Question

我有一个数据文件（.csv），其中每个观察点是333个区域之一。每个区都有一个ID，如1101,1102，.......其次，我有另一个数据文件（.csv），其中每个观察是112,975个城镇之一，包括人口数据。城镇数据有一个district_ID字段。每个区有大约300个城镇。因此，有一个区域district_ID == 1101和大约300个城镇district_ID == 1101。

我想在我的分区数据集中创建一个区级人口变量。这意味着将多个城镇观测与每个单一区域观测相匹配，并对城镇级人口进行求和。

谢谢！

Answer 1

data.table解决方案：

#some example data
set.seed(42)
districts <- data.frame(district_ID=1:10,whatever=rnorm(10))
towns <- data.frame(town=1:100,district_ID=rep(1:10,each=10),
                    population=rpois(100,sample(c(1e3,1e4,1e5))))

library(data.table)
districts <- data.table(districts,key="district_ID")
towns <- data.table(towns,key="district_ID")

#calculate district population
temp <- towns[,list(district_pop=sum(population)),by=district_ID]
#merge result with districts data.table
districts <- merge(districts,temp)

#    district_ID    whatever district_pop
# 1:           1  1.37095845       434886
# 2:           2 -0.56469817       334084
# 3:           3  0.36312841       342241
# 4:           4  0.63286260       433224
# 5:           5  0.40426832       334039
# 6:           6 -0.10612452       342810
# 7:           7  1.51152200       433362
# 8:           8 -0.09465904       333810
# 9:           9  2.01842371       342035
# 10:          10 -0.06271410       432302

Answer 2

编辑：使用更大的数据集进行基准测试。

使用tapply函数计算每个区的人口：

districtdata$population<-
  tapply(towndata$population,towndata$district_ID,sum)[districts$district_ID]

一些基准测试，只是为了好玩：

fn1<-function(districts,towns) 
{
  districts$population<-
       tapply(towns$population,towns$district_ID,sum)[districts$district_ID]

  districts
}
fn2<-function(districts,towns) #Roland's data.table approach:
{ 
  districts <- data.table(districts,key="district_ID")
  towns <- data.table(towns,key="district_ID")
  temp<-towns[,list(district_pop=sum(population)),by=district_ID]
  merge(districts,temp)
}



set.seed(42)
districts <- data.frame(district_ID=1:300,whatever=rnorm(300))
towns <- data.frame(town=1:100000,district_ID=rep(1:300,each=300),
                    population=rpois(300000,sample(c(1e3,1e4,1e5))))

microbenchmark(fn1(districts,towns),fn2(districts,towns))
Unit: milliseconds
                  expr       min        lq    median        uq       max neval
 fn1(districts, towns) 215.29266 231.47103 243.72353 265.28280 355.43895   100
 fn2(districts, towns)  20.03636  27.51046  36.11116  58.56448  88.70766   100

Answer 3

怎么样：

aggregate(population ~ district_ID, towns, sum)

（基于Roland的合成数据）

将许多字段匹配（和求和）到R中的一个

3 个答案: