Round values of coordinates within a "shapely.geometry.polygon.Polygon" in Python

Question

I met a problem of rounding coordinates in a shapely.geometry.polygon.Polygon in Python.

Here is an example to reproduce the problem:

# Load packages
import numpy as np
import pandas as pd
from shapely.geometry import Polygon,Point
Suppose there are a list of lat and a list of lon and their resolutions
lat = [28.45,28.55,28.65,28.75,28.85]
lon = [76.85,76.95,77.05,77.15,77.25,77.35]
res_lat = 0.1
res_lon = 0.1
List all the combinations of the given lat and lon
So we have a dataframe representing grid centres
def expand_grid(input_list_of_lat,input_list_of_lon):
    """
    List all combinations of the given list of latitudes and the given list of longitudes.
    """
    points = [(A,B) for A in input_list_of_lat for B in input_list_of_lon]
    points = np.array(points)
    points_lat = points[:,0]
    points_lon = points[:,1]
    coordinates_of_full_grid = pd.DataFrame({'lat': points_lat,'lon': points_lon})
    return coordinates_of_full_grid
test_grid_df = expand_grid(lat,lon)
For each grid, create a "shapely.geometry.polygon.Polygon"
def get_grid_polygon(centre_lat,centre_lon,res_lat,res_lon):
    # Create a list of tuples (lon,lat) to store information of grid corners

    grid_corners = [Point(centre_lon-res_lon/2,centre_lat-res_lat/2), # bottom-left
                    Point(centre_lon+res_lon/2,centre_lat-res_lat/2), # bottom-right
                    Point(centre_lon+res_lon/2,centre_lat+res_lat/2), # top-right
                    Point(centre_lon-res_lon/2,centre_lat+res_lat/2)] # top-left
    # Convert the corners to a list of points covering the area
    point_list = [grid_corners[0],grid_corners[1],grid_corners[2],grid_corners[3],grid_corners[0]]
    # Create the polygon based on the above list of points
    polygon = Polygon([[p.x, p.y] for p in point_list])
    return polygon
Apply the above function to the sample data frame
test_grid_df['polygon'] = np.nan
for i in range(test_grid_df.shape[0]):
    test_grid_df['polygon'][i] = get_grid_polygon(centre_lat = test_grid_df['lat'][i],
                                                  centre_lon = test_grid_df['lon'][i],
                                                  res_lat = res_lat,
                                                  res_lon = res_lon)
Check the output
Then you will see that the precisions of coordinates within the polygons are a mess
How to round them while keeping the dtype as "shapely.geometry.polygon.Polygon"?
print(test_grid_df)
print(type(test_grid_df['polygon'][0]))

I have seen similar questions, but answers on those posts either do not work or do not solve my problem.

For example, I have seen that the code below can round the coordinates:

test_grid_df['polygon'][i] = shapely.wkt.dumps(test_grid_df['polygon'][i], rounding_precision=3)

But the dtype is also changed from shapely.geometry.polygon.Polygon to str. Then I can not use it to plot or do other stuff with the str, I need it to be still stored as shapely.geometry.polygon.Polygon after rounding the coordinates.

I have tried many ways, but no matter how I provide the coordinates, and or how I generate the "polygon", as long as I generate the "polygon", the precision of some coordinates will be ruined, and I will get values like "76.000000000001" or "73.9999999999".

https://gis.stackexchange.com/questions/321518/rounding-coordinates-to-five-decimals-in-geopandas/321531#321531 — BERA, Jan 18 '23 at 09:25
@BERA, Thanks, I managed to round values in geopandas.geoseries.GeoSeries using your solution, but I still can not round values in Polygons, I will try again when I have time. — Jeremy, Jan 18 '23 at 10:20

score 0 · Answer 1 · answered May 04 '23 at 20:48

The standard way to round coordinates to a certain precision in shapely is using shapely.set_precision, like this:

# Load packages
import numpy as np
import pandas as pd
import shapely
from shapely.geometry import Polygon, Point
Suppose there are a list of lat and a list of lon and their resolutions
lat = [28.45, 28.55, 28.65, 28.75, 28.85]
lon = [76.85, 76.95, 77.05, 77.15, 77.25, 77.35]
res_lat = 0.1
res_lon = 0.1
List all the combinations of the given lat and lon
So we have a dataframe representing grid centres
def expand_grid(input_list_of_lat, input_list_of_lon):
    """
    List all combinations of the given list of latitudes and the given list of longitudes.
    """
    points = [(A, B) for A in input_list_of_lat for B in input_list_of_lon]
    points = np.array(points)
    points_lat = points[:, 0]
    points_lon = points[:, 1]
    coordinates_of_full_grid = pd.DataFrame({"lat": points_lat, "lon": points_lon})
    return coordinates_of_full_grid
test_grid_df = expand_grid(lat, lon)
For each grid, create a "shapely.geometry.polygon.Polygon"
def get_grid_polygon(centre_lat, centre_lon, res_lat, res_lon):
    # Create a list of tuples (lon,lat) to store information of grid corners
    grid_corners = [
        Point(centre_lon - res_lon / 2, centre_lat - res_lat / 2),  # bottom-left
        Point(centre_lon + res_lon / 2, centre_lat - res_lat / 2),  # bottom-right
        Point(centre_lon + res_lon / 2, centre_lat + res_lat / 2),  # top-right
        Point(centre_lon - res_lon / 2, centre_lat + res_lat / 2),
    ]  # top-left
    # Convert the corners to a list of points covering the area
    point_list = [
        grid_corners[0],
        grid_corners[1],
        grid_corners[2],
        grid_corners[3],
        grid_corners[0],
    ]
    # Create the polygon based on the above list of points
    polygon = Polygon([[p.x, p.y] for p in point_list])
    return polygon
Apply the above function to the sample data frame
test_grid_df["polygon"] = np.nan
for i in range(test_grid_df.shape[0]):
    test_grid_df["polygon"][i] = get_grid_polygon(
        centre_lat=test_grid_df["lat"][i],
        centre_lon=test_grid_df["lon"][i],
        res_lat=res_lat,
        res_lon=res_lon,
    )
Check the output
Then you will see that the precisions of coordinates within the polygons are a mess
How to round them while keeping the dtype as "shapely.geometry.polygon.Polygon"?
print(test_grid_df)
print(type(test_grid_df["polygon"][0]))
Round coordinates to 3 decimals
test_grid_df.polygon = shapely.set_precision(test_grid_df.polygon, grid_size=0.001)
print(test_grid_df)
print(type(test_grid_df["polygon"][0]))

Result:

    lat    lon                                            polygon
0   28.45  76.85  POLYGON ((76.8 28.5, 76.9 28.5, 76.9 28.4, 76....
1   28.45  76.95  POLYGON ((76.9 28.5, 77 28.5, 77 28.4, 76.9 28...
2   28.45  77.05  POLYGON ((77 28.5, 77.1 28.5, 77.1 28.4, 77 28...
3   28.45  77.15  POLYGON ((77.1 28.5, 77.2 28.5, 77.2 28.4, 77....
4   28.45  77.25  POLYGON ((77.2 28.5, 77.3 28.5, 77.3 28.4, 77....
5   28.45  77.35  POLYGON ((77.3 28.5, 77.4 28.5, 77.4 28.4, 77....
6   28.55  76.85  POLYGON ((76.8 28.6, 76.9 28.6, 76.9 28.5, 76....
7   28.55  76.95  POLYGON ((76.9 28.6, 77 28.6, 77 28.5, 76.9 28...
8   28.55  77.05  POLYGON ((77 28.6, 77.1 28.6, 77.1 28.5, 77 28...
9   28.55  77.15  POLYGON ((77.1 28.6, 77.2 28.6, 77.2 28.5, 77....
10  28.55  77.25  POLYGON ((77.2 28.6, 77.3 28.6, 77.3 28.5, 77....

Round values of coordinates within a "shapely.geometry.polygon.Polygon" in Python

Suppose there are a list of lat and a list of lon and their resolutions

List all the combinations of the given lat and lon

So we have a dataframe representing grid centres

For each grid, create a "shapely.geometry.polygon.Polygon"

Apply the above function to the sample data frame

Check the output

Then you will see that the precisions of coordinates within the polygons are a mess

How to round them while keeping the dtype as "shapely.geometry.polygon.Polygon"?

1 Answers1