问题描述
我将与上述问题相对应的2个数据帧的简化集放在此处：
ss <- structure(list(country = structure(c(1L,2L,3L,4L,5L,6L,7L,8L,9L,10L,1L,10L),.Label = c("a","b","c","d","e","f","g","h","k","v"),class = "factor"),year = c(1961L,1962L,1963L,1961L,1963L),x = c(19L,23L,24L,16L,28L,29L,20L,14L,21L,30L,12L,17L,25L,26L,13L,11L,22L,27L,15L,18L),y = c(23L,19L,18L,30L),z = c(22L,19L)),class = "data.frame",row.names = c(NA,-30L))
和
zz <- structure(list(country = structure(c(1L,1L),.Label = "w",year = 1961:1963,x = c(2L,3L),y = c(3L,2L),z = 1:3),-3L))
数据框ss代表10个国家/地区3年的数据。并且，数据框zz代表相应年份的世界数据。
是否有任何方法可以使用诸如ss(for each each group as country)/zz之类的条件，从而可以提取每个国家的价值作为与世界数据的比率。我的意思是，ss的前两列也应保留。
我们可以避免使用只能添加到更多编码行的dplyr和tidverse重塑数据。
谢谢。
解决方法
使用match。
cbind(ss[1:2],ss[-(1:2)] / zz[match(ss$year,zz$year),-(1:2)])
# country year x y z
# 1 a 1961 9.5000000 7.666667 22.000000
# 2 b 1962 4.0000000 20.000000 2.000000
# 3 c 1963 1.0000000 14.000000 7.666667
# 4 d 1961 11.5000000 2.333333 16.000000
# 5 e 1962 24.0000000 4.000000 14.500000
# 6 f 1963 5.3333333 12.500000 4.666667
# 7 g 1961 14.0000000 1.666667 11.000000
# 8 h 1962 9.0000000 8.000000 6.500000
# 9 k 1963 9.6666667 5.000000 9.000000
# 10 v 1961 10.0000000 4.333333 26.000000
# 11 a 1962 14.0000000 9.000000 2.500000
# 12 b 1963 7.0000000 0.500000 4.000000
# 13 c 1961 15.0000000 7.000000 2.000000
# 14 d 1962 1.0000000 11.000000 4.500000
# 15 e 1963 4.0000000 13.000000 3.333333
# 16 f 1961 8.5000000 5.333333 25.000000
# 17 g 1962 25.0000000 27.000000 3.500000
# 18 h 1963 8.6666667 1.000000 7.000000
# 19 k 1961 6.5000000 9.666667 6.000000
# 20 v 1962 8.0000000 24.000000 10.000000
# 21 a 1963 0.6666667 1.500000 1.000000
# 22 b 1961 3.5000000 5.000000 30.000000
# 23 c 1962 10.0000000 6.000000 9.000000
# 24 d 1963 3.6666667 9.500000 2.666667
# 25 e 1961 3.0000000 4.666667 1.000000
# 26 f 1962 22.0000000 22.000000 12.000000
# 27 g 1963 9.0000000 6.000000 5.666667
# 28 h 1961 2.5000000 6.000000 15.000000
# 29 k 1962 15.0000000 17.000000 14.000000
# 30 v 1963 6.0000000 15.000000 6.333333
这也可以使用data.table包中的一行来完成：
as.data.table(ss)[zz,.(country,year,x = x/i.x,y = y/i.y,z = z/i.z),on = .(year)]
# country year x y z
# 1: a 1961 9.5000000 7.666667 22.000000
# 2: d 1961 11.5000000 2.333333 16.000000
# 3: g 1961 14.0000000 1.666667 11.000000
# 4: v 1961 10.0000000 4.333333 26.000000
# 5: c 1961 15.0000000 7.000000 2.000000
# 6: f 1961 8.5000000 5.333333 25.000000
# 7: k 1961 6.5000000 9.666667 6.000000
# 8: b 1961 3.5000000 5.000000 30.000000
# 9: e 1961 3.0000000 4.666667 1.000000
# 10: h 1961 2.5000000 6.000000 15.000000
# 11: b 1962 4.0000000 20.000000 2.000000
# 12: e 1962 24.0000000 4.000000 14.500000
# 13: h 1962 9.0000000 8.000000 6.500000
# 14: a 1962 14.0000000 9.000000 2.500000
# 15: d 1962 1.0000000 11.000000 4.500000
# 16: g 1962 25.0000000 27.000000 3.500000
# 17: v 1962 8.0000000 24.000000 10.000000
# 18: c 1962 10.0000000 6.000000 9.000000
# 19: f 1962 22.0000000 22.000000 12.000000
# 20: k 1962 15.0000000 17.000000 14.000000
# 21: c 1963 1.0000000 14.000000 7.666667
# 22: f 1963 5.3333333 12.500000 4.666667
# 23: k 1963 9.6666667 5.000000 9.000000
# 24: b 1963 7.0000000 0.500000 4.000000
# 25: e 1963 4.0000000 13.000000 3.333333
# 26: h 1963 8.6666667 1.000000 7.000000
# 27: a 1963 0.6666667 1.500000 1.000000
# 28: d 1963 3.6666667 9.500000 2.666667
# 29: g 1963 9.0000000 6.000000 5.666667
# 30: v 1963 6.0000000 15.000000 6.333333
# country year x y z