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.


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).

import ogr, 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.


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 '16 at 0:08
  • 1
    The result is the same but last I checked this was a tiny bit faster. Sep 17 '16 at 17:03
  • 2
    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 '20 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

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

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)

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