问题
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$Latitude+1), fun = distHaversine)
Error in .pointsToMatrix(x) : Wrong length for a vector, should be 2
Although I'm very unsure about the syntax and if this is the right way to fill the arguments of the function. I'm very new to R so I hope I can get some help.
Any input is appreciated!
回答1:
I think you're almost there already. Assuming you want store the distance between the previous recording (n
) and the current recording (n+1
) at n+1
, you can use:
library(geosphere)
date <- c("2018-06-18 03:01.00","2018-06-18 03:06.00","2018-06-18 03:08.00","2018-06-18 03:12.00","2018-06-18 03:16.00","2018-06-18 03:20.00")
latitude <- c(-2.434901,-2.434598,-2.434726,-2.434816,-2.434613,-2.434511)
longitude <- c(34.85359,34.85387,34.85382,34.85371,34.85372,34.85376)
datagps <- data.frame(date,lat,lon)
datagps$length <- distm(x=datagps[,2:3], fun = distHaversine)[,1]
That gives the first result 0, the rest as the distance between the consecutive points
回答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
回答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
来源:https://stackoverflow.com/questions/55217286/calculating-distance-between-two-gps-locations-in-a-data-frame-using-distm-in