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我有以下格式的数据（行数：~ 100 万）

head(dt)
   pickup_longitude pickup_latitude dropoff_longitude dropoff_latitude
1:        -74.00394        40.74289         -73.99337         40.73425
2:        -73.97386        40.75219         -73.95870         40.77253
3:        -73.95441        40.76442         -73.97078         40.75835
4:        -73.96234        40.76722         -73.97551         40.75687
5:        -74.00466        40.70743         -73.99937         40.72152
6:        -73.99557        40.71602         -73.99997         40.74332

library(geosphere)
dt = data.table(pickup_longitude = c(-74.00394, -73.97386, -73.95441, -73.96234, -74.00466, -73.99557),
            pickup_latitude = c(40.74289, 40.75219, 40.76442, 40.76722, 40.70743, 40.71602), 
            dropoff_longitude = c(-73.99337, -73.95870, -73.97078, -73.97551, -73.99937, -73.99997),
            dropoff_latitude = c(40.73425, 40.77253, 40.75835, 40.75687, 40.72152, 40.74332))
dt[, distance := apply(dt, 1, function(t) distm(x = c(t[1], t[2]), y = c(t[3], t[4])))]

我已经使用了上面的代码，因为包中apply的函数没有向量化。但是，上面的代码需要很多时间。distmgeosphereapply

我也试过：

dt[, distance := distm(x = c(pickup_longitude, pickup_latitude), y = c(dropoff_longitude, dropoff_latitude)), by = 1:nrow(dt)]

还有什么可以更好更快地计算距离的方法？

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1 回答 1

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我试过这个。

dt[, distance := distHaversine(matrix(c(pickup_longitude, pickup_latitude), ncol = 2),
                        matrix(c(dropoff_longitude, dropoff_latitude), ncol = 2))]

这工作得很好。

于 2017-05-19T10:13:55.097 回答