Calculating distance between two GPS locations in a data frame using distm () in R

Question

This question has been asked previously but never with the following arrangement of the data. Below is a sample of it:

> head(datagps)
   Date & Time [Local]  Latitude Longitude
1: 2018-06-18 03:01:00 -2.434901  34.85359
2: 2018-06-18 03:06:00 -2.434598  34.85387
3: 2018-06-18 03:08:00 -2.434726  34.85382
4: 2018-06-18 03:12:00 -2.434816  34.85371
5: 2018-06-18 03:16:00 -2.434613  34.85372
6: 2018-06-18 03:20:00 -2.434511  34.85376

As you can see, I've a Date & Time [Local] column where GPS positions are registered every 4 min on average. I would like to calculate the distance (in meters) between two consecutive recordings and store this measure in a new column Step. I've been trying to implement distm() to my data:

> datagps$Step<-distm(c(datagps$Longitude, datagps$Latitude), c(datagps$Longitude+1, datagps$Latitude+1), fun = distHaversine)
Error in .pointsToMatrix(x) : Wrong length for a vector, should be 2

Although I'm very unsure about the syntax and if this is the right way to fill the arguments of the function. I'm very new to R so I hope I can get some help.

Any input is appreciated!

Answer 1

I think you're almost there already. Assuming you want store the distance between the previous recording (n) and the current recording (n+1) at n+1, you can use:

library(geosphere)
date <- c("2018-06-18 03:01.00","2018-06-18 03:06.00","2018-06-18 03:08.00","2018-06-18 03:12.00","2018-06-18 03:16.00","2018-06-18 03:20.00")
latitude <- c(-2.434901,-2.434598,-2.434726,-2.434816,-2.434613,-2.434511)  
longitude <- c(34.85359,34.85387,34.85382,34.85371,34.85372,34.85376)
datagps <- data.frame(date,lat,lon)

datagps$length <- distm(x=datagps[,2:3], fun = distHaversine)[,1]

That gives the first result 0, the rest as the distance between the consecutive points

Answer 2

If you take a look at the function's documentation you'll see:

library(geosphere)
?distm

x longitude/latitude of point(s). Can be a vector of two numbers, a matrix of 2 columns (first one is longitude, second is latitude) or a SpatialPoints* object

y Same as x. If missing, y is the same as x

This means that you can use both a matrix or a vector.

One approach could be:

res <- distm(as.matrix(df1[,c("Longitude","Latitude")]), fun = distHaversine)

res

#         [,1]     [,2]     [,3]     [,4]     [,5]     [,6]
#[1,]  0.00000 45.90731 32.15371 16.36018 35.16947 47.35305
#[2,] 45.90731  0.00000 15.29559 30.09289 16.76621 15.60347
#[3,] 32.15371 15.29559  0.00000 15.81292 16.79079 24.84658
#[4,] 16.36018 30.09289 15.81292  0.00000 22.62521 34.40483
#[5,] 35.16947 16.76621 16.79079 22.62521  0.00000 12.19500
#[6,] 47.35305 15.60347 24.84658 34.40483 12.19500  0.00000

Answer 3

solution using sf-package

sample data

library(data.table)
dt1 <- data.table::fread( 'DateTime, Latitude, Longitude
2018-06-18 03:01:00, -2.434901,  34.85359
2018-06-18 03:06:00, -2.434598,  34.85387
2018-06-18 03:08:00, -2.434726,  34.85382
2018-06-18 03:12:00, -2.434816,  34.85371
2018-06-18 03:16:00, -2.434613,  34.85372
2018-06-18 03:20:00, -2.434511,  34.85376')

setDF(dt1)

code

library(sf)
#create spatial points object
dt1.sf <- st_as_sf( x= dt1, 
                    coords = c("Longitude", "Latitude"),
                    crs = "+proj=longlat +datum=WGS84")
#calculate distances
st_distance(dt1.sf)

output

# Units: [m]
#          [,1]     [,2]     [,3]     [,4]     [,5]     [,6]
# [1,]  0.00000 45.74224 32.07520 16.32379 34.97450 47.08749
# [2,] 45.74224  0.00000 15.20702 29.96245 16.76520 15.56348
# [3,] 32.07520 15.20702  0.00000 15.77068 16.72801 24.69270
# [4,] 16.32379 29.96245 15.77068  0.00000 22.47452 34.18116
# [5,] 34.97450 16.76520 16.72801 22.47452  0.00000 12.12446
# [6,] 47.08749 15.56348 24.69270 34.18116 12.12446  0.00000