How to apply a set of functions to each group of a

问题:

I need to reshape data.frame in R in one step. In short, change of values of objects (x1 to x6) is visible row by row (from 1990 to 1995):

> 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

I am able to do reshaping step by step, does anybody know how do it in one step?

Original data Table 1 - see that minimal value from all timeseries is "0"

Step1: Table 2 - rescale each timeseries that each would have minimal value equal "0". All times fall down on x-axes.

Step2: Table 3 - apply diff() function on each timeline.

Step3: Table 4 - apply sort() function on each timeseries.

I hope the pictures are clear enough for understanding each step.

So final table looks like this:

> 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))

回答1:

It sounds like you want to apply a set of functions to each group of a grouping variable. There are many ways to do this in R (from base R by and tapply to add-on packages like plyr, data.table, and dplyr). I've been learning how to use package dplyr, and came up with the following solution.

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

回答2:

Using data.table, this is simply:

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

Note that your 'step 1' is unnecessary as your second step is calculating difference and it wouldn't have any effect (and is therefore skipped here).