python - for loops, indexing, taking the min value of a list -

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i have 2 shapefiles: of lakes , of cities. need find closest city each lake , add name of city lake shapefile. have:

for lake in lake_cursor:     lake_geom = lake.shape     city_dist_list = [] #create city dis list = list of dist 1 lake each city     cityid in range(0, city_length-1):         #obtaining x , y both cities , lakes         cityx = citylist_x_coor[cityid]         cityy = citylist_y_coor[cityid]         lakex = lake_geom.centroid.x         lakey = lake_geom.centroid.y         #calculate distance         dist = math.sqrt((cityx-lakex)**2 + (cityy-lakey)**2)         #add dist city dist list         city_dist_list.append(dist)     closest = min(city_dist_list)     closestid = city_dist_list.index(closest)      lake.city_name = citylist_city_name[closestid]     lake.x_coor = citylist_x_coor[closestid]     lake.y_coor = citylist_y_coor[closestid]     print closest

but keep getting error message starting @ lake.city_name. python shell isn't telling me wrong - ideas? , how fix well? thanks!

the traceback shows:

traceback (most recent call last):   file "c:\users\xxx\xxx\xxx.py", line 71, in <module>     lake.city_name = citylist_city_name[closestid]   file "c:\program files\arcgis\desktop10.1\arcpy\arcpy\arcobjects\_base.py", line 35, in __setattr__     return setattr(self._arc_object, attr, ao) runtimeerror: error 999999: error executing function.

without seeing rest of code, it's hard tell certain, sure looks traceback trying update in searchcursor. searchcursor's rows don't have setters. make sure using updatecursor, , don't forget lake_cursor.updaterow(lake) each lake update. below simple example of how use updatecursor.

import arcpy fc = 'c:/temp/temp.shp' rows = arcpy.updatecursor(fc) row in rows:     print row.mapname     row.mapname = 'some value'     rows.updaterow(row)  del row, rows

this example uses regular arcpy.updatecursor. if have arcgis 10.1, can use arcpy.da.updatecursor, run lot faster. read these updatecursors here , here.


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