How to geocode a large number of addresses?

How to geocode a large number of addresses?

By : user3861827
Date : November 21 2020, 07:01 PM
hope this fix your issue When doing a large number of requests to Google geocoding service you need to throttle the requests as responses start failing. To give you a better idea here is a snippet from a Drupal (PHP) module that I wrote.
code :

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iPhone iOS5 CLGeocoder how to geocode a large (200) set of addresses?

iPhone iOS5 CLGeocoder how to geocode a large (200) set of addresses?

By : Hardy
Date : March 29 2020, 07:55 AM
will help you You can't immediately geocode large sets. iOS throttles you. I have seen that iOS limits you to 50 geocodes at a time, with the "time" factor being an unknown.
I've had a similar problem, but as I needed only to present the geocoding data in a sequence to the user that takes time, I queued all my geocodings.
How to geocode a set of 200 addresses at once

How to geocode a set of 200 addresses at once

By : Abhishek Singh
Date : March 29 2020, 07:55 AM
around this issue I have a collection of addresses, and I needed to convert them into co-ordinates. What I did was to divide the collection into chunks of maximum 40 contacts(you may go up to 50, but not beyond that in a minute). So a group with over 40 contacts had to have a delay in between so that all the addresses would be processed and converted into co-ordinates before the mapview loads.
I need to load the mapview based on user requirements. In my case, the mapview could be loaded again and gain, irrespective of the contacts list in the group and irrespective of the time factor. If the user intends to load the mapview and reload it again under 1 minute I needed to develop a logic for that. For that I decided to save the state of the mapview, if the user decided to reload the same group of addresses under 1 minute.
Geocode two or more addresses

Geocode two or more addresses

By : user3726582
Date : March 29 2020, 07:55 AM
To fix the issue you can do There is a mistake in your code in the destinationGeocoder you must use address instead of destination the geocode property is always address
code :
 destinationGeocoder.geocode( { 'address': destination }, function(results, status) {
        var addr_type = results[0].types[0];    // type of address inputted that was geocoded
        if ( status == google.maps.GeocoderStatus.OK ) 
            ShowLocation( results[0].geometry.location, destination, addr_type );
            alert("Geocode was not successful for the following reason: " + status);        
How do I reverse geocode a large number of points in python?

How do I reverse geocode a large number of points in python?

By : M.Fatih
Date : March 29 2020, 07:55 AM
it fixes the issue If you want to avoid for example the heavy-duty machinery of a PostGIS database, it could be of interest to employ the rtree package as (as the documentation states) "cheapo spatial database". The idea would be mostly as follows:
code :
#!/usr/bin/env python
from itertools import product
from random import uniform, sample, seed
from rtree import index
from shapely.geometry import Point, Polygon, box, shape
from shapely.affinity import translate


#generate random polygons, in your case, the polygons are stored
#in geo_data['features']
P = Polygon([(0, 0), (0.5, 0), (0.5, 0.5), (0, 0.5), (0, 0)])
polygons = []
for dx, dy in product(range(0, 100), range(0, 100)):
    polygons.append(translate(P, dx, dy))

#construct the spatial index and insert bounding boxes of all polygons
idx = index.Index()
for pid, P in enumerate(polygons):
    idx.insert(pid, P.bounds)

delta = 0.5
for i in range(0, 1000):
    #generate random points
    x, y = uniform(0, 10), uniform(0, 10)
    pnt = Point(x, y)

    #create a region around the point of interest
    bounds = (x-delta, y-delta, x+delta, y+delta)

    #also possible, but much slower
    #bounds = pnt.buffer(delta).bounds

    #the index tells us which polygons are worth checking, i.e.,
    #the bounding box of which intersects with the region constructed in previous step
    for candidate in idx.intersection(bounds):
        P = polygons[candidate]

        #test only these candidates
        if P.contains(pnt):
            print(pnt, P)
Storing Large Number of IP Addresses in Memory

Storing Large Number of IP Addresses in Memory

By : Neha Kanodia
Date : March 29 2020, 07:55 AM
help you fix your problem Update: If you stored the entire 4 Billion IPv4 addresses as a single array then you could represent time as an individual short.
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