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我有一个 ID 和坐标的数据框。我需要计算我所有 id 之间的地理距离，删除彼此相距太远的那些，然后继续我的分析。

我有 30k 个 id，这将生成一个 30k x 30k 矩阵。这是一个示例：

 latitude longitude        id
-23.52472 -46.47785 917_62346
-23.62010 -46.69345 244_42975
-23.61636 -46.48148 302_75289
-23.53826 -46.46756 917_96304
-23.58266 -46.54495 302_84126
-23.47005 -46.70921 233_97098
-23.49235 -46.49342 917_62953
-23.52226 -46.72710 244_42245
-23.64853 -46.72237 635_90928
-23.49640 -46.61215  244_2662

x2 = structure(list(latitude = c(-23.5247247, -23.6200954, -23.6163624, 
-23.5382557, -23.5826609, -23.4700519, -23.4923465, -23.5222581, 
-23.6485288, -23.4964047), longitude = c(-46.4778499, -46.6934512, 
-46.4814794, -46.4675563, -46.5449536, -46.7092093, -46.4934192, 
-46.7270957, -46.7223717, -46.6121477), id = c("917_62346", "244_42975", 
"302_75289", "917_96304", "302_84126", "233_97098", "917_62953", 
"244_42245", "635_90928", "244_2662")), .Names = c("latitude", 
"longitude", "id"), row.names = c(12041L, 18549L, 13641L, 28386L, 
9380L, 6064L, 12724L, 21671L, 18939L, 3396L), class = "data.frame")

首先，我尝试直接使用它，使用geosphere包：

library(geosphere)
library(data.table)
d.matrix <- distm(cbind(x2$longitude, x2$latitude))

由于内存问题，这不起作用，Error: cannot allocate vector of size 15.4 Gb. 我的第二次尝试是先生成所有的pairwise组合，然后与原始数据集合并得到lats和lons，然后计算距离，比如

dis.long <- expand.grid(x2$id, x2$id)
dis.long <- merge(dis.long, x2, by.x = "Var1", by.y = "id")
dis.long <- merge(dis.long, x2, by.x = "Var2", by.y = "id")
dis.long <- dis.long[ , dist_km2 := distGeo(matrix(c(longitude.x, latitude.x), ncol = 2), 
                                        matrix(c(longitude.y, latitude.y), ncol = 2))/1000]

但是，expand_grid 内存不足。这对我来说很奇怪，因为生成的矩阵将是 900mi 行乘以 2 列，而且我已经处理了更大的数据集（比如 200 mi x 50 矩阵）。

另一个观察结果是，我已经尝试使用诸如new_id = seq(1L,30000L,1L), 之类的 id 来查看整数是否可以解决它，但是当我尝试扩展时，我遇到了同样的内存问题。

我目前在这些配置下，除了 16gb Ram 桌面

> sessionInfo()
R version 3.3.3 (2017-03-06)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows >= 8 x64 (build 9200)

locale:
[1] LC_COLLATE=English_United States.1252  LC_CTYPE=English_United States.1252   
[3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C                          
[5] LC_TIME=English_United States.1252    

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] xlsx_0.5.7        xlsxjars_0.6.1    rJava_0.9-8       geosphere_1.5-5   sp_1.2-5          haven_1.0.0      
[7] stringr_1.2.0     data.table_1.10.4

谁能告诉我如何计算这些距离？为什么我不能生成这个特定的 expand.grid 同时能够构造更大的对象？

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1 回答 1

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您不需要比较all-vs-all，包括自我比较和方向比较（AB！= BA）；因此你应该尝试combn而不是expand.grid

您的数据

x2 = structure(list(latitude = c(-23.5247247, -23.6200954, -23.6163624, 
-23.5382557, -23.5826609, -23.4700519, -23.4923465, -23.5222581, 
-23.6485288, -23.4964047), longitude = c(-46.4778499, -46.6934512, 
-46.4814794, -46.4675563, -46.5449536, -46.7092093, -46.4934192, 
-46.7270957, -46.7223717, -46.6121477), id = c("917_62346", "244_42975", 
"302_75289", "917_96304", "302_84126", "233_97098", "917_62953", 
"244_42245", "635_90928", "244_2662")), .Names = c("latitude", 
"longitude", "id"), row.names = c(12041L, 18549L, 13641L, 28386L, 
9380L, 6064L, 12724L, 21671L, 18939L, 3396L), class = "data.frame")

展开.grid

OP <- function(df) {
            x3 = expand.grid(df$id, df$id)
            Y <- merge(x3, df, by.x = "Var1", by.y = "id")
            Y <- merge(Y, df, by.x = "Var2", by.y = "id")
            return(Y)
      }

与组合

CP <- function(df) {
            Did = as.data.frame(t(combn(df$id, 2)))
            Z <- merge(Did, df, by.x = "V1", by.y = "id")
            Z <- merge(Z, df, by.x = "V2", by.y = "id")
            return(Z)
      }

比较

dis.long <- OP(x2)
object.size(dis.long)
# 7320 bytes

new <- CP(x2)
object.size(new)
# 5016 bytes

更大的例子

num <- 5e2
bigx <- data.frame(latitude=rnorm(num)*-23, longitude=rnorm(num)*-46, id=1:num)

bigdl <- OP(bigx)
object.size(bigdl)
# 10001224 bytes

bignew <- CP(bigx)
object.size(bignew)
# 4991224 bytes

大约一半大小

于 2017-09-01T19:47:18.023 回答