我的数据具有以下形式：
Individual Year X2011Int X2010Int X2009Int X2008Int X2007Int
1 2011 10.36703 18.07049 4789.338 51.3443 423.235
2 2010 10.36703 18.07049 4789.338 51.3443 423.235
3 2009 10.36703 18.07049 4789.338 51.3443 423.235
4 2008 10.36703 18.07049 4789.338 51.3443 423.235
其中X2009Int，X2008Int，X2007Int列表示在指定时间段内特定事件的严重性。我的目标是创建一个函数，我可以循环遍历列以创建以下内容：
Individual Year T T-1 T-2
1 2011 =X2011Int =X2010Int =X2009Int
2 2010 =X2010Int =X2009Int =X2008Int
3 2009 =X2009Int =X2008Int =X2007Int
4 2008 =X2008Int =X2007Int =X2006Int
为了进一步说明2009年个人3的情况。在$ T栏下我需要X2009Int中的事件强度值，$ T-1列=来自X2008Int的值，$ T-2 = X2007Int，$ T -3 = X2006Int等（年份范围从2012-1980）
关键点在我的循环中：
for(i in unique(d1$Year)){
print(i)
d1$T[d1$Year == i] <- TOY2[TOY2$Year == i,colnames(TOY2) == i] #placed the i value in the T column.
d1$T.1[d1$Year == (i-1)] <- TOY2[TOY2$Year == (i-1),colnames(TOY2) == (i-1)]
d1$T.2[d1$Year == i-2] <- TOY2[TOY2$Year == i-2,colnames(TOY2) == i-2]
d1$T.3[d1$Year == i-3] <- TOY2[TOY2$Year == i-3,colnames(TOY2) == i-3]
}
第一次迭代（$T）列工作正常但后来没有按预期执行（即，我的原始数据帧中的错误值被粘贴到列中）
第二个问题是我是否能够使用lapply执行相同的任务，如果是这样的话？
非常感谢任何帮助！
编辑：＆gt; dput(head(TOY2))
structure(list(Individual = 1201:1206, Wave = c(6L, 6L, 6L, 6L,
6L, 6L), Country = c(32L, 32L, 32L, 32L, 32L, 32L), Year = c(2011,
2011, 2011, 2011, 2011, 2011), `2000` = c(45.19665424, 45.19665424,
45.19665424, 45.19665424, 45.19665424, 45.19665424), `2001` = c(176.2932337,
176.2932337, 176.2932337, 176.2932337, 176.2932337, 176.2932337
), `2002` = c(9.601447666, 9.601447666, 9.601447666, 9.601447666,
9.601447666, 9.601447666), `2003` = c(259.2992188, 259.2992188,
259.2992188, 259.2992188, 259.2992188, 259.2992188), `2004` = c(4.357976722,
4.357976722, 4.357976722, 4.357976722, 4.357976722, 4.357976722
), `2005` = c(1.955436508, 1.955436508, 1.955436508, 1.955436508,
1.955436508, 1.955436508), `2006` = c(1.865651073, 1.865651073,
1.865651073, 1.865651073, 1.865651073, 1.865651073), `2007` = c(61.65472296,
61.65472296, 61.65472296, 61.65472296, 61.65472296, 61.65472296
), `2008` = c(34.62974414, 34.62974414, 34.62974414, 34.62974414,
34.62974414, 34.62974414), `2009` = c(32.96903414, 32.96903414,
32.96903414, 32.96903414, 32.96903414, 32.96903414), `2010` = c(6.761739867,
6.761739867, 6.761739867, 6.761739867, 6.761739867, 6.761739867
), `2011` = c(0, 0, 0, 0, 0, 0), `2012` = c(12.05299366, 12.05299366,
12.05299366, 12.05299366, 12.05299366, 12.05299366)), .Names = c("Individual",
"Wave", "Country", "Year", "2000", "2001", "2002", "2003", "2004",
"2005", "2006", "2007", "2008", "2009", "2010", "2011", "2012"
), row.names = 1201:1206, class = "data.frame")
预期数据框d1的第一行应如下所示：
Individual Wave Country Year T T.1 T.2 T.3
1201 6 32 2011 0 6.76174 32.96903 34.62974
T值对应原始$2011 df中的TOY2值。
T-1值对应$2010中的TOY2值
T-2值对应$2009中的TOY2值
等。
答案 0 :(得分：2)
如果我理解你的要求，下面的黑客攻击（使用.ts和dplyr）应该有效。我更改了tidyr的示例输入，以便为每个d1使用不同的值。我相信这表明你想要的更好（假设我正确地解释了你的问题）。
注意：这为OP尝试执行的操作提供了另一种方法，并假设不需要Year解决方案。
lapply可能有一种更简单的方法，但这似乎有效。
答案 1 :(得分：2)
考虑reshape使用aggregate使用融合（从长到长）。下面使用您的dput(head(TOY2))：
library(reshape2)
mdf <- melt(TOY2, id.vars=c("Individual", "Wave", "Country", "Year"))
mdf$variable <- as.numeric(as.character(mdf$variable)) # CONVERT TO NUMERIC
# CREATE T COLUMNS
for(i in 0:11){
mdf[paste0('T-', i)] <- ifelse((mdf$Year - mdf$variable == i), mdf$value, 0)
}
mdf$variable <- NULL # REMOVE MELT COLS
mdf$value <- NULL # REMOVE MELT COLS
aggdf <- aggregate(. ~ Individual + Wave + Country + Year, mdf, FUN=max)
# Individual Wave Country Year T-0 T-1 T-2 T-3 T-4 T-5
# 1 1201 6 32 2011 0 6.76174 32.96903 34.62974 61.65472 1.865651
# 2 1202 6 32 2011 0 6.76174 32.96903 34.62974 61.65472 1.865651
# 3 1203 6 32 2011 0 6.76174 32.96903 34.62974 61.65472 1.865651
# 4 1204 6 32 2011 0 6.76174 32.96903 34.62974 61.65472 1.865651
# 5 1205 6 32 2011 0 6.76174 32.96903 34.62974 61.65472 1.865651
# 6 1206 6 32 2011 0 6.76174 32.96903 34.62974 61.65472 1.865651
# T-6 T-7 T-8 T-9 T-10 T-11
# 1 1.955437 4.357977 259.2992 9.601448 176.2932 45.19665
# 2 1.955437 4.357977 259.2992 9.601448 176.2932 45.19665
# 3 1.955437 4.357977 259.2992 9.601448 176.2932 45.19665
# 4 1.955437 4.357977 259.2992 9.601448 176.2932 45.19665
# 5 1.955437 4.357977 259.2992 9.601448 176.2932 45.19665
# 6 1.955437 4.357977 259.2992 9.601448 176.2932 45.19665
答案 2 :(得分：2)
这是矩阵子集的一个很好的例子。
col_index <- match(toy$year, names(toy))
toy$T_0 <- toy[cbind(1:nrow(toy), col_index - 0)]
现在我们可以把它放在一个函数
中val_find <- function(ind) {
col_index <- match(toy$year, names(toy))
toy[cbind(1:nrow(toy), col_index - ind)]
}
toy[,paste0("T_", 0:8)] <- sapply(0:8, val_find)
id <- 1:12
wave <- 6
country <- gl(3, 4, labels=LETTERS[1:3])
year <- rep(c(2011,2012,2010), each=4)
dates <- setNames(as.data.frame(matrix(1:144, 12, 12)), as.character(2001:2012))
toy <- cbind(id, wave, country, year, dates)
#Try function
toy[,paste0("T_", 0:8)] <- sapply(0:8, val_find)
toy
# id wave country year 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 T_0 T_1 T_2 T_3 T_4 T_5 T_6 T_7
# 1 1 6 A 2011 1 13 25 37 49 61 73 85 97 109 121 133 121 109 97 85 73 61 49 37
# 2 2 6 A 2011 2 14 26 38 50 62 74 86 98 110 122 134 122 110 98 86 74 62 50 38
# 3 3 6 A 2011 3 15 27 39 51 63 75 87 99 111 123 135 123 111 99 87 75 63 51 39
# 4 4 6 A 2011 4 16 28 40 52 64 76 88 100 112 124 136 124 112 100 88 76 64 52 40
# 5 5 6 B 2012 5 17 29 41 53 65 77 89 101 113 125 137 137 125 113 101 89 77 65 53
# 6 6 6 B 2012 6 18 30 42 54 66 78 90 102 114 126 138 138 126 114 102 90 78 66 54
# 7 7 6 B 2012 7 19 31 43 55 67 79 91 103 115 127 139 139 127 115 103 91 79 67 55
# 8 8 6 B 2012 8 20 32 44 56 68 80 92 104 116 128 140 140 128 116 104 92 80 68 56
# 9 9 6 C 2010 9 21 33 45 57 69 81 93 105 117 129 141 117 105 93 81 69 57 45 33
# 10 10 6 C 2010 10 22 34 46 58 70 82 94 106 118 130 142 118 106 94 82 70 58 46 34
# 11 11 6 C 2010 11 23 35 47 59 71 83 95 107 119 131 143 119 107 95 83 71 59 47 35
# 12 12 6 C 2010 12 24 36 48 60 72 84 96 108 120 132 144 120 108 96 84 72 60 48 36