使用data.table上的geoosphere distm函数计算距离

本文介绍了使用data.table上的geoosphere distm函数计算距离的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我创建了一个包含6列的data.table。我的data.table有一个列包含两个位置：位置1和位置2.我试图使用distm函数来计算每行上的位置之间的距离，创建第7列。 geosphere包中的distM包需要针对每个lat / long组合计算两个不同的向量。我的代码下面不工作，所以我想弄清楚如何提供向量到函数。 LOC_1_ID LOC1_LAT_CORD LOC1_LONG_CORD LOC_2_ID LOC2_LAT_CORD LOC2_LONG_CORD 1 35.68440 -80.48090 70624 34.86752 -82.46632 6 35.49770 -80.62870 70624 34.86752 -82.46632 10 35.66042 -80.50053 70624 34.86752 -82.46632 假设res包含data.table，以下代码不起作用。 res [，DISTANCE：= distm（c（LOC1_LAT_CORD，LOC1_LONG_CORD），c（LOC2_LAT_CORD，LOC2_LONG_CORD），fun = distHaversine）* 0.000621371] 如果我要拉出每个向量的函数工作正常。 loc1 loc2 distm（loc1，loc2 ，fun = distHaversine） 真的，我的问题是如何应用函数来选择数据中的列。该函数需要向量作为参数。解决方案 distm fucntion生成一组点的距离矩阵。如果你只是比较每一行上的点，并添加一列，你确定这是你想要的功能吗？ 听起来你想要 distHaversine 或 distGeo library（data.table） library（geosphere） dt< - read.table（text =LOC_1_ID LOC1_LAT_CORD LOC1_LONG_CORD LOC_2_ID LOC2_LAT_CORD LOC2_LONG_CORD 1 35.68440 -80.48090 70624 34.86752 -82.46632 6 35.49770 -80.62870 70624 34.86752 -82.46632 10 35.66042 -80.50053 70624 34.86752 -82.46632，header = T） setDT（dt） dt [ ，distance_hav：= distHaversine（matrix（c（LOC1_LONG_CORD，LOC1_LAT_CORD），ncol = 2）， matrix（c（LOC2_LONG_CORD，LOC2_LAT_CORD），ncol = 2））] ＃LOC_1_ID LOC1_LAT_CORD LOC1_LONG_CORD LOC_2_ID LOC2_LAT_CORD LOC2_LONG_CORD distance_hav ＃1：1 35.68440 -80.48090 70624 34.86752 -82.46632 202046.3 ＃2：6 35.49770 -80.62870 70624 34.86752 -82.46632 181310.0 ＃3：10 35.66042 -80.50053 70624 34.86752 -82.46632 199282.1 I've created a data.table in that has 6 columns. My data.table has a columns compairing two locations: Location 1 and Location 2. I'm trying to use the distm function to calculate the distance between the locations on each row, creating a 7th column. The distm package in the geosphere package requires two different vectors for each lat/long combo to be calculated against. My code below does not work, so I'm trying to figure out how to provide vectors to the function.LOC_1_ID LOC1_LAT_CORD LOC1_LONG_CORD LOC_2_ID LOC2_LAT_CORD LOC2_LONG_CORD 1 35.68440 -80.48090 70624 34.86752 -82.46632 6 35.49770 -80.62870 70624 34.86752 -82.4663210 35.66042 -80.50053 70624 34.86752 -82.46632Assuming res holds the data.table the below code does not work. res[,DISTANCE := distm(c(LOC1_LAT_CORD, LOC1_LONG_CORD),c(LOC2_LAT_CORD, LOC2_LONG_CORD), fun=distHaversine)*0.000621371]If I were to pull out each vector the function works fine.loc1 <- res[LOC1_ID == 1,.(LOC1_LAT_CORD, LOC1_LONG_CORD)]loc2 <- res[LOC2_ID==70624,.(LOC2_LAT_CORD, LOC2_LONG_CORD)]distm(loc1, loc2, fun=distHaversine)Really, my question is how to apply functions to select columns within a data.table when that function requires vectors as parameters. 解决方案 The distm fucntion generates a Distance matrix of a set of points. Are you sure this is the function you want if you're just comparing the points on each row, and adding one column?It sounds like you actually want either distHaversine or distGeolibrary(data.table)library(geosphere)dt <- read.table(text = "LOC_1_ID LOC1_LAT_CORD LOC1_LONG_CORD LOC_2_ID LOC2_LAT_CORD LOC2_LONG_CORD1 35.68440 -80.48090 70624 34.86752 -82.466326 35.49770 -80.62870 70624 34.86752 -82.4663210 35.66042 -80.50053 70624 34.86752 -82.46632", header = T)setDT(dt)dt[, distance_hav := distHaversine(matrix(c(LOC1_LONG_CORD, LOC1_LAT_CORD), ncol = 2), matrix(c(LOC2_LONG_CORD, LOC2_LAT_CORD), ncol = 2))]# LOC_1_ID LOC1_LAT_CORD LOC1_LONG_CORD LOC_2_ID LOC2_LAT_CORD LOC2_LONG_CORD distance_hav# 1: 1 35.68440 -80.48090 70624 34.86752 -82.46632 202046.3# 2: 6 35.49770 -80.62870 70624 34.86752 -82.46632 181310.0# 3: 10 35.66042 -80.50053 70624 34.86752 -82.46632 199282.1 这篇关于使用data.table上的geoosphere distm函数计算距离的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！