This site returns

X: -11705274.6374
Y: 4826473.6922

when you search with the first key value of 000090, as an example. I assume that this is a spatial reference.

I am looking for instructions, or examples, of how to convert this to latitude and longitude using Python.

9 Answers 9


The simplest way to transform coordinates in Python is pyproj, i.e. the Python interface to PROJ.4 library. In fact:

from pyproj import Proj, transform

inProj = Proj(init='epsg:3857')
outProj = Proj(init='epsg:4326')
x1,y1 = -11705274.6374,4826473.6922
x2,y2 = transform(inProj,outProj,x1,y1)
print x2,y2

returns -105.150271116 39.7278572773

EDIT based on Marc's comment:

pyproj 2.4 gives a FutureWarning about deprecated Proj initialization with the init= syntax. The updated syntax is identical but without the init=. Like this:

inProj = Proj('epsg:3857')
outProj = Proj('epsg:4326')

By default the site you linked to uses the Spatial Reference System EPSG 3857 (WGS84 Web Mercator). I found this information here.

You can either specify another Spatial Reference System by entering the desired EPSG into the form under Spatial Reference or you can convert the returned coordinates with Python.

For instance you can use the GDAL Python bindings to convert this point from the projected coordinate system (EPSG 3857) to a geographic coordinate system (EPSG 4326).

from osgeo import ogr
from osgeo import osr  

pointX = -11705274.6374 
pointY = 4826473.6922

# Spatial Reference System
inputEPSG = 3857
outputEPSG = 4326

# create a geometry from coordinates
point = ogr.Geometry(ogr.wkbPoint)
point.AddPoint(pointX, pointY)

# create coordinate transformation
inSpatialRef = osr.SpatialReference()

outSpatialRef = osr.SpatialReference()

coordTransform = osr.CoordinateTransformation(inSpatialRef, outSpatialRef)

# transform point

# print point in EPSG 4326
print point.GetX(), point.GetY()

This returns for your point the coordinates of -105.150271116 39.7278572773.

  • @Thomas it is returning 39.7278572773 -105.150271116, why is it flipping coordinates?
    – Ayaz49
    Mar 27 at 10:23

When using pyproj, note the differences from various releases in how it is used to transform data. Here are a few examples using new/old capabilities based on the question:

Using pyproj >= 2.2.0

import pyproj
print(pyproj.__version__)  # 2.4.1
print(pyproj.proj_version_str) # 6.2.1

proj = pyproj.Transformer.from_crs(3857, 4326, always_xy=True)

x1, y1 = (-11705274.6374, 4826473.6922)
x2, y2 = proj.transform(x1, y1)
print((x2, y2))  # (-105.15027111593008, 39.72785727727918)

Using pyproj <= 1.9.6

import pyproj
print(pyproj.__version__)  # 1.9.6
print(pyproj.proj_version_str) # 4.9.3

inProj = pyproj.Proj(init='epsg:3857')
outProj = pyproj.Proj(init='epsg:4326')

x1, y1 = (-11705274.6374, 4826473.6922)
x2, y2 = pyproj.transform(inProj, outProj, x1, y1)
print((x2, y2))  # (-105.15027111593008, 39.72785727727918)

There are a few considerations with the different versions of PROJ/pyproj:

  • transform for older versions always return the same axis order of "x, y" or "longitude, latitude", whereas PROJ 6+ is the order as defined by EPSG, unless an option like always_xy=True is specified
  • Proj is limited to converting between geographic and projection coordinates within one datum, whereas the newer Transformer takes into account datum shifts, and is recommended for newer pyproj use

afalciano has the right answer but wanted to include a variant usage of pyproj.

It does require you know the proj4 string and is a tiny bit faster.

import pyproj
p = pyproj.Proj("+proj=merc +lon_0=0 +k=1 +x_0=0 +y_0=0 +a=6378137 +b=6378137 +towgs84=0,0,0,0,0,0,0 +units=m +no_defs")
lon, lat = p(x, y, inverse=True)
print lat, lon
  • 2
    You don't need the proj4 string, substitute the second line for p = pyproj.Proj(init='epsg:3857') and the result is the same. Sep 15, 2016 at 0:08
  • 1
    The result is the same but last I checked this was a tiny bit faster. Sep 17, 2016 at 17:03
  • 3
    Even though it was not the intention of this answer, it is particularly useful when you have a totally custom projection, which is not listed in the EPSG-repository.
    – Andreas L.
    Jan 13, 2020 at 13:51

The output is not a spatial/coordinate reference system, it's a pair of coordinates. You need to know what the spatial reference is to reproject the coordinates.

However, that's not required in this case. Just pass an appropriate output spatial reference to the service and it will return the coordinates in Lon/Lat.

Here is the page with output coordinates in Lon/Lat format using the WGS-84 geographic spatial reference system (EPSG 4326).


Tried the code suggested by Marcel Wilson and it is faster:

from pyproj import Proj, transform
import time
import pyproj

# Test 1 - 0.0006158 s
inProj = Proj(init='epsg:3857')
outProj = Proj(init='epsg:4326')
x1,y1 = -11705274.6374,4826473.6922
x2,y2 = transform(inProj,outProj,x1,y1)
print('%.7f' % (end-start))

# Test 2 - 0.0000517 s --- factor 11,9
x,y = -11705274.6374,4826473.6922
p = pyproj.Proj("+proj=merc +lon_0=0 +k=1 +x_0=0 +y_0=0 +a=6378137 +b=6378137 +towgs84=0,0,0,0,0,0,0 +units=m +no_defs")
lon, lat = p(x, y, inverse=True)
print(lat, lon)
print('%.7f' % (end-start))

39.72785727727918 -105.15027111593008
39.72785727727918 -105.15027111593008

Please note that the transform function of pyproj accepts also arrays, which is quite useful when it comes to dataframes

import pandas as pd
from pyproj import Proj, transform

df = pd.DataFrame({'x': [-11705274.6374]*100, 
                   'y': [4826473.6922]*100})
inProj, outProj = Proj(init='epsg:3857'), Proj(init='epsg:4326')
df['x2'], df['y2'] = transform(inProj, outProj, df['x'].tolist(), df['y'].tolist())

I found this post when looking for ways of doing this within QGIS. As described here, the method used looks like so:

def convertProjection(self,x,y,from_crs,to_crs):
    crsSrc = QgsCoordinateReferenceSystem(from_crs)
    crsDest = QgsCoordinateReferenceSystem(to_crs)
    xform = QgsCoordinateTransform(crsSrc, crsDest)
    pt = xform.transform(QgsPoint(x,y))
    return pt.x, pt.y

# Remove the "EPSG:" part
from_crs = 3857
to_crs = 4326
x = -11705274.6374    
y = 4826473.6922
lon, lat = self.convertProjection(x,y,from_crs, to_crs)

convert esri extent bbox to lat long

  1. use esri javascipt api module ( client side projection, fast, no need ajax, no need internet, projection done in your browser, extremely fast)

    first include css, js

link(rel='stylesheet',href='https://js.arcgis.com/4.17/esri/themes/light/main.css') script(src='https://js.arcgis.com/4.17/')

This is my working example: enter image description here

  1. proj4js ( need ajax, internet to read projection string from website, due to website does not allow jsonp, not allow cors, you have to setup your own proxy to bypass CORS problem, not easy for normal user )

But if you already known from what to what, then you can just copy past projection string, no internet needed, all projectin done in your browser.

If you don't know project from what to what, projection wkid is dynamic generated in the runtime, then like I said previously, you must need ajax, need internet to read projection string from website on the fly, in the run time. It is not easy, because you have to bypass CORS problem.

Here is my working example: enter image description here

  1. use Arcgis.com hosted web service ( need internet, very slow, not reliable, if ESRI server is crowded, you might get it very slow. Every ajax request can only conver one pair of lat lng) enter image description here

enter image description here

Here is how to project on website enter image description here

you can get xmin, ymin from here enter image description here

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