Manhattan Distance for two geolocations

2019-03-21 08:42发布

Let's say I have two locations represented by latitude and longitude. Location 1 : 37.5613 , 126.978 Location 2 : 37.5776 , 126.973

How can I calculate the distance using Manhattan distance ?

Edit : I know the formula for calculating Manhattan distance like stated by Emd4600 on the answer which is |x1-x2| - |y1-y2| but I think it's for Cartesian. If it is can be applied that straight forward |37.5613-37.5776| + |126.978-126.973| what is the distance unit of the result ?

标签: distance
1条回答
等我变得足够好
2楼-- · 2019-03-21 09:14

Given a plane with p1 at (x1, y1) and p2 at (x2, y2), it is, the formula to calculate the Manhattan Distance is |x1 - x2| + |y1 - y2|. (that is, the difference between the latitudes and the longitudes). So, in your case, it would be:

|126.978 - 126.973| + |37.5613 - 37.5776| = 0.0213

EDIT: As you have said, that would give us the difference in latitude-longitude units. Basing on this webpage, this is what I think you must do to convert it to the metric system. I haven't tried it, so I don't know if it's correct:

First, we get the latitude difference:

Δφ = |Δ2 - Δ1|
Δφ = |37.5613 - 37.5776| = 0.0163

Now, the longitude difference:

Δλ = |λ2 - λ1|
Δλ = |126.978 - 126.973| = 0.005

Now, we will use the haversine formula. In the webpage it uses a = sin²(Δφ/2) + cos φ1 ⋅ cos φ2 ⋅ sin²(Δλ/2), but that would give us a straight-line distance. So to do it with Manhattan distance, we will do the latitude and longitude distances sepparatedly.

First, we get the latitude distance, as if longitude was 0 (that's why a big part of the formula got ommited):

a = sin²(Δφ/2)
c = 2 ⋅ atan2( √a, √(1−a) )
latitudeDistance = R ⋅ c // R is the Earth's radius, 6,371km

Now, the longitude distance, as if the latitude was 0:

a = sin²(Δλ/2)
c = 2 ⋅ atan2( √a, √(1−a) )
longitudeDistance = R ⋅ c // R is the Earth's radius, 6,371km

Finally, just add up |latitudeDistance| + |longitudeDistance|.

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