD", "FRANKLIN", "TROY", "ATWATER", "MECHANIC", "SUFFIELD", "WINDHAM", "JACKSON", "HAMBDEN", "WASHINGTON", "WHETSTONE", "URBAN", "URBAN", "URBAN", "MANTUA", "HUNTSBURG", "MUNSON", "SOUTHINGTON", "RANDOLPH", "PRAIRIE", "MILTON", "SUGAR CREEK", "CLARK", "CRAWFORD", "THOMPSON", "POLK", "NEWTON", "URBAN", "BETHLEHEM", "FRANKLIN", "LAWRENCE", "SPRINGFIELD", "WARREN", "HOWLAND", "FARMINGTON", "URBAN", "BRACEVILLE", "BAZETTA", "JACKSON", "BRISTOL", "URBAN", "PERRY", "CANAAN", "CHAMPION", "BLOOMFIELD", "BERLIN", "BROOKFIELD", "LAWRENCE", "MIDDLEFIELD", "BURTON", "LEROY", "HAMILTON"] mxd = arcpy.mapping.MapDocument("CURRENT") df = arcpy.mapping.ListDataFrames(mxd, "Layers")[0] lyr = arcpy.mapping.ListLayers(mxd, "ohioGrids", df)[0] for lyr in arcpy.mapping.ListLayers(mxd, "*"): if lyr.name == "ohioGrids": for a in counties: b = "%s" % (a) # #print ("%s" % (b)) for c in townships: d = "%s" % (c) # #print ("%s" % (d)) arcpy.SelectLayerByAttribute_management("ohioGrids", "CLEAR_SELECTION") cnty = '"COUNTY_CD" = \'%s\'' % b seperator = " AND " twpship = '"TOWNSHIP_N" = \'%s\'' % c lyr.definitionQuery = cnty + seperator + twpship result1 = arcpy.GetCount_management("ohioGrids") #print('{} has {} records'.format("ohioGrids", result[0])) if result1 > 0: #arcpy.RefreshActiveView() # GetExtent lyr = arcpy.mapping.ListLayers(mxd, 'ohioGrids', df)[0] ext = lyr.getExtent() df.extent = ext # SelectLayerByLocation selection = arcpy.SelectLayerByLocation_management ("Meters", "WITHIN", "ohioGrids") result = arcpy.GetCount_management(selection) count = int(result.getOutput(0)) if count > 0: print "COUNTY: " + b + " & " + "TOWNSHIP: " + d print count arcpy.SelectLayerByAttribute_management("ohioGrids", "CLEAR_SELECTION") else: arcpy.SelectLayerByAttribute_management("ohioGrids", "CLEAR_SELECTION") ‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍

JoshuaBixby · Answer

Beyond a discussion on lists versus dictionaries, something isn't right. The code is pretty straightforward and simple, so I doubt restructuring the Python data structures will result in much of a change. I suspect your problem is this line: selection = arcpy.SelectLayerByLocation_management ("Meters", "WITHIN", "ohioGrids")‍‍ Put a print statement with the date and time before and after that operation, see how long it is taking.

DavidPike · Answer

scanned through it, can this not be done with a spatial join?

danretired · Answer

It is probably slow because you aren't working with dictionaries but two mutually exclusive lists.

If you had a dictionary, then there would be a "key" which you could query and it would return all the possible values for that key.  Right now, you pick something from list "A" and cycle through list "B" even though it is likely that list "B" contains entries that couldn't exist in "A".  There may be overlaps where some values may be associated with multiple keys, but the would be taken of with a true dictionary structure.