Convert LineString / MultiLineString geometries to lat lon - python

I am using this Mapillary endpoint: https://tiles.mapillary.com/maps/vtp/mly1_public/2/{zoom_level}/{x}/{y}?access_token={} and getting such responses back (see photo). Also, here is the Mapillary documentation.
It is not quite clear to me what the nested coordinate lists in the response represent. By the looks of it, I initially thought it may have to do with pixel coordinates. But judging by the context (the API documentation) and the endpoint I am using, I would say that is not the case. Also, I am not sure if the json response you see in the picture is valid geojson. Some online formatters did not accept it as valid.
I would like to find the bounding box of the "sequence". For context, that would be the minimal-area rectangle defined by two lat, lon positions that fully encompasses the geometry of the so-called "sequence"; and a "sequence" is basically a series of photos taken during a vehicle/on-foot trip, together with the metadata associated with the photos (metadata is available using another endpoint, but that is just for context).
My question is: is it possbile to turn the coordinates you see in the pictures into (lat,lon)? Having those, it would be easy for me to find the bounding box of the sequence. And if so, how? Also, please notice that some of the nested lists are of type LineString while others are MultiLineString (which I read about the difference here: help.arcgis.com, hope this helps)
Minimal reproducible code snippet:
import json
import requests
import mercantile
import mapbox_vector_tile as mvt
ACCESS_TOKEN = 'XXX' # can be provided from here: https://www.mapillary.com/dashboard/developers
z6_tiles = list(mercantile.tiles( #us_west_coast_bbox
west=-125.066423,
south=42.042594,
east=-119.837770,
north=49.148042,
zooms=6
))
# pprint(z6_tiles)
vector_tiles_url = 'https://tiles.mapillary.com/maps/vtp/mly1_public/2/{}/{}/{}?access_token={}'
for tile in z6_tiles:
res = requests.get(vector_tiles_url.format(tile.z,tile.x,tile.y,ACCESS_TOKEN))
res_json = mvt.decode(res.content)
with open('idea.json','w+') as f:
json.dump(res_json, f, indent=4)

I think this get_normalized_coordinates is the solution I was looking for. Please take this with a grain of salt, as I did not fully test it yet. Will try to and then I will update my answer. Also, please be cautious, because for tiles closer to either the South or the North Pole, the Z14_TILE_DMD_WIDTH constant will not be the one you see, but something more like: 0.0018958715374282065.
Z14_TILE_DMD_WIDTH = 0.02197265625
Z14_TILE_DMD_HEIGHT = 0.018241950298914844
def get_normalized_coordinates(bbox: mercantile.LngLatBbox,
target_lat: int,
target_lon: int,
extent: int=4096): # 4096 is Mapillary's default
"""
Returns lon,lat tuple representing real position on world map of a map feature.
"""
min_lon, min_lat, _, _ = bbox
return min_lon + target_lon / extent * Z14_TILE_DMD_WIDTH,
min_lat + target_lat / extent * Z14_TILE_DMD_HEIGHT
And if you are wondering how I came with the constants that you see, I simply iterated over the list of tiles that I am interested in and checked to make sure they all have the same width/height size (this might have not been the case, keeping in mind what I mentioned above about tiles closer to one of the poles - I think this is called "distortion", not sure). Also, for context: these tiles I iterated over are within this bbox: (-125.024414, 31.128199, -108.896484, 49.152970) (min_lon, min_lat, max_lon, max_lat; US west coast) which I believe is also why all the tiles have the same width/height sizes.
set_test = set()
for tile in relevant_tiles_set:
curr_bbox = mercantile.bounds(list_relevant_tiles_set[i])
dm_width_diff: float = curr_bbox.east - curr_bbox.west
dm_height_diff: float = curr_bbox.north - curr_bbox.south
set_test.add((dm_width_diff, dm_height_diff))
set_test
output:
{(0.02197265625, 0.018241950298914844}
UPDATE: forgot to mention that you actually do not need to compute those WIDTH, HEIGHT constants. You just replace those with (max_lon - min_lon) and (max_lat - min_lat) respectively. What I did with those constants was something for testing purposes only

Related

Convert UK grid coordinates (X,Y) to latitude and longitude

I have a list of position coordinates given in UK grid reference format (X,Y) and I would like to convert them into latitude and longitude.
I am using OSGridConverter (python library) but is not converting it correctly. For example, one input for one location is X = 517393.6563 and Y = 194035.5469.
from OSGridConverter import grid2latlong
l=grid2latlong('TG 517393.6563 194035.5469')
The above, gives me an error: OSGridConverter error: Invalid grid reference
Therefore, which is wrong, I try:
>>>l=grid2latlong('TG 517393 194035')
>>>(l.latitude,l.longitude)
(52.71367793063314, 1.7297510074170983)
Which ends with a location out in UK, which is not correct. Most probably is due data formats, but I am not sure about how to solve it.
You should probably use something like pyproj:
import pyproj
crs_british = pyproj.Proj(init='EPSG:27700')
crs_wgs84 = pyproj.Proj(init='EPSG:4326')
long, lat = pyproj.transform(crs_british, crs_wgs84, 517393.6563, 194035.5469)
print(lat, long)
In your case this will give: 51.63289090467179, -0.3052119183057834
Firstly, a definition:
OS Grid Grid references (OSGGRs) refer to grid references prepended with the two letter grid identifiers, as explained rather well here.
As you found OSGGRs arn't floating point numbers. See the link above.
Moving on, the numbers for OSGGRs have their origin at the south westerly point of the grid - which is very different from the origin of the OSGB BNG projection (coords are easting and northings). See above link for great explanation.
You seem to be taking the BGGRs numeric portion as an Easting/Northing, which is incorrect.
So, armed with this knowledge, we can use OSGridConverter to convert from OSGGRs to eastings and northings using a conversion to lat long as an intermediary. But note the authors comments on errors at the bottom of the PyPI package page
import OSGridConverter
cvt_wgs84 = OSGridConverter.grid2latlong('SN5117013189')
print(cvt_wgs84.latitude, cvt_wgs84.longitude)
51.79749210128498 -4.160451113839529
cvt_EN = OSGridConverter.latlong2grid(cvt_wgs84.latitude, cvt_wgs84.longitude)
print(cvt_EN.E,cvt_EN.N)
251116 213191
# Back to OSGGR, however, note the error of tens of meters.
str(cvt_EN)
'SN 51116 13191'
I did a quick 'by-eye' check by-map on the agreement between the OSGGR coords and the lat/long, and the error is in meters. May be acceptable to some.

Get an object's position in another object's coordinate system

Is there a way in MEL or Python in Maya to get one object's position in the coordinate system of another object? I have a camera in a scene that may be rotated in any direction and am trying to measure the distance in its local Z axis to the vertices of various objects in the scene. This obviously needs to be fast, since it will likely be run thousands of times across the scene.
In Maxscript the command would be something like
" in coordsys $camera "
but I have yet to find something like this in Maya. If there's no direct command to do this, does anyone have a way to calculate it using matrix math?
There is no one liner similar to the MXS idiom -- and no easy way to do it in mel. However in Python you can do this fairly easily.
First you need to get the matrix for the coordinate system you want as an MMatrix, which is part of the OpenMaya api. Then get the position you want to check as an MPoint, which is another api class. Here's the cheap way to get them (there are faster methods but they're much wordier):
from maya.api.OpenMaya import MVector, MMatrix, MPoint
import maya.cmds as cmds
def world_matrix(obj):
"""'
convenience method to get the world matrix of <obj> as a matrix object
"""
return MMatrix( cmds.xform(obj, q=True, matrix=True, ws=True))
def world_pos(obj):
"""'
convenience method to get the world position of <obj> as an MPoint
"""
return MPoint( cmds.xform(obj, q=True, t=True, ws=True))
Once you have the matrix and the point, the relative position is simply point times the inverse of the matrix:
relative_position = world_pos('pSphere1') * world_matrix('pCube1').inverse()
print relative_position
# (0.756766, -0.0498943, 3.38499, 1)
The result will be an MPoint, which has 4 numbers (x, y, z and w); the 4th will always be 1 so you can just ignore it, although the math needs it to account for scales and shears.
Use this MEL script to calculate the distance from camera1 to nurbsSphere1 primitive:
vector $p1 = `getAttr camera1.translate`;
vector $p2 = `getAttr nurbsSphere1.translate`;
vector $result = $p1 - $p2;
print (mag($result))
Printed result must be like this:
# MEL 40.1965
Or use this Python script to calculate the distance from camera1 to nurbsSphere1 primitive:
import maya.cmds as cmds
import math
distance = math.sqrt(pow((float)(cmds.getAttr("nurbsSphere1.tx") - cmds.getAttr("camera1.tx")),2) +
pow((float)(cmds.getAttr("nurbsSphere1.ty") - cmds.getAttr("camera1.ty")),2) +
pow((float)(cmds.getAttr("nurbsSphere1.tz") - cmds.getAttr("camera1.tz")),2) )
print(distance)
Printed result must be like this:
# Python 40.1964998512

Count number of points in multipolygon shapefile using Python

I have a polygon shapefile of the U.S. made up of individual states as their attribute values. In addition, I have arrays storing latitude and longitude values of point events that I am also interested in. Essentially, I would like to 'spatial join' the points and polygons (or perform a check to see which polygon [i.e., state] each point is in), then sum the number of points in each state to find out which state has the most number of 'events'.
I believe the pseudocode would be something like:
Read in US.shp
Read in lat/lon points of events
Loop through each state in the shapefile and find number of points in each state
print 'Here is a list of the number of points in each state: '
Any libraries or syntax would be greatly appreciated.
Based on what I can tell, the OGR library is what I need, but I am having trouble with the syntax:
dsPolygons = ogr.Open('US.shp')
polygonsLayer = dsPolygons.GetLayer()
#Iterating all the polygons
polygonFeature = polygonsLayer.GetNextFeature()
k=0
while polygonFeature:
k = k + 1
print "processing " + polygonFeature.GetField("STATE") + "-" + str(k) + " of " + str(polygonsLayer.GetFeatureCount())
geometry = polygonFeature.GetGeometryRef()
#Read in some points?
geomcol = ogr.Geometry(ogr.wkbGeometryCollection)
point = ogr.Geometry(ogr.wkbPoint)
point.AddPoint(-122.33,47.09)
point.AddPoint(-110.11,33.33)
#geomcol.AddGeometry(point)
print point.ExportToWkt()
print point
numCounts=0.0
while pointFeature:
if pointFeature.GetGeometryRef().Within(geometry):
numCounts = numCounts + 1
pointFeature = pointsLayer.GetNextFeature()
polygonFeature = polygonsLayer.GetNextFeature()
#Loop through to see how many events in each state
I like the question. I doubt I can give you the best answer, and definitely can't help with OGR, but FWIW I'll tell you what I'm doing right now.
I use GeoPandas, a geospatial extension of pandas. I recommend it — it's high-level and does a lot, giving you everything in Shapely and fiona for free. It is in active development by twitter/#kajord and others.
Here's a version of my working code. It assumes you have everything in shapefiles, but it's easy to generate a geopandas.GeoDataFrame from a list.
import geopandas as gpd
# Read the data.
polygons = gpd.GeoDataFrame.from_file('polygons.shp')
points = gpd.GeoDataFrame.from_file('points.shp')
# Make a copy because I'm going to drop points as I
# assign them to polys, to speed up subsequent search.
pts = points.copy()
# We're going to keep a list of how many points we find.
pts_in_polys = []
# Loop over polygons with index i.
for i, poly in polygons.iterrows():
# Keep a list of points in this poly
pts_in_this_poly = []
# Now loop over all points with index j.
for j, pt in pts.iterrows():
if poly.geometry.contains(pt.geometry):
# Then it's a hit! Add it to the list,
# and drop it so we have less hunting.
pts_in_this_poly.append(pt.geometry)
pts = pts.drop([j])
# We could do all sorts, like grab a property of the
# points, but let's just append the number of them.
pts_in_polys.append(len(pts_in_this_poly))
# Add the number of points for each poly to the dataframe.
polygons['number of points'] = gpd.GeoSeries(pts_in_polys)
The developer tells me that spatial joins are 'new in the dev version', so if you feel like poking around in there, I'd love to hear how that goes! The main problem with my code is that it's slow.
import geopandas as gpd
# Read the data.
polygons = gpd.GeoDataFrame.from_file('polygons.shp')
points = gpd.GeoDataFrame.from_file('points.shp')
# Spatial Joins
pointsInPolygon = gpd.sjoin(points, polygons, how="inner", op='intersects')
# Add a field with 1 as a constant value
pointsInPolygon['const']=1
# Group according to the column by which you want to aggregate data
pointsInPolygon.groupby(['statename']).sum()
**The column ['const'] will give you the count number of points in your multipolygons.**
#If you want to see others columns as well, just type something like this :
pointsInPolygon = pointsInPolygon.groupby('statename').agg({'columnA':'first', 'columnB':'first', 'const':'sum'}).reset_index()
[1]: https://geopandas.org/docs/user_guide/mergingdata.html#spatial-joins
[2]: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.groupby.html

pyephem FixedObject() for given RA/Dec

I'm looking to determine the alt/az of (un-famous) stars at given RA/Dec at specific times from Mauna Kea. I'm trying to compute these parameters using pyephem, but the resulting alt/az don't agree with other sources. Here's the calculation for HAT-P-32 from Keck:
import ephem
telescope = ephem.Observer()
telescope.lat = '19.8210'
telescope.long = '-155.4683'
telescope.elevation = 4154
telescope.date = '2013/1/18 10:04:14'
star = ephem.FixedBody()
star._ra = ephem.degrees('02:04:10.278')
star._dec = ephem.degrees('+46:41:16.21')
star.compute(telescope)
print star.alt, star.az
which returns -28:43:54.0 73:22:55.3, though according to Stellarium, the proper alt/az should be: 62:26:03 349:15:13. What am I doing wrong?
EDIT: Corrected latitude and longitude, which were formerly reversed.
First, you've got long and latitude backwards; second, you need to provide the strings in hexadecimal form; and third, you need to provide the RA as hours, not degrees:
import ephem
telescope = ephem.Observer()
# Reversed longitude and latitude for Mauna Kea
telescope.lat = '19:49:28' # from Wikipedia
telescope.long = '-155:28:24'
telescope.elevation = 4154.
telescope.date = '2013/1/18 00:04:14'
star = ephem.FixedBody()
star._ra = ephem.hours('02:04:10.278') # in hours for RA
star._dec = ephem.degrees('+46:41:16.21')
star.compute(telescope)
This way, you get:
>>> print star.alt, star.az
29:11:57.2 46:43:19.6
PyEphem always uses UTC for time, so that programs operate the same and give the same output wherever they are run. You simply need to convert the date you are using to UTC, instead of using your local time zone, and the results agree fairly closely with Stellarium; use:
telescope.date = '2013/1/18 05:04:14'
The result is this alt/az:
62:27:19.0 349:26:19.4
To know where the small remaining difference comes from, I would have to look into how the two programs handle each step of their computation; but does this get you close enough?

Python Module for Distance between UK postcodes

I need to calculate distances between UK postcodes.
I don't want to use a web api.
Does a python module/ library exist for this?
Or do I have to put together something of my own using data from the OrdnanceSurvey?
Thanks,
1/You can use any rest geolocation api eg google maps, that would provide you accurate distance based on the pincodes.
2/ You can use any updated database which has post codes and latitude/longitude information, and use that information to calculate distance between the two points.
Helpful links:
i) http://blog.acmultimedia.co.uk/2008/03/uk-post-code-distance-calculator-using-phpmysql/
ii) Django - how can I find the distance between two locations?
This question is a bit old now but maybe worth pointing out the difference between polar coordinates (Long/Lat) used by Google and SatNavs and Eastings & Northings provided by the Ordnance Survey which shows how far a given postcode centroid is from a given datum point somewhere off to the South West of the Scilly Isles.
If you just want to work out distances it's a one-liner using Pythagoras. EG Stonehenge to House of Parliament
>>> # Coordinates of Stonehenge and Westminster in E/Northings
... s = (412183, 142346)
... p = (530268, 179545)
... d = ((s[0] - p[0]) ** 2 + (s[1] - p[1]) ** 2) ** 0.5
... print (d)
123805.62517914927
I was looking for a library to convert from polar to Cartesian coordinates which is relatively tricky even on a perfect sphere which the world most certainly isn't. I'm currently trying, and mostly failing, to get my head around this:
https://scipython.com/book/chapter-2-the-core-python-language-i/additional-problems/converting-between-an-os-grid-reference-and-longitudelatitude/
I don't know of a directly usable module, but you could use GRASS or QGIS, both of which support python scripting so the functionality can be used as python modules. You would still need to figure out how to do it manually in either of these tools though, but that's not really very difficult.
See below for simple code using the postcodes module for python along with the google maps distance matrix API.
from postcodes import PostCoder
import requests
import json
import pprint
pc = PostCoder()
origin_postcode = str.upper(raw_input("Enter origin PostCode:"))
dest_postcode = str.upper(raw_input("Enter destination PostCode:"))
origin = str(pc.get('%s' % origin_postcode)['geo']['lat'])+','+str(pc.get('%s' % origin_postcode)['geo']['lng'])
dest = str(pc.get('%s' % dest_postcode)['geo']['lat'])+','+str(pc.get('%s' % dest_postcode)['geo']['lng'])
data = requests.get('https://maps.googleapis.com/maps/api/distancematrix/json?units=imperial&origins=%s&destinations=%s&key=AIzaSyDfTip2PrdaRkF1muCLP8REAk3FsLjmBrU' % (origin, dest))
new = json.loads(data.text)
miles = str(new['rows'][0]['elements'][0]['distance']['text']).replace('mi','miles')
duration = str((new['rows'][0]['elements'][0]['duration']['text']))
print '\n\nThe distance from %s to %s is %s, taking approximately %s (driving).' % (origin_postcode, dest_postcode, miles, duration)

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