Question

我有一个100X100相关矩阵，邮政编码作为列名和行名。我还有一个数据框，其中包含所有zipcdes的纬度和经度，以及一个根据lat和long计算距离的函数。

以下是相关矩阵的片段

            08846       48186       90621      92602       92701       92702       92703      92705      92706      92712
08846  1.00000000 -0.18704668  0.17631080 -0.0195590 -0.08640209 -0.09109788 -0.04251868 -0.1586506 -0.0778115 -0.0572327
48186 -0.18704668  1.00000000 -0.09365048  0.1616530  0.20468051  0.17682056  0.18009911  0.1417840  0.1958971  0.1938676
90621  0.17631080 -0.09365048  1.00000000  0.5880756  0.75200501  0.74694849  0.76071605  0.6593806  0.7640519  0.7657806
92602 -0.01955900  0.16165299  0.58807565  1.0000000  0.88187818  0.88947447  0.89310793  0.9615530  0.8926566  0.8926482
92701 -0.08640209  0.20468051  0.75200501  0.8818782  1.00000000  0.99314798  0.98011569  0.9294281  0.9827633  0.9886139
92702 -0.09109788  0.17682056  0.74694849  0.8894745  0.99314798  1.00000000  0.98791442  0.9470895  0.9853157  0.9933086
92703 -0.04251868  0.18009911  0.76071605  0.8931079  0.98011569  0.98791442  1.00000000  0.9321385  0.9938496  0.9981231
92705 -0.15865058  0.14178399  0.65938061  0.9615530  0.92942815  0.94708954  0.93213849  1.0000000  0.9268797  0.9357917
92706 -0.07781150  0.19589706  0.76405191  0.8926566  0.98276329  0.98531570  0.99384961  0.9268797  1.0000000  0.9948550
92712 -0.05723270  0.19386757  0.76578065  0.8926482  0.98861389  0.99330864  0.99812312  0.9357917  0.9948550  1.0000000

这是邮政编码表的片段

    zip       city state latitude longitude
1 00210 Portsmouth    NH  43.0059  -71.0132
2 00211 Portsmouth    NH  43.0059  -71.0132
3 00212 Portsmouth    NH  43.0059  -71.0132
4 00213 Portsmouth    NH  43.0059  -71.0132
5 00214 Portsmouth    NH  43.0059  -71.0132
6 00215 Portsmouth    NH  43.0059  -71.0132

这是函数taht计算距离bwteen lat和long。

Calc_Dist <- function (long1, lat1, long2, lat2)
{
  rad <- pi/180
  a1 <- lat1 * rad
  a2 <- long1 * rad
  b1 <- lat2 * rad
  b2 <- long2 * rad
  dlon <- b2 - a2
  dlat <- b1 - a1
  a <- (sin(dlat/2))^2 + cos(a1) * cos(b1) * (sin(dlon/2))^2
  c <- 2 * atan2(sqrt(a), sqrt(1 - a))
  R <- 6378.145
  d <- R * c
  return(d)
}

我的目标是将相关矩阵子集化为仅包含相距500英里以上的邮政编码（现在距离计算输出以千米为单位但可以很容易地改变，现在并不重要）。使用更大的相关矩阵（~10000 x 10000），我可能不得不做得越好。有什么建议吗？

提前致谢，本

Answer 1

您必须使用该距离功能至关重要吗？我认为dist应该更有效率。

#Making your zip.table a data.table helps us with speed
library(reshape)
library(data.table)
setDT(zip.table) 

#Calculate distance matrix and put into table form
setorder(zip.dist,zip)
zip.dist <- dist(zip.table[,.(longitude=abs(longitude),latitude)])
zip.dist <- as.matrix(zip.dist)
zip.dist <- melt(zip.dist)[melt(upper.tri(zip.dist))$value,]
setDT(zip.dist)
setnames(zip.dist,c("zip1", "zip2", "distance"))

#Do a very similar procedure with your correlation matrix
#It is important that you sorted your zip.table by zip before applying `cor`
zip.corr <- as.matrix(zip.corr)
zip.corr <- melt(zip.corr)[melt(upper.tri(zip.corr))$value,]
setDT(zip.corr)
setnames(zip.corr,c("zip1", "zip2", "cor"))

#Subset zip.dist to only include zip codes more than 500 miles apart
zip.dist <- zip.dist[distance*69 > 500] #69 mile ~ 1 degreen lat/lon

#Merge together
setkey(zip.dist,zip1,zip2)
setkey(zip.corr,zip1,zip2)
result.table <- zip.dist[zip.corr, nomatch=0]

由于这些地方都非常接近，我不认为你使用欧几里德距离会损失很多。特别是因为它是一个大县内的一个纬度/经度。

子集矩阵基于数据帧的内容

1 个答案: