我需要一步重塑R 中的data.frame 。 简而言之,对象(x1到x6)的值的变化是逐行可见的(从1990年到1995年):
> tab1[1:10, ] # raw data see plot for tab1
id value year
1 x1 7 1990
2 x1 10 1991
3 x1 11 1992
4 x1 7 1993
5 x1 3 1994
6 x1 1 1995
7 x2 6 1990
8 x2 7 1991
9 x2 9 1992
10 x2 5 1993
我能够一步一步地重塑,有人知道如何一步完成吗?
原始数据 表1 - 看到所有时间序列的最小值是" 0"
步骤1: 表2 - 重新缩放每个每个具有最小值等于" 0"的时间序列。 所有时间都落在x轴上。
步骤2:
表3 - 在每个时间轴上应用diff()
函数。
步骤3:
表4 - 在每个时间序列上应用sort()
函数。
我希望这些图片足够清晰,以便了解每一步。
所以决赛桌看起来像这样:
> tab4[1:10, ]
id value time
1 x1 -4 1
2 x1 -4 2
3 x1 -2 3
4 x1 1 4
5 x1 3 5
6 x2 -4 1
7 x2 -3 2
8 x2 1 3
9 x2 1 4
10 x2 2 5
# Source data:
tab1 <- data.frame(id = rep(c("x1","x2","x3","x4","x5","x6"), each = 6),
value = c(7,10,11,7,3,1,6,7,9,5,2,3,11,9,7,9,1,
0,1,2,2,4,7,4,2,3,1,6,4,2,3,5,4,3,5,6),
year = rep(c(1990:1995), times = 6))
tab2 <- data.frame(id = rep(c("x1","x2","x3","x4","x5","x6"), each = 6),
value = c(6,9,10,6,2,0,4,5,7,3,0,1,11,9,7,9,1,0,
0,1,1,3,6,3,1,2,0,5,3,1,0,2,1,0,2,3),
year = rep(c(1990:1995), times = 6))
tab3 <- data.frame(id = rep(c("x1","x2","x3","x4","x5","x6"), each = 5),
value = c(3,1,-4,-4,-2,1,2,-4,-3,1,-2,-2,2,-8,-1,
1,0,2,3,-3,1,-2,5,-2,-2,2,-1,-1,2,1),
time = rep(c(1:5), times = 6))
tab4 <- data.frame(id = rep(c("x1","x2","x3","x4","x5","x6"), each = 5),
value = c(-4,-4,-2,1,3,-4,-3,1,1,2,-8,-2,-2,-1,2,
-3,0,1,2,3,-2,-2,-2,1,5,-1,-1,1,2,2),
time = rep(c(1:5), times = 6))
答案 0 :(得分:3)
使用data.table
,这只是:
require(data.table) ## 1.9.2
ans <- setDT(tab1)[, list(value=diff(value)), by=id] ## aggregation
setkey(ans, id,value)[, time := seq_len(.N), by=id] ## order + add 'time' column
请注意,您的步骤1&#39;是不必要的,因为你的第二步是计算差异,它不会产生任何影响(因此在此处被忽略)。
答案 1 :(得分:2)
听起来您想要将一组函数应用于分组变量的每个组。 R中有很多种方法(从基础R by
和tapply
到plyr
,data.table
和dplyr
等附加软件包)。我一直在学习如何使用包dplyr
,并提出了以下解决方案。
require(dplyr)
tab4 = tab1 %>%
group_by(id) %>% # group by id
mutate(value = value - min(value), value = value - lag(value)) %>% # group min to 0, difference lag 1
na.omit %>% # remove NA caused by lag 1 differencing
arrange(id, value) %>% # order by value within each id
mutate(time = 1:length(value)) %>% # Make a time variable from 1 to 5 based on current order
select(-year) # remove year column to match final OP output