I have the following (simplified) Model:
class Zone(gismodels.Model):
name = gismodels.CharField()
poly = gismodels.PolygonField()
I want to create and save a polygon that represents a circle, based upon a given point and radius.
The only way I can figure out how to achieve this, is to call the postgis ST_Buffer function using raw SQL. I'm really hoping that there is another way.
Is it possible to access the GEOS buffer methods?
Yes, it is possible to use the geos buffer method:
>>> from django.contrib.gis import geos
>>> center = geos.Point(5, 5)
>>> radius = 2
>>> circle = center.buffer(radius)
>>> circle
<Polygon object at 0x1029d8370>
The radius here is in the same units as the coordinates of the points. This will work for some coordinate systems like UTM, but not as well for others.
Also, while this is appropriate for constructing a circular geometry, the PostGIS documentation notes that for doing radius searches ST_DWithin is more efficient.
I spent a ridiculous amount of time trying to get this working. Since this is the number one google search result, here's what worked for me:
radius_km = radius*1.609 # convert miles to km
point = target.geolocation # a django PointField using SRID 4326
# re-project point to a flat coordinate system
# so we can use meters instead of degrees below,
# AND get an actual circle instead of oval
point.transform(6347)
poly = point.buffer(radius_km*1000) # get a circular polygon from radius
poly.transform(4326)# re-project the resulting polygon back
Bonus: If you're doing this so you can get a circle on a google static map, grab polyline:
import polyline
import ast
geo = ast.literal_eval(poly.geojson) # turn the text into a dict
points = geo['coordinates'][0]
pl = polyline.encode(points, geojson=True, precision=5) # make a polyline out of the polygon for google
map_url += '&path=color:0x00000000%7Cfillcolor:0x0000AA33%7Cweight:1%7Cenc:' + pl
Related
I have a similar question to this one. Using geodjango, I want to draw a circle on a map with a certain radius in km. However, the suggested solution
a) does not use km but instead degrees, and
b) becomes an oval further north or south.
Here is what I do:
from django.contrib.gis import geos
lat = 49.17
lng = -123.96
center = geos.Point(lng, lat)
radius = 0.01
circle = center.buffer(radius)
# And I then use folium to show a map on-screen:
map = folium.Map(
location=[lat,lng],
zoom_start=14,
attr="Mapbox"
)
folium.GeoJson(
circle.geojson,
name="geojson",
).add_to(map)
The result is this:
How can I
a) draw a circle that is always 3 km in radius, independent from the position on the globe, and
b) ensure this is a circle and not an oval at all latitudes?
Here is the Code
from django.contrib.gis import geos
import folium
lat = 49.17
lng = -123.96
center = geos.Point(x=lng, y=lat, srid=4326)
center.transform(3857) # Transform Projection to Web Mercator
radius = 3000 # now you can use meters
circle = center.buffer(radius)
circle.transform(4326) # Transform back to WGS84 to create geojson
# And I then use folium to show a map on-screen:
map = folium.Map(
location=[lat,lng],
zoom_start=14,
attr="Mapbox"
)
geojson = folium.GeoJson(
circle.geojson,
name="geojson",
)
geojson.add_to(map)
Explanation
This problem occurs due to Map Projections.
Lat/Long Coordinates are represented by the Map Projection WGS84. The Values are in degrees.
The map you see in folium has another map projection (Web Mercator). It tries to represent the world as a plane, which produces distortions to the north and south. The coordinate values are in meters.
On a globe your created circle would look completely round, but because folium uses another projection it gets distorted.
It is also important to know that every projection is represented by a number (EPSG Code). With this epsg codes, you can transform your coordinates from one projection into another.
Web Mercator -> EPSG 3857
WGS84 -> EPSG 4326
With my Code you now get a round circle in folium for Web Mercator, but be aware that it would look oval and distorted, when looking at it on a globe.
This is just a very easy explanation. You might have a look at Map Projections to better understand the problem.
This guide gives a good overview:
Map Projections
try this
folium.Circle(
radius=3000,
location=[lat,lng],
popup="Whatever name",
color="#3186cc",
fill=True,
fill_color="#3186cc",
).add_to(m)
I've saved user's coordinates in the User model. Post model has latitude, longitude and radius field. Only the users in that vicinity(of Post) will be able to see that post. I don't know how to use filter() here so I used the following approach:
post=Posts.objects.all()
for a in post:
distance= geopy.distance.geodesic((lat1,lng1), (a.latitude, a.longitude)).km
print(distance)
if distance < a.radius:
p.append(a)
else:
continue
Here, lat1 and lng1 are the coordinates of current User. Suggest if there is any better way as this seems very inefficient.
Depending on your requirements, you could use a square instead of a circle. Pre-calculate the x-max, x-min, y-max and y-min boundaries for your square and then do a simple User.filter(lat__gt=lat_min, user.lng__gt=lng_min, user.lat__lt=lat_max ... lookup in the database.
In a past project, I used this:
def get_latlng_bounderies(lat, lng, distance):
"""
Return min/max lat/lng values for a distance around a latlng.
:lat:, :lng: the center of the area.
:distance: in km, the "radius" around the center point.
:returns: Two corner points of a square that countains the circle,
lat_min, lng_min, lat_max, lng_max.
"""
gc = great_circle(kilometers=distance)
p0 = gc.destination((lat, lng), 0)
p90 = gc.destination((lat, lng), 90)
p180 = gc.destination((lat, lng), 180)
p270 = gc.destination((lat, lng), 270)
ret = p180[0], p270[1], p0[0], p90[1]
return ret
Its not a circle, so its not exact around the "corners" of the square, but its much faster, because its a simple float comparision in the database.
I have the following sample of handwriting taken with three different writing instruments:
Looking at the writing, I can tell that there is a distinct difference between the first two and the last one. My goal is to determine an approximation of the stroke thickness for each letter, allowing me to group them based on being thin or thick.
So far, I have tried looking into stroke width transform, but I have struggled to translate it to my example.
I am able to preprocess the image such that I am just left with just the contours of the test in question. For example, here is thick from the last line:
I suggest detecting contours with cv::findContours as you are doing and then compare bounding rectangle area and contour area. The thicker writing the greater coefficent (contourArea/boundingRectArea) will be.
This approach will help you. This will calcuate the stroke width.
from skimage.feature import peak_local_max
from skimage import img_as_float
def adaptive_thresholding(image):
output_image = cv2.adaptiveThreshold(image,255,cv2.ADAPTIVE_THRESH_GAUSSIAN_C,cv2.THRESH_BINARY,21,2)
return output_image
def stroke_width(image):
dist = cv2.distanceTransform(cv2.subtract(255,image), cv2.DIST_L2, 5)
im = img_as_float(dist)
coordinates = peak_local_max(im, min_distance=15)
pixel_strength = []
for element in coordinates:
x = element[0]
y = element[1]
pixel_strength.append(np.asarray(dist)[x,y])
mean_pixel_strength = np.asarray(pixel_strength).mean()
return mean_pixel_strength
image = cv2.imread('Small3.JPG', 0)
process_image = adaptive_thresholding(image)
stroke_width(process_image)
A python implementation for this might go something like this, using Stroke Width Transform implementation of SWTloc.
Full Disclosure: I am the author of this library.
EDIT : Post v2.0.0
Transforming The Image
import swtloc as swt
imgpath = 'images/path_to_image.jpeg'
swtl = swt.SWTLocalizer(image_paths=imgpath)
swtImgObj = swtl.swtimages[0]
# Perform SWT Transformation with numba engine
swt_mat = swtImgObj.transformImage(auto_canny_sigma=1.0, gaussian_blurr=False,
minimum_stroke_width=3, maximum_stroke_width=50,
maximum_angle_deviation=np.pi/3)
Localize Letters
localized_letters = swtImgObj.localizeLetters()
Plot Histogram of Each Letters Strokes Widths
import seaborn as sns
import matplotlib.pyplot as plt
all_sws = []
for letter_label, letter in localized_letters.items():
all_sws.append(letter.stroke_widths_mean)
sns.displot(all_sws, bins=31)
From the distribution plot, it can be inferred that there might be three fontsize of the text available in the image - [3, 15, 27]
I have a SVG drawing (from a Building Map) and I want to rotate the complete document 90 degrees clockwise. Now, the drawing orientation is portrait, the idea is to have a landscape orientation.
Besides of this, I would like to scale the complete document (so including all elements).
For now, I could not find the possibilties for doing this on the web. So that is why I am asking overhere. My questions are:
Is it possible?
If yes, in what way can this be done? And who wants to help with this issue.
I managed to rotate an SVG figure with svgutils module. It can be installed with pip.
>>> import svgutils
>>> svg = svgutils.transform.fromfile('camera.svg')
>>> originalSVG = svgutils.compose.SVG('camera.svg')
>>> originalSVG.rotate(90)
>>> originalSVG.move(svg.height, 10)
<svgutils.compose.SVG object at 0x7f11dc78fb10>
>>> figure = svgutils.compose.Figure(svg.height, svg.width, originalSVG)
>>> figure.save('svgNew.svg')
Note that width and height attributes must be specified in original svg file in svg tag
Reference I used
Actually, this method didn't do anything with elements except wrapping them all with g tag with transform attribute. But it seems that with this module you can access each and every element in SVG tree and do whatever you want with them.
Scaling an SVG is also easy:
>>> originalSVG.scale(2)
<svgutils.compose.SVG object at 0x7f11dc78fb10>
>>> figure = svgutils.compose.Figure(float(svg.height) * 2, float(svg.width) * 2, originalSVG)
>>> figure.save('svgNew.svg')
In my model I have a polygon field defined via
polygon = models.PolygonField(srid=4326, geography=True, null=True, blank=True)
When I want to determine the area of the polygon, I call
area_square_degrees = object.polygon.area
But how can I convert the result in square degrees into m2 with GeoDjango?
This answer does not work, since area does not have a method sq_m. Is there any built-in conversion?
You need to transform your data to the correct spatial reference system.
area_square_local_units = object.polygon.transform(srid, clone=False).area
In the UK you might use the British National Grid SRID of 27700 which uses meters.
area_square_meters = object.polygon.transform(27700, clone=False).area
You may or may not want to clone the geometry depending on whether or not you need to do anything else with it in its untransformed state.
Docs are here https://docs.djangoproject.com/en/1.8/ref/contrib/gis/geos/
I have struggled a lot with this, since i could'nt find a clean solution. The trick is you have to use the postgis capabilities (and and thus its only working with postgis..):
from django.contrib.gis.db.models.functions import Area
loc_obj = Location.objects.annotate(area_=Area("poly")).get(pk=??)
# put the primary key of the object
print(loc_obj.area_) # distance object, result should be in meters, but you can change to the unit you want, e.g .mi for miles etc..
The models.py:
class Location(models.Model):
poly = gis_models.PolygonField(srid=4326, geography=True)
It's i think the best way to do it if you have to deal with geographic coordinates instead of projections. It does handle the curve calculation of the earth, and the result is precise even with big distance/area
I needed an application to get the area of poligons around the globe and if I used an invalid country/region projection I got the error OGRException: OGR failure
I ended using an OpenLayers implementation
using the 4326 projection (is the default projection) to avoid concerning about every country/region specific projection.
Here is my code:
import math
def getCoordsM2(coordinates):
d2r = 0.017453292519943295 # Degrees to radiant
area = 0.0
for coord in range(0, len(coordinates)):
point_1 = coordinates[coord]
point_2 = coordinates[(coord + 1) % len(coordinates)]
area += ((point_2[0] - point_1[0]) * d2r) *\
(2 + math.sin(point_1[1] * d2r) + math.sin(point_2[1] * d2r))
area = area * 6378137.0 * 6378137.0 / 2.0
return math.fabs(area)
def getGeometryM2(geometry):
area = 0.0
if geometry.num_coords > 2:
# Outer ring
area += getCoordsM2(geometry.coords[0])
# Inner rings
for counter, coordinates in enumerate(geometry.coords):
if counter > 0:
area -= getCoordsM2(coordinates)
return area
Simply pass your geometry to getGeometryM2 function and you are done!
I use this function in my GeoDjango model as a property.
Hope it helps!
If Its earths surface area that you are talking about, 1 square degree has 12,365.1613 square km. So multiple your square degree and multiply by 10^6 to convert to meters.