CODEWITHSUNDEEP
pythonpandasgeopy
ScalaBoy
ScalaBoy

Reputation: 3392

How to check that a point is inside the given radius?

I have the following code that takes very long time to execute. The pandas DataFrames df and df_plants are very small (less than 1Mb). I wonder if there is any way to optimise this code:

import pandas as pd
import geopy.distance
import re

def is_inside_radius(latitude, longitude, df_plants, radius):
    if (latitude != None and longitude != None):
        lat = float(re.sub("[a-zA-Z]", "", str(latitude)))
        lon = float(re.sub("[a-zA-Z]", "", str(longitude)))
        for index, row in df_plants.iterrows():
            coords_1 = (lat, lon)
            coords_2 = (row["latitude"], row["longitude"])
            dist = geopy.distance.distance(coords_1, coords_2).km
            if dist <= radius:
                return 1
    return 0

df["inside"] = df.apply(lambda row: is_inside_radius(row["latitude"],row["longitude"],df_plants,10), axis=1)

I use regex to process latitude and longitude in df because the values contain some errors (characters) which should be deleted.

The function is_inside_radius verifies if row[latitude] and row[longitude] are inside the radius of 10 km from any of the points in df_plants.

Upvotes: 4

Views: 3490

Answers (2)

yoonghm
yoonghm

Reputation: 4625

Can you try this?

import pandas as pd
from geopy import distance
import re

def is_inside_radius(latitude, longitude, df_plants, radius):
  if (latitude != None and longitude != None):
    lat = float(re.sub("[a-zA-Z]", "", str(latitude)))
    lon = float(re.sub("[a-zA-Z]", "", str(longitude)))
    coords_1 = (lat, lon)

    for row in df_plants.itertuples():
      coords_2 = (row["latitude"], row["longitude"])
      if distance.distance(coords_1, coords_2).km <= radius:
        return 1
  return 0

df["inside"] = df.map(
                    lambda row: is_inside_radius(
                      row["latitude"],
                      row["longitude"],
                      df_plants,
                      10),
                    axis=1)

From https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.iterrows.html#pandas-dataframe-iterrows, pandas.DataFrame.itertuples() returns namedtuples of the values which is generally faster than pandas.DataFrame.iterrows(), and preserve dtypes across returned rows.

Upvotes: 4

Dominique
Dominique

Reputation: 17493

I've encountered such a problem before, and I see one simple optimisation: try to avoid the floating point calculation as much a possible, which you can do as follows:
Imagine:
You have a circle, defined by Mx and My (center coordinates) and R (radius).
You have a point, defined by is coordinates X and Y.

If your point (X,Y) is not even within the square, defined by (Mx, My) and size 2*R, then it will also not be within the circle, defined by (Mx, My) and radius R.
In pseudo-code:

function is_inside(X,Y,Mx,My,R):
  if (abs(Mx-X) >= R) OR (abs(My-Y) >= R)
  then return false
  else:
    // and only here you perform the floating point calculation

Upvotes: 2

Related Questions