Using zip function into distance calculating program?

Question

I try to calculate distance between points from gpx file using harvesine formula. I stuck in moment when I should use 'lats' and 'lons' for formula which I found in internet. I want to use that two list as pairs, so zip function would be ok. But how i should do that?

import xml.etree.ElementTree as ET
import math
f = ('St_Louis_Zoo_sample.gpx')
p = ET.parse(f)
root = p.getroot()
lats=[]
lons=[]

for mainElement in root.findall('{http://www.topografix.com/GPX/1/1}wpt'):    
        y = float(mainElement.attrib['lat'])
        x = float(mainElement.attrib['lon'])
        lats.append(y)
        lons.append(x)

def distance(origin, destination):
    lat1, lon1 = origin
    lat2, lon2 = destination
    radius = 6371 # km

    dlat = math.radians(lat2-lat1)
    dlon = math.radians(lon2-lon1)
    a = math.sin(dlat/2) * math.sin(dlat/2) + math.cos(math.radians(lat1)) \
        * math.cos(math.radians(lat2)) * math.sin(dlon/2) * math.sin(dlon/2)
    c = 2 * math.atan2(math.sqrt(a), math.sqrt(1-a))
    d = radius * c

    return d

Please edit your question to provide more background information on where you got this code and what you are trying to do. Titles should be topic summaries, written last, and should not duplicate the only content in the qyestion. — Vince, Feb 18 '17 at 15:19
The question is a continuation of How to properly get coordinates from gpx file in python — gene, Feb 18 '17 at 15:49

score 4 · Accepted Answer · edited Apr 13 '17 at 12:34

Look at How to calculate distances in a point sequence?, you need to iterate by pair of points to compute the distances

  points =  zip(lats,lons)
  # or directly with a list comprehension
  points = [(float(mainElement.attrib['lat']), float(mainElement.attrib['lon'])) for mainElement in tree.findall('{http://www.topografix.com/GPX/1/1}wpt')]
  print points
  [(38.63473, -90.29408), (38.63368, -90.28679), (38.63408, -90.29323), (38.63533, -90.29019), (38.63677, -90.28976), (38.63496, -90.28948), (38.63421, -90.29458), (38.63633, -90.29083), (38.63395, -90.28715), (38.63347, -90.28769)]

  def pair(list):
      '''Iterate over pairs in a list -> pair of points '''
      for i in range(1, len(list)):
             yield list[i-1], list[i]
  for first, second in pair(points):
        print first, second, distance(first, second)
  (38.63473, -90.29408) (38.63368, -90.28679) 0.643881089995
  (38.63368, -90.28679) (38.63408, -90.29323) 0.561144390742
  (38.63408, -90.29323) (38.63533, -90.29019) 0.29839986488
  (38.63533, -90.29019) (38.63677, -90.28976) 0.164418862328
  (38.63677, -90.28976) (38.63496, -90.28948) 0.202726888151
  (38.63496, -90.28948) (38.63421, -90.29458) 0.450763798782
  (38.63421, -90.29458) (38.63633, -90.29083) 0.402073386107
  (38.63633, -90.29083) (38.63395, -90.28715) 0.414976855973
  (38.63395, -90.28715) (38.63347, -90.28769) 0.0710547282793

You can also use the itertools standard module

  import itertools
  for first, second in itertools.izip(points, points[1:]):
        print first, second, distance(first, second)
  (38.63473, -90.29408) (38.63368, -90.28679) 0.643881089995
  (38.63368, -90.28679) (38.63408, -90.29323) 0.561144390742
  (38.63408, -90.29323) (38.63533, -90.29019) 0.29839986488
  ....

Using zip function into distance calculating program?

1 Answers1