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我正在尝试计算 1981 年至 1986 年间每年（1982...1985 年）和每个年龄的人口数量的调查间估计。我的数据集中的复杂因素是我有 52 个省和大约 600 个 ZONA91OK（区），并且每个 PROVINCE 有不同数量的区。

为了获得包含每个缺失年份、国籍、省和每个地区 (ZONA91OK) 信息的向量，我想应用的公式如下：

前任。1982 年

1982 年，年龄组 10-14 的值：x(10,1982)=[(x(10,1981)-x(15,1986))/5]-x(10,1982)
x(15,1982)= [(x(15,1981)-x(20,1986))/5]-x(15,1982)
x(20,1982)=[(x(20,1981)-x(25,1986))/ 5]-x(20,1982)
...
x(55,1982)=[(x(55,1981)-x(55,1986))/5]-x(55,1982) -异常-

非常感谢您对这个问题的任何帮助！

这是可重现的样本（整个数据库的一个子集，因为它非常大） 

mydata<-structure(list(YEAR = c(1981, 1981, 1981, 1981, 1981, 1981, 1981, 
1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 
1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 1981, 
1981, 1981, 1981, 1981, 1981, 1981, 1981, 1986, 1986, 1986, 1986, 
1986, 1986, 1986, 1986, 1986, 1986, 1986, 1986, 1986, 1986, 1986, 
1986, 1986, 1986, 1986, 1986, 1986, 1986, 1986, 1986, 1986, 1986, 
1986, 1986, 1986, 1986, 1986, 1986, 1986, 1986, 1986, 1986), 
PROVINCE = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), 
ZONA91OK = c(101, 101, 101, 101, 101, 101, 101, 101, 101, 
102, 102, 102, 102, 102, 102, 102, 102, 102, 1036, 1036, 
1036, 1036, 1036, 1036, 1036, 1036, 1036, 1059, 1059, 1059, 
1059, 1059, 1059, 1059, 1059, 1059, 101, 101, 101, 101, 101, 
101, 101, 101, 101, 102, 102, 102, 102, 102, 102, 102, 102, 
102, 1036, 1036, 1036, 1036, 1036, 1036, 1036, 1036, 1036, 
1059, 1059, 1059, 1059, 1059, 1059, 1059, 1059, 1059), AGE5 = c(10, 
15, 20, 25, 30, 35, 40, 45, 50, 10, 15, 20, 25, 30, 35, 40, 
45, 50, 10, 15, 20, 25, 30, 35, 40, 45, 50, 10, 15, 20, 25, 
30, 35, 40, 45, 50, 10, 15, 20, 25, 30, 35, 40, 45, 50, 10, 
15, 20, 25, 30, 35, 40, 45, 50, 10, 15, 20, 25, 30, 35, 40, 
45, 50, 10, 15, 20, 25, 30, 35, 40, 45, 50), NATIONALITY = structure(c(9L, 
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L), .Label = c("España", 
"UE-15 y PD", "Resto Europa", "Magreb", "África Sub-sahariana", 
"Latinoamérica", "Asia", "Resto del Mundo", "No computable"
), class = "factor"), FREQUENCY = c(993.8141, 994.907, 894.0322, 
845.8348, 659.6786, 577.2588, 540.6329, 684.9917, 673.6348, 
910.511, 1068.9258, 936.9949, 763.547, 643.4404, 572.72, 
536.6591, 665.975, 768.6866, 967.694100000002, 980.340100000001, 
811.637500000001, 746.058500000001, 769.820600000001, 722.398000000001, 
730.371600000001, 690.084600000001, 501.9178, 8243.04149999997, 
7785.02419999994, 7505.78429999991, 7464.74579999992, 7663.47079999997, 
6700.90559999997, 5203.31959999996, 5582.66059999997, 4837.30459999996, 
869.1754, 982.7461, 945.5031, 904.2817, 813.7127, 663.955, 
577.2896, 544.1257, 689.9815, 780.3824, 879.7538, 1025.5724, 
882.475, 716.0049, 627.3571, 579.4372, 525.4546, 679.9666, 
1035.6544, 952.521599999999, 962.537599999999, 832.3296, 
733.1696, 726.1568, 704.1248, 700.1136, 667.0624, 9023.05139999993, 
8285.31719999994, 8080.95919999994, 8175.28479999993, 7786.53429999994, 
7796.56439999994, 6842.11639999996, 5239.83509999998, 5616.95939999997
)), .Names = c("YEAR", "PROVINCE", "ZONA91OK", "AGE5", "NATIONALITY", 
"FREQUENCY"), row.names = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 
10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L, 22L, 
23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, 
36L, 8173L, 8174L, 8175L, 8176L, 8177L, 8178L, 8179L, 8180L, 
8181L, 8182L, 8183L, 8184L, 8185L, 8186L, 8187L, 8188L, 8189L, 
8190L, 8191L, 8192L, 8193L, 8194L, 8195L, 8196L, 8197L, 8198L, 
8199L, 8200L, 8201L, 8202L, 8203L, 8204L, 8205L, 8206L, 8207L, 
8208L), class = "data.frame")
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2 回答 2

2

可能最容易?ddply从plyr包中使用。这应该让你开始......

require(plyr)
df <- ddply(mydata, c("PROVINCE", "NATIONALITY", "ZONA91OK"), function(x){
  x1981 <- x[x[,"YEAR"]==1981,]
  x1986 <- x[x[,"YEAR"]==1986,]
  #1982
  x1982 <- x1981
  x1982[,"YEAR"] <- 1982
  # This seems strange. Should probably be "+" instead of "-"
  x1982[,"FREQUENCY"] <- (x1981[,"FREQUENCY"]-x1986[,"FREQUENCY"])/5 - x1981[,"FREQUENCY"]
  # Add additional years here...
  rbind(x1981, x1982, x1986)
})
# perhaps reorder
df[order(df[,"YEAR"]),]
于 2013-10-03T12:49:45.417 回答
1

如果你能保证没有丢失数据——也就是说，每一年都包含完全相同的行数，以及相同的省份、年龄、国籍和地区的组合，那么这里有一个简单的解决方案：

df<-mydata[with(mydata,order(YEAR,NATIONALITY,PROVINCE,ZONA91OK,AGE5)),]
splitdata<-split(df,df$YEAR)
for(i in 1982:1985){
  chi<-as.character(i)
  splitdata[[chi]]<-splitdata$`1981`
  splitdata[[chi]]$YEAR<-i
  splitdata[[chi]]$FREQUENCY<-splitdata$`1986`$FREQUENCY*(i-1981)/5+
    splitdata$`1981`$FREQUENCY*(1986-i)/5
}

newdata<-do.call(rbind,splitdata)
newdata

编辑时：ddply@shadow 发布的方法是一种更“正确”的 R 方法，使用ddply它可以更轻松地处理丢失的数据。但是，如果您没有丢失数据，那么 split 方法似乎更有效：

Unit: milliseconds
        expr      min       lq    median        uq      max neval
 splitmethod 2.573737 2.638159  2.676954  2.735601 235.2619   100
 ddplymethod 9.812680 9.989192 10.148241 17.128667 243.7309   100
于 2013-10-03T13:00:04.160 回答