How can I do an approximate search for a "latitude, longitude" coordinate value in a "file.txt" list in Python?
37.04508, -95.57605
37.04278, -95.58895 37.04369, -95.58592 37.04369, -95.58582 37.04376, -95.58557 37.04376, -95.58546 37.04415, -95.58429 37.0443, -95.5839 37.04446, -95.58346 37.04461, -95.58305 37.04502, -95.58204 37.04516, -95.58184 37.04572, -95.58139 37.0459, -95.58127 37.04565, -95.58073 37.04546, -95.58033 37.04516, -95.57948 37.04508, -95.57914 37.04494, -95.57842 37.04483, -95.5771 37.0448, -95.57674 37.04474, -95.57606 37.04467, -95.57534 37.04462, -95.57474 37.04458, -95.57396 37.04454, -95.57274 37.04452, -95.57233 37.04453, -95.5722 37.0445, -95.57164 37.04448, -95.57122 37.04444, -95.57054 37.04432, -95.56845 37.04432, -95.56834 37.04424, -95.5668 37.04416, -95.56545 37.044, -95.56251 37.04396, -95.5618
37.04508, -95.57914
Line 17
Any help will be greatly appreciated! Thank you.
What you could do is compute the distance between each coordinate then check if that is the closest:
from math import radians, cos, sin, asin, sqrt
# Taken from https://stackoverflow.com/questions/4913349/haversine-formula-in-python-bearing-and-distance-between-two-gps-points
def compute_distance(lon1, lat1, lon2, lat2):
lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
# haversine formula
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
c = 2 * asin(sqrt(a))
r = 6371 # Radius of earth in kilometers. Use 3956 for miles
return c * r
def search_closest(to_find, coordinates):
closest_coord = (0, 0)
closest_distance = compute_distance(coordinates[0][1], coordinates[0][0], to_find[1], to_find[0])
for coord in coordinates:
current_distance = compute_distance(coord[1], coord[0], to_find[1], to_find[0])
if closest_distance > current_distance:
closest_coord = coord
closest_distance = current_distance
return closest_coord
if __name__ == "__main__":
# Placeholder for files.txt content
coordinates = [
(37.04278, -95.58895),
(37.04369, -95.58592),
(37.04369, -95.58582),
(37.04376, -95.58557),
(37.04376, -95.58546),
(37.04415, -95.58429),
(37.0443, -95.5839),
(37.04446, -95.58346),
(37.04461, -95.58305),
(37.04502, -95.58204),
(37.04516, -95.58184),
(37.04572, -95.58139),
(37.0459, -95.58127),
(37.04565, -95.58073),
(37.04546, -95.58033),
(37.04516, -95.57948),
(37.04508, -95.57914),
(37.04494, -95.57842),
(37.04483, -95.5771),
(37.0448, -95.57674),
(37.04474, -95.57606),
(37.04467, -95.57534),
(37.04462, -95.57474),
(37.04458, -95.57396),
(37.04454, -95.57274),
(37.04452, -95.57233),
(37.04453, -95.5722),
(37.0445, -95.57164),
(37.04448, -95.57122),
(37.04444, -95.57054),
(37.04432, -95.56845),
(37.04432, -95.56834),
(37.04424, -95.5668),
(37.04416, -95.56545),
(37.044, -95.56251),
(37.04396, -95.5618)
]
to_find = (37.04508, -95.57605)
closest = search_closest(to_find, coordinates)
print(closest)
Edit: Used Haversine to compute distance
Used a different approach to fixatd but this works aswell opening the txt file you requested.
import sys, os
import math
coords = open('coords.txt').read().split("\n")
x=[]
y=[]
for r in coords:
row = r.split(", ")
x.append(row[0])
y.append(row[1])
lowest = None
currentval = None
store = None
value = (37.04508, -95.57605)
for i in range(len(x)):
currentval = (math.sqrt((((float(x[i]) - value[0])**2) + ((float(y[i]) - value[1])**2))) * 111000)
if i == 0:
lowest = currentval
if currentval < lowest:
lowest = currentval
store = (float(x[i]), float(y[i]))
else:
continue
print (store)
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