2

I am looking for a faster way to achieve the operation below. The dataset contains > 1M rows but I have provided a simplified example to illustrate the task --

To create the data table --

dt <- data.table(name=c("john","jill"), a1=c(1,4), a2=c(2,5), a3=c(3,6), 
      b1=c(10,40), b2=c(20,50), b3=c(30,60))

colGroups <- c("a","b")   # Columns starting in "a", and in "b"

Original Dataset
-----------------------------------
name    a1   a2   a3   b1   b2   b3
john    1    2    3    10   20   30
jill    4    5    6    40   50   60

The above dataset is transformed such that 2 new rows are added for each unique name and in each row, the values are left shifted for each group of columns independently (in this example I have used a columns and b columns but there are many more)

Transformed Dataset
-----------------------------------
name    a1   a2   a3   b1   b2   b3
john    1    2    3    10   20   30  # First Row for John
john    2    3    0    20   30    0  # "a" values left shifted, "b" values left shifted
john    3    0    0    30   0     0  # Same as above, left-shifted again

jill    4    5    6    40   50   60  # Repeated for Jill
jill    5    6    0    50   60    0 
jill    6    0    0    60    0    0

And so on. My dataset is extremely large, which is why I am trying to see if there is an efficient way to implement this.

Thanks in advance.

4

3 回答 3

5

更新：（更快）的解决方案是按如下方式使用索引（在 1e6*7 上大约需要 4 秒）：

ll <- vector("list", 3)
ll[[1]] <- copy(dt[, -1])
d_idx <- seq(2, ncol(dt), by=3)
for (j in 1:2) {
    tmp <- vector("list", 2)
    for (i in seq_along(colGroups)) {
        idx <- ((i-1)*3+2):((i*3)+1)
        cols <- setdiff(idx, d_idx[i]:(d_idx[i]+j-1))
        # ..cols means "look up one level"
        tmp[[i]] <- cbind(dt[, ..cols], data.table(matrix(0, ncol=j)))
    }
    ll[[j+1]] <- do.call(cbind, tmp)
}
ans <- cbind(data.table(name=dt$name), rbindlist(ll))
setkey(ans, name)

第一次尝试（旧）： 非常有趣的问题。我会使用melt.data.tableand dcast.data.table（从 1.8.11 开始）来处理它，如下所示：

require(data.table)
require(reshape2)
# melt is S3 generic, calls melt.data.table, returns a data.table (very fast)
ans <- melt(dt, id=1, measure=2:7, variable.factor=FALSE)[, 
                    grp := rep(colGroups, each=nrow(dt)*3)]
setkey(ans, name, grp)
ans <- ans[, list(variable=c(variable, variable[1:(.N-1)], 
          variable[1:(.N-2)]), value=c(value, value[-1],
     value[-(1:2)]), id2=rep.int(1:3, 3:1)), list(name, grp)]
# dcast in reshape2 is not yet a S3 generic, have to call by full name
ans <- dcast.data.table(ans, name+id2~variable, fill=0L)[, id2 := NULL]

对具有相同列数的 1e6 行进行基准测试：

require(data.table)
require(reshape2)
set.seed(45)
N <- 1e6
dt <- cbind(data.table(name=paste("x", 1:N, sep="")), 
               matrix(sample(10, 6*N, TRUE), nrow=N))
setnames(dt, c("name", "a1", "a2", "a3", "b1", "b2", "b3"))
colGroups = c("a", "b")

system.time({
ans <- melt(dt, id=1, measure=2:7, variable.factor=FALSE)[, 
                    grp := rep(colGroups, each=nrow(dt)*3)]
setkey(ans, name, grp)
ans <- ans[, list(variable=c(variable, variable[1:(.N-1)], 
          variable[1:(.N-2)]), value=c(value, value[-1],
     value[-(1:2)]), id2=rep.int(1:3, 3:1)), list(name, grp)]
ans <- dcast.data.table(ans, name+id2~variable, fill=0L)[, id2 := NULL]

})

#   user  system elapsed 
# 45.627   2.197  52.051
于 2013-10-03T23:57:34.733 回答
1

您可以追加行，然后按组向上移动列。由于每组的总列数是固定的，因此您迭代每个组号。

## Add in the extra rows
dt <- dt[, rbindlist(rep(list(.SD), 3)), by=name]


### ASSUMING A FIXED NUMBER PER COLGROUP
N <- 3

colsShifting <- as.vector(sapply(colGroups, paste0, 2:N))

for (i in (2:N)-1 ) {
  current <- colsShifting[ (i) +  ( (N-1) * (seq_along(colGroups)-1) )]
  dt[, c(current) := {
              .NN <- .N; 
              .CROP <- .SD[1:(.NN-i)]  ## These lines are only for clean code. You can put it all into the `rbindlist` line
              rbindlist(list(.CROP, as.data.table(replicate(ncol(.SD), rep(0, i),simplify=FALSE ))))
            } 
      , .SDcols=current
      , by=name]
  }

这使：

dt
#     name a1 a2 a3 b1 b2 b3
#  1: john  1  2  3 10 20 30
#  2: john  1  2  0 10 20  0
#  3: john  1  0  0 10  0  0
#  4: jill  4  5  6 40 50 60
#  5: jill  4  5  0 40 50  0
#  6: jill  4  0  0 40  0  0
于 2013-10-03T23:07:13.543 回答
1

只需编辑所选答案的@Arun (s) 代码。在这里提供，因为我无法在评论部分发帖。

#Parameterized version of @Arun (author) code (in the selected answer)

#Shifting Columns in R
#--------------------------------------------
N = 5  # SET - Number of unique names
set.seed(5)
colGroups <- c("a","b") # ... (i) # SET colGroups
totalColsPerGroup <- 10 # SET Cols Per Group
numColsToLeftShift <- 8 # SET Cols to Shift

lenColGroups <- length(colGroups) # ... (ii)

# From (i) and (ii)
totalCols = lenColGroups * totalColsPerGroup


dt <- cbind(data.table(name=paste("x", 1:N, sep="")), 
            matrix(sample(5, totalCols*N, TRUE), nrow=N)) # Change 5 if needed

ll <- vector("list", numColsToLeftShift)
ll[[1]] <- copy(dt[, -1, with=FALSE])
d_idx <- seq(2, ncol(dt), by=totalColsPerGroup)
for (j in 1:(numColsToLeftShift)) {
  tmp <- vector("list", 2)
  for (i in seq_along(colGroups)) {
    idx <- ((i-1)*totalColsPerGroup+2):((i*totalColsPerGroup)+1) #OK
    tmp[[i]] <- cbind(dt[, setdiff(idx, d_idx[i]:(d_idx[i]+j-1)), 
                         with=FALSE], data.table(matrix(0, ncol=j)))

  }      
  ll[[j+1]] <- do.call(cbind, tmp)

}
ans <- cbind(data.table(name=dt$name), rbindlist(ll))
setkey(ans, name)

--

于 2013-10-04T14:42:24.833 回答