行减去不同列表R中的行

如何计算不同列表中不同行之间的差异？ 不同的列表有不同的维度。 我使用的代码如下

names(ri1)
[1] "Sedol" "code"  "ri"    "date"
ri1<-ri1[order(ri1$Sedol,ri1$date),]
sri<-split(ri1,ri1$Sedol)
ri1$r<-as.vector(sapply(seq_along(sri), function(x) diff(c(0, sri[[x]][,3]))))

我想要这样的结果

---

注意：

r=r（t+1）/r（t）-1

您应该结合使用

头部

和

尾部

，如下所示：

r.fun <- function(ri) c(0, tail(ri, -1) / head(ri, -1) - 1)
lapply(sri1, transform, r = r.fun(ri))

或

编辑：如果希望输出如示例中所示为XX%，请将

r.fun

替换为：

r.fun <- function(ri) paste0(round(100 * c(0, tail(ri, -1) / head(ri, -1) - 1)), "%")

---

r.fun使用diff和lappy可以得到

# I generate some data
  dat1 <- data.frame(date = seq(1990,1999,length.out=5),ri = seq(1,10,length.out=5))
  dat2 <- data.frame(date = seq(1990,1999,length.out=5),ri=seq(1,5,length.out=5))
# I put the data.frame in a list 
  ll <- list(dat1,dat2)
 # I use lapply:
  ll <- lapply(ll,function(dat){
    # I apply the formula you give in a vector version
    # maybe you need only diff in percent?
    dat$r <- round(c(0,diff(dat$ri))/dat$ri*100)
    dat
  })
ll
[[1]]
     date    ri  r
1 1990.00  1.00  0
2 1992.25  3.25 69
3 1994.50  5.50 41
4 1996.75  7.75 29
5 1999.00 10.00 22

[[2]]
     date ri  r
1 1990.00  1  0
2 1992.25  2 50
3 1994.50  3 33
4 1996.75  4 25
5 1999.00  5 20

#我生成了一些数据
dat1你确定你的r公式吗？它没有给出结果您可以通过使用
dput
提供
ri1
使示例重现。请参阅
date  ri
1990  1
1991  2

date  ri  r
1990  1   0%
1991  2   100%
1992  3   100%

date  ri  r
1990  1   0%
1991  2   100%
1992  3   100%
1993  4   100%

date  ri   r
1990  1    0%
1991  2    100%

r.fun <- function(ri) c(0, tail(ri, -1) / head(ri, -1) - 1)
lapply(sri1, transform, r = r.fun(ri))

transform(ri1, r = ave(ri, Sedol, FUN = r.fun))

library(plyr)
ddply(ri1, "Sedol", transform, r = r.fun(ri))

r.fun <- function(ri) paste0(round(100 * c(0, tail(ri, -1) / head(ri, -1) - 1)), "%")

# I generate some data
  dat1 <- data.frame(date = seq(1990,1999,length.out=5),ri = seq(1,10,length.out=5))
  dat2 <- data.frame(date = seq(1990,1999,length.out=5),ri=seq(1,5,length.out=5))
# I put the data.frame in a list 
  ll <- list(dat1,dat2)
 # I use lapply:
  ll <- lapply(ll,function(dat){
    # I apply the formula you give in a vector version
    # maybe you need only diff in percent?
    dat$r <- round(c(0,diff(dat$ri))/dat$ri*100)
    dat
  })
ll
[[1]]
     date    ri  r
1 1990.00  1.00  0
2 1992.25  3.25 69
3 1994.50  5.50 41
4 1996.75  7.75 29
5 1999.00 10.00 22

[[2]]
     date ri  r
1 1990.00  1  0
2 1992.25  2 50
3 1994.50  3 33
4 1996.75  4 25
5 1999.00  5 20