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() inSpatialRef.ImportFromEPSG(inputEPSG) outSpatialRef = osr.SpatialReference() outSpatialRef.ImportFromEPSG(outputEPSG) coordTransform = osr.CoordinateTransformation(inSpatialRef, outSpatialRef) # transform point point.Transform(coordTransform) # print point in EPSG 4326 print point.GetX(), point.GetY() 

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

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 

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 start=time.time() inProj = Proj(init='epsg:3857') outProj = Proj(init='epsg:4326') x1,y1 = -11705274.6374,4826473.6922 x2,y2 = transform(inProj,outProj,x1,y1) end=time.time() print(y2,x2) print('%.7f' % (end-start)) # Test 2 - 0.0000517 s --- factor 11,9 start=time.time() 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) end=time.time() print(lat, lon) print('%.7f' % (end-start)) ----------------- 39.72785727727918 -105.15027111593008 0.0006158 39.72785727727918 -105.15027111593008 0.0000517 

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) 

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()) 

This is a shorter version of ustroetz's answer since the ogr.wkbPoint geometry is not required for a point.

from osgeo import osr pointX = -11705274.6374 pointY = 4826473.6922 pointZ = 0.0 # Spatial Reference System inputEPSG = 3857 outputEPSG = 4326 # create coordinate transformation inSpatialRef = osr.SpatialReference() inSpatialRef.ImportFromEPSG(inputEPSG) outSpatialRef = osr.SpatialReference() outSpatialRef.ImportFromEPSG(outputEPSG) coordTransform = osr.CoordinateTransformation(inSpatialRef, outSpatialRef) # transform point, height z is optional x, y, z = coordTransform.TransformPoint(pointX, pointY, pointZ) # print point in EPSG 4326 print(x, y) 

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:

2. 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:

3. 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)

Here is how to project on website

you can get xmin, ymin from here