Develop Reference

project local coordinates to global GPS with reference point

I have a bunch of shapes (e.g. shapely LineStrings or Polygons) in a geopandas GeoDataFrame.
The shapes specify coordinates in a local 200x200 meters grid, i.e. all coordinates are between (0, 0) and (200, 200).
I now would like to "place" these lines globally.
For this, I want to specify a GPS Point (with a given lat/lon) as a reference.
My first (naive) approach would be to use geographiclib, take all shapes' coords (in local X/Y) and apply the following transformation and "recreate" the shape:
# Convert coordinates to GPS location
from shapely.geometry import LineString
from geographiclib.geodesic import Geodesic
geod = Geodesic.WGS84 # the base geodesic (i.e. the world)
origin = (48.853772345870176, 2.350983211585546) # this is somewhere in Paris, for example
def local_to_latlong(x, y, orientation=0, scale=1):
""" Two step process.
- First walk x meters to east from origin.
- Then, from that point, walk y meters north from origin.
Optional:
- orientation allows to "spin" the coordinates
- scale allows to grow/shrink the distances
"""
go_X = geod.Direct(*origin, orientation + 90, x * scale) # x is East-coordinate
go_Y = geod.Direct(go_X["lat2"], go_X["lon2"], orientation + 0, y * scale) # y is North-coordinate
return go_Y["lat2"], go_Y["lon2"]
original_line = LineString([(0,0), (100,100), (200,100)])
global_line = LineString([local_to_latlong(x, y) for y, x in original_line.coords])
However, I hope that this is not the smartest way to do it, and that there are smarter ways out there...
I would like to apply such a transformation onto any shape within a GeoDataFrame. Ideally, it would work using a "to_crs", but I am not sure how to transform the shapes so they are "in reference to a origin" and which crs to use.

given your origin is EPSG:4326, you can estimate the UTM zone
with this you can get UTM zone coordinates of origin
translate your custom 200x200 metre zone into co-ordinates of UTM zone
finally use to_crs() to transform into EPSG:4326
import shapely.geometry
import geopandas as gpd
import pandas as pd
import numpy as np
# generate some polygons (squares), where grid is 200*200
gdf = gpd.GeoDataFrame(
geometry=pd.DataFrame(
np.repeat(np.sort(np.random.randint(0, 200, [20, 2]), axis=1), 2, axis=1)
).apply(lambda d: shapely.geometry.box(*d), axis=1)
)
# chage to linestrings, clearer when we plot
gdf["geometry"] = gdf["geometry"].exterior
origin = (2.350983211585546, 48.853772345870176) # this is somewhere in Paris, for example
# work out utm crs of point. utm is in metres
gdf_o = gpd.GeoDataFrame(geometry=[shapely.geometry.Point(origin)], crs="EPSG:4326")
crs = gdf_o.estimate_utm_crs()
# where is origin in utm zone
xo,yo = gdf_o.to_crs(crs).loc[0,"geometry"].xy
# translate custom zone to co-ordinates of utm zone
# assume point is center of 200x200 grid (hence subtract 100)
gdf_gps = gdf["geometry"].translate(xoff=xo[0]-100, yoff=yo[0]-100).set_crs(crs).to_crs("epsg:4326")
# plot on map to show it has worked...
m = gdf_gps.explore()
m = gdf_o.explore(m=m, color="red", marker_kwds={"radius":20})
m

Python: using polygons to create a mask on a given 2d grid

I have some polygons (Canadian provinces), read in with GeoPandas, and want to use these to create a mask to apply to gridded data on a 2-d latitude-longitude grid (read from a netcdf file using iris). An end goal would be to only have data for a given province remaining, with the rest of the data masked out. So the mask would be 1's for grid boxes within the province, and 0's or NaN's for grid boxes outside the province.
The polygons can be obtained from the shapefile here:
https://www.dropbox.com/s/o5elu01fetwnobx/CAN_adm1.shp?dl=0
The netcdf file I am using can be downloaded here:
https://www.dropbox.com/s/kxb2v2rq17m7lp7/t2m.20090815.nc?dl=0
I imagine there are two approaches here but I am struggling with both:
1) Use the polygon to create a mask on the latitude-longitude grid so that this can be applied to lots of datafiles outside of python (preferred)
2) Use the polygon to mask the data that have been read in and extract only the data inside the province of interest, to work with interactively.
My code so far:
import iris
import geopandas as gpd
#read the shapefile and extract the polygon for a single province
#(province names stored as variable 'NAME_1')
Canada=gpd.read_file('CAN_adm1.shp')
BritishColumbia=Canada[Canada['NAME_1'] == 'British Columbia']
#get the latitude-longitude grid from netcdf file
cubelist=iris.load('t2m.20090815.nc')
cube=cubelist[0]
lats=cube.coord('latitude').points
lons=cube.coord('longitude').points
#create 2d grid from lats and lons (may not be necessary?)
[lon2d,lat2d]=np.meshgrid(lons,lats)
#HELP!
Thanks very much for any help or advice.
UPDATE: Following the great solution from #DPeterK below, my original data can be masked, giving the following:

It looks like you have started well! Geometries loaded from shapefiles expose various geospatial comparison methods, and in this case you need the contains method. You can use this to test each point in your cube's horizontal grid for being contained within your British Columbia geometry. (Note that this is not a fast operation!) You can use this comparison to build up a 2D mask array, which could be applied to your cube's data or used in other ways.
I've written a Python function to do the above – it takes a cube and a geometry and produces a mask for the (specified) horizontal coordinates of the cube, and applies the mask to the cube's data. The function is below:
def geom_to_masked_cube(cube, geometry, x_coord, y_coord,
mask_excludes=False):
"""
Convert a shapefile geometry into a mask for a cube's data.
Args:
* cube:
The cube to mask.
* geometry:
A geometry from a shapefile to define a mask.
* x_coord: (str or coord)
A reference to a coord describing the cube's x-axis.
* y_coord: (str or coord)
A reference to a coord describing the cube's y-axis.
Kwargs:
* mask_excludes: (bool, default False)
If False, the mask will exclude the area of the geometry from the
cube's data. If True, the mask will include *only* the area of the
geometry in the cube's data.
.. note::
This function does *not* preserve lazy cube data.
"""
# Get horizontal coords for masking purposes.
lats = cube.coord(y_coord).points
lons = cube.coord(x_coord).points
lon2d, lat2d = np.meshgrid(lons,lats)
# Reshape to 1D for easier iteration.
lon2 = lon2d.reshape(-1)
lat2 = lat2d.reshape(-1)
mask = []
# Iterate through all horizontal points in cube, and
# check for containment within the specified geometry.
for lat, lon in zip(lat2, lon2):
this_point = gpd.geoseries.Point(lon, lat)
res = geometry.contains(this_point)
mask.append(res.values[0])
mask = np.array(mask).reshape(lon2d.shape)
if mask_excludes:
# Invert the mask if we want to include the geometry's area.
mask = ~mask
# Make sure the mask is the same shape as the cube.
dim_map = (cube.coord_dims(y_coord)[0],
cube.coord_dims(x_coord)[0])
cube_mask = iris.util.broadcast_to_shape(mask, cube.shape, dim_map)
# Apply the mask to the cube's data.
data = cube.data
masked_data = np.ma.masked_array(data, cube_mask)
cube.data = masked_data
return cube
If you just need the 2D mask you could return that before the above function applies it to the cube.
To use this function in your original code, add the following at the end of your code:
geometry = BritishColumbia.geometry
masked_cube = geom_to_masked_cube(cube, geometry,
'longitude', 'latitude',
mask_excludes=True)
If this doesn't mask anything it might well mean that your cube and geometry are defined on different extents. That is, your cube's longitude coordinate runs from 0°–360°, and if the geometry's longitude values run from -180°–180°, then the containment test will never return True. You can fix this by changing the extents of your cube with the following:
cube = cube.intersection(longitude=(-180, 180))

I found an alternative solution to the excellent one posted by #DPeterK above, which yields the same result. It uses matplotlib.path to test if points are contained within the exterior coordinates described by the geometries loaded from a shape file. I am posting this because this method is ~10 times faster than that given by #DPeterK (2:23 minutes vs 25:56 minutes). I'm not sure what is preferable: an elegant solution, or a speedy, brute force solution. Perhaps one can have both?!
One complication with this method is that some geometries are MultiPolygons - i.e. the shape consists of several smaller polygons (in this case, the province of British Columbia includes islands off of the west coast, which can't be described by the coordinates of the mainland British Columbia Polygon). The MultiPolygon has no exterior coordinates but the individual polygons do, so these each need to be treated individually. I found that the neatest solution to this was to use a function copied from GitHub (https://gist.github.com/mhweber/cf36bb4e09df9deee5eb54dc6be74d26), which 'explodes' MultiPolygons into a list of individual polygons that can then be treated separately.
The working code is outlined below, with my documentation. Apologies that it is not the most elegant code - I am relatively new to Python and I'm sure there are lots of unnecessary loops/neater ways to do things!
import numpy as np
import iris
import geopandas as gpd
from shapely.geometry import Point
import matplotlib.path as mpltPath
from shapely.geometry.polygon import Polygon
from shapely.geometry.multipolygon import MultiPolygon
#-----
#FIRST, read in the target data and latitude-longitude grid from netcdf file
cubelist=iris.load('t2m.20090815.minus180_180.nc')
cube=cubelist[0]
lats=cube.coord('latitude').points
lons=cube.coord('longitude').points
#create 2d grid from lats and lons
[lon2d,lat2d]=np.meshgrid(lons,lats)
#create a list of coordinates of all points within grid
points=[]
for latit in range(0,241):
for lonit in range(0,480):
point=(lon2d[latit,lonit],lat2d[latit,lonit])
points.append(point)
#turn into np array for later
points=np.array(points)
#get the cube data - useful for later
fld=np.squeeze(cube.data)
#create a mask array of zeros, same shape as fld, to be modified by
#the code below
mask=np.zeros_like(fld)
#NOW, read the shapefile and extract the polygon for a single province
#(province names stored as variable 'NAME_1')
Canada=gpd.read_file('/Users/ianashpole/Computing/getting_province_outlines/CAN_adm_shp/CAN_adm1.shp')
BritishColumbia=Canada[Canada['NAME_1'] == 'British Columbia']
#BritishColumbia.geometry.type reveals this to be a 'MultiPolygon'
#i.e. several (in this case, thousands...) if individual polygons.
#I ultimately want to get the exterior coordinates of the BritishColumbia
#polygon, but a MultiPolygon is a list of polygons and therefore has no
#exterior coordinates. There are probably many ways to progress from here,
#but the method I have stumbled upon is to 'explode' the multipolygon into
#it's individual polygons and treat each individually. The function below
#to 'explode' the MultiPolygon was found here:
#https://gist.github.com/mhweber/cf36bb4e09df9deee5eb54dc6be74d26
#---define function to explode MultiPolygons
def explode_polygon(indata):
indf = indata
outdf = gpd.GeoDataFrame(columns=indf.columns)
for idx, row in indf.iterrows():
if type(row.geometry) == Polygon:
#note: now redundant, but function originally worked on
#a shapefile which could have combinations of individual polygons
#and MultiPolygons
outdf = outdf.append(row,ignore_index=True)
if type(row.geometry) == MultiPolygon:
multdf = gpd.GeoDataFrame(columns=indf.columns)
recs = len(row.geometry)
multdf = multdf.append([row]*recs,ignore_index=True)
for geom in range(recs):
multdf.loc[geom,'geometry'] = row.geometry[geom]
outdf = outdf.append(multdf,ignore_index=True)
return outdf
#-------
#Explode the BritishColumbia MultiPolygon into its constituents
EBritishColumbia=explode_polygon(BritishColumbia)
#Loop over each individual polygon and get external coordinates
for index,row in EBritishColumbia.iterrows():
print 'working on polygon', index
mypolygon=[]
for pt in list(row['geometry'].exterior.coords):
print index,', ',pt
mypolygon.append(pt)
#See if any of the original grid points read from the netcdf file earlier
#lie within the exterior coordinates of this polygon
#pth.contains_points returns a boolean array (true/false), in the
#shape of 'points'
path=mpltPath.Path(mypolygon)
inside=path.contains_points(points)
#find the results in the array that were inside the polygon ('True')
#and set them to missing. First, must reshape the result of the search
#('points') so that it matches the mask & original data
#reshape the result to the main grid array
inside=np.array(inside).reshape(lon2d.shape)
i=np.where(inside == True)
mask[i]=1
print 'fininshed checking for points inside all polygons'
#mask now contains 0's for points that are not within British Columbia, and
#1's for points that are. FINALLY, use this to mask the original data
#(stored as 'fld')
i=np.where(mask == 0)
fld[i]=np.nan
#Done.

Fix up shapely polygon object when discontinuous after map projection

This demo program (intended to be run in an IPython notebook; you need matplotlib, mpl_toolkits.basemap, pyproj, and shapely) is supposed to plot increasingly large circles on the surface of the Earth. It works correctly as long as the circle does not cross over one of the poles. If that happens, the result is complete nonsense when plotted on a map (see below cell 2)
If I plot them "in a void" instead of on a map (see below cell 3) the results are correct in the sense that, if you removed the horizontal line going from +180 to -180 longitude, the rest of the curve would indeed delimit the boundary between the interior and exterior of the desired circle. However, they are wrong in that the polygon is invalid (.is_valid is False), and much more importantly, the nonzero-winding-number interior of the polygon does not enclose the correct region of the map.
I believe this is happening because shapely.ops.transform is blind to the coordinate singularity at +180==-180 longitude. The question is, how do I detect the problem and repair the polygon, so that it does enclose the correct region of the map? In this case, an appropriate fixup would be to replace the horizontal segment from (X,+180) -- (X,-180) with three lines, (X,+180) -- (+90,+180) -- (+90,-180) -- (X,-180); but note that if the circle had gone over the south pole, the fixup lines would need to go south instead. And if the circle had gone over both poles, we'd have a valid polygon again but its interior would be the complement of what it should be. I need to detect all of these cases and handle them correctly. Also, I do not know how to "edit" a shapely geometry object.
Downloadable notebook: https://gist.github.com/zackw/e48cb1580ff37acfee4d0a7b1d43a037
## cell 1
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.basemap import Basemap
import pyproj
from shapely.geometry import Point, Polygon, MultiPolygon
from shapely.ops import transform as sh_transform
from functools import partial
wgs84_globe = pyproj.Proj(proj='latlong', ellps='WGS84')
def disk_on_globe(lat, lon, radius):
aeqd = pyproj.Proj(proj='aeqd', ellps='WGS84', datum='WGS84',
lat_0=lat, lon_0=lon)
return sh_transform(
partial(pyproj.transform, aeqd, wgs84_globe),
Point(0, 0).buffer(radius)
)
## cell 2
def plot_poly_on_map(map_, pol):
if isinstance(pol, Polygon):
map_.plot(*(pol.exterior.xy), '-', latlon=True)
else:
assert isinstance(pol, MultiPolygon)
for p in pol:
map_.plot(*(p.exterior.xy), '-', latlon=True)
plt.figure(figsize=(14, 12))
map_ = Basemap(projection='cyl', resolution='c')
map_.drawcoastlines(linewidth=0.25)
for rad in range(1,10):
plot_poly_on_map(
map_,
disk_on_globe(40.439, -79.976, rad * 1000 * 1000)
)
plt.show()
## cell 3
def plot_poly_in_void(pol):
if isinstance(pol, Polygon):
plt.plot(*(pol.exterior.xy), '-')
else:
assert isinstance(pol, MultiPolygon)
for p in pol:
plt.plot(*(p.exterior.xy), '-', latlon=True)
plt.figure()
for rad in range(1,10):
plot_poly_in_void(
disk_on_globe(40.439, -79.976, rad * 1000 * 1000)
)
plt.show()
(The sunlit region shown at http://www.die.net/earth/rectangular.html is an example of what a circle that crosses a pole should look like when projected onto an equirectangular map, as long as it's not an equinox today.)

Manually fixing up the projected polygon turns out not to be that bad.
There are two steps: first, find all segments of the polygon that cross the coordinate singularity at longitude ±180, and replace them with excursions to either the north or south pole, whichever is nearest; second, if the resulting polygon doesn't contain the origin point, invert it. Note that both steps must be carried out whether or not shapely thinks the projected polygon is "invalid"; depending on where the starting point is, it may cross one or both poles without being invalid.
This probably isn't the most efficient way to do it, but it works.
import pyproj
from shapely.geometry import Point, Polygon, box as Box
from shapely.ops import transform as sh_transform
from functools import partial
wgs84_globe = pyproj.Proj(proj='latlong', ellps='WGS84')
def disk_on_globe(lat, lon, radius):
"""Generate a shapely.Polygon object representing a disk on the
surface of the Earth, containing all points within RADIUS meters
of latitude/longitude LAT/LON."""
aeqd = pyproj.Proj(proj='aeqd', ellps='WGS84', datum='WGS84',
lat_0=lat, lon_0=lon)
disk = sh_transform(
partial(pyproj.transform, aeqd, wgs84_globe),
Point(0, 0).buffer(radius)
)
# Fix up segments that cross the coordinate singularity at longitude ±180.
# We do this unconditionally because it may or may not create a non-simple
# polygon, depending on where the initial point was.
boundary = np.array(disk.boundary)
i = 0
while i < boundary.shape[0] - 1:
if abs(boundary[i+1,0] - boundary[i,0]) > 180:
assert (boundary[i,1] > 0) == (boundary[i,1] > 0)
vsign = -1 if boundary[i,1] < 0 else 1
hsign = -1 if boundary[i,0] < 0 else 1
boundary = np.insert(boundary, i+1, [
[hsign*179, boundary[i,1]],
[hsign*179, vsign*89],
[-hsign*179, vsign*89],
[-hsign*179, boundary[i+1,1]]
], axis=0)
i += 5
else:
i += 1
disk = Polygon(boundary)
# If the fixed-up polygon doesn't contain the origin point, invert it.
if not disk.contains(Point(lon, lat)):
disk = Box(-180, -90, 180, 90).difference(disk)
assert disk.is_valid
assert disk.boundary.is_simple
assert disk.contains(Point(lon, lat))
return disk
The other problem -- mpl_toolkits.basemap.Basemap.plot producing garbage -- is not corrected by fixing up the polygon as above. However, if you manually project the polygon into map coordinates and then draw it using a descartes.PolygonPatch, that works, as long as the projection has a rectangular boundary, and that's enough of a workaround for me. (I think it would work for any projection if one added a lot of extra points along all straight lines at the map boundary.)
%matplotlib inline
from matplotlib import pyplot as plt
from mpl_toolkits.basemap import Basemap
from descartes import PolygonPatch
plt.figure(figsize=(14, 12))
map_ = Basemap(projection='cea', resolution='c')
map_.drawcoastlines(linewidth=0.25)
for rad in range(3,19,2):
plt.gca().add_patch(PolygonPatch(
sh_transform(map_,
disk_on_globe(40.439, -79.976, rad * 1000 * 1000)),
alpha=0.1))
plt.show()

Why are Basemap south polar stereographic map projection coordinates not agreeing with those of data sets in the same projection?

Some satellite based earth observation products provide latitude/longitude information while others provide the X/Y coordinates within a given grid projection (and there are also some having both, see example).
My approach in the second case is to set up a Basemap map which has the same parameters (projection, ellipsoid, origin of map) as given by the data provider in a way that the given X/Y values equal the Basemap coordinates. However if I do so the geolocation does not agree with other data sets including the Basemap coastline.
I have experienced this with three different data sets from different trustworthy sources. For the minimal example I use Landsat data provided by the U.S. Geological Survey which includes both, X/Y coordinates of a South Polar Stereographic grid and the corresponding lat/lon coordinates for all four corners of the image.
From a Landsat metafile we get (ID: LC82171052016079LGN00):
CORNER_UL_LAT_PRODUCT = -66.61490 CORNER_UL_LON_PRODUCT = -61.31816
CORNER_UR_LAT_PRODUCT = -68.74325 CORNER_UR_LON_PRODUCT = -58.04533
CORNER_LL_LAT_PRODUCT = -67.68721 CORNER_LL_LON_PRODUCT = -67.01109
CORNER_LR_LAT_PRODUCT = -69.94052 CORNER_LR_LON_PRODUCT = -64.18581
CORNER_UL_PROJECTION_X_PRODUCT = -2259300.000
CORNER_UL_PROJECTION_Y_PRODUCT = 1236000.000
CORNER_UR_PROJECTION_X_PRODUCT = -1981500.000
CORNER_UR_PROJECTION_Y_PRODUCT = 1236000.000
CORNER_LL_PROJECTION_X_PRODUCT = -2259300.000
CORNER_LL_PROJECTION_Y_PRODUCT = 958500.000
CORNER_LR_PROJECTION_X_PRODUCT = -1981500.000
CORNER_LR_PROJECTION_Y_PRODUCT = 958500.000
...
GROUP = PROJECTION_PARAMETERS MAP_PROJECTION = "PS" DATUM = "WGS84"
ELLIPSOID = "WGS84" VERTICAL_LON_FROM_POLE = 0.00000 TRUE_SCALE_LAT =
-71.00000 FALSE_EASTING = 0 FALSE_NORTHING = 0 GRID_CELL_SIZE_PANCHROMATIC = 15.00 GRID_CELL_SIZE_REFLECTIVE = 30.00
GRID_CELL_SIZE_THERMAL = 30.00 ORIENTATION = "NORTH_UP"
RESAMPLING_OPTION = "CUBIC_CONVOLUTION" END_GROUP =
PROJECTION_PARAMETERS
By using Basemap with the right map projection we should be able to derive the corner lat/lon values from the X/Y values:
import numpy as np
from mpl_toolkits.basemap import Basemap
m=Basemap(resolution='h',projection='spstere', ellps='WGS84', boundinglat=-60,lon_0=180, lat_ts=-71)
x_crn=np.array([-2259300,-1981500,-2259300,-1981500])# upper left, upper right, lower left, lower right
y_crn=np.array([1236000, 1236000, 958500, 958500])# upper left, upper right, lower left, lower right
x0, y0= m(0, -90)
#Basemap coordinates at the south pole
#note that (0,0) of the Basemap is in a corner of the map,
#while other data sets use the south pole.
#This is easy to take into account:
lon_crn, lat_crn = m(x0-x_crn, y0-y_crn, inverse=True)
print 'lon_crn: '+str(lon_crn)
print 'lat_crn: '+str(lat_crn)
Which returns:
lon_crn: [-61.31816102 -58.04532791 -67.01108782 -64.1858106 ]
lat_crn: [-67.23548626 -69.3099076 -68.28071626 -70.47651326]
As you can see the longitudes agree to the given precision with those from the metafile, but the latitudes are to low.
I can approximate the latitudes by:
lat_crn=(lat_crn+90.)*1.0275-90.
But this is really not satisfying.
This is how the image is located if using the X/Y corner coordinates from the metafile (in red the Basemap drawcoastlines()):
and this is how it looks like using the corner lat/lon:
In this case I can simply use the lat/lon coordinates, but as mentioned before there are datasets (like this) which is provided by X/Y coordinates only, which makes it very important to rely on the Basemap projection. I know that there are other modules to re-project the data as a potential workaround, but it should work without other modules and a re-projection could introduce errors itself.
As this problem appears with different data sets I like to believe that it is a bug in the Basemap module, but I might also make the same mistake again and again or have wrong expectations.

I did some experimentation and it seems like changing lat_ts has no effect with projection='spstere'. In fact, it seems as if the projection latitude is implicitly assumed to be lat_ts=-90. regardless of what value you assign.
I had more success using projection='stere' instead, so that you would construct the Basemap in your example as follows:
m=Basemap(width=5400000., height=5400000., projection='stere',
ellps='WGS84', lon_0=180., lat_0=-90., lat_ts=-71.)
You may prefer to set the latitude and longitude of the corners instead of the width and height of the plot for your application.

Find the intersection between two geographical data points

I have two pairs of lat/lon (expressed in decimal degrees) along with their radius (expressed in meters). What I am trying to achieve is to find if an intersect between these two points exits (of course, it is obvious that this doesn't hold here but the plan is to try this algorithm in many other data points). In order to check this I am using Shapely's intersects() function. My question however is how should I deal with the different units? Should I make some sort of transformation \ projection first (same units for both lat\lon and radius)?
48.180759,11.518950,19.0
47.180759,10.518950,10.0
EDIT:
I found this library here (https://pypi.python.org/pypi/utm) which seems helpfull. However, I am not 100% sure if I apply it correctly. Any ideas?
X = utm.from_latlon(38.636782, 21.414384)
A = geometry.Point(X[0], X[1]).buffer(30.777)
Y = utm.from_latlon(38.636800, 21.414488)
B = geometry.Point(Y[0], Y[1]).buffer(23.417)
A.intersects(B)
SOLUTION:
So, I finally managed to solve my problem. Here are two different implementations that both solve the same problem:
X = from_latlon(48.180759, 11.518950)
Y = from_latlon(47.180759, 10.518950)
print(latlonbuffer(48.180759, 11.518950, 19.0).intersects(latlonbuffer(47.180759, 10.518950, 19.0)))
print(latlonbuffer(48.180759, 11.518950, 100000.0).intersects(latlonbuffer(47.180759, 10.518950, 100000.0)))
X = from_latlon(48.180759, 11.518950)
Y = from_latlon(47.180759, 10.518950)
print(geometry.Point(X[0], X[1]).buffer(19.0).intersects(geometry.Point(Y[0], Y[1]).buffer(19.0)))
print(geometry.Point(X[0], X[1]).buffer(100000.0).intersects(geometry.Point(Y[0], Y[1]).buffer(100000.0)))

Shapely only uses the Cartesian coordinate system, so in order to make sense of metric distances, you would need to either:
project the coordinates into a local projection system that uses distance units in metres, such as a UTM zone.
buffer a point from (0,0), and use a dynamic azimuthal equidistant projection centered on the lat/lon point to project to geographic coords.
Here's how to do #2, using shapely.ops.transform and pyproj
import pyproj
from shapely.geometry import Point
from shapely.ops import transform
from functools import partial
WGS84 = pyproj.Proj(init='epsg:4326')
def latlonbuffer(lat, lon, radius_m):
proj4str = '+proj=aeqd +lat_0=%s +lon_0=%s +x_0=0 +y_0=0' % (lat, lon)
AEQD = pyproj.Proj(proj4str)
project = partial(pyproj.transform, AEQD, WGS84)
return transform(project, Point(0, 0).buffer(radius_m))
A = latlonbuffer(48.180759, 11.518950, 19.0)
B = latlonbuffer(47.180759, 10.518950, 10.0)
print(A.intersects(B)) # False
Your two buffered points don't intersect. But these do:
A = latlonbuffer(48.180759, 11.518950, 100000.0)
B = latlonbuffer(47.180759, 10.518950, 100000.0)
print(A.intersects(B)) # True
As shown by plotting the lon/lat coords (which distorts the circles):