R 拆分不同的长度值并绑定到列

我有一个相当大的数据集（大约100k个观测值），类似于：

data <- data.frame(
                 ID = seq(1, 5, 1),
                 Values = c("1,2,3", "4", " ", "4,1,6,5,1,1,6", "0,0"), 
                 stringsAsFactors=F)
data
  ID        Values
1  1         1,2,3
2  2             4
3  3              
4  4 4,1,6,5,1,1,6
5  5           0,0

最佳尝试是strsplit+

rbind

：

df <- data.frame(do.call(
                        "rbind",
                        strsplit(as.character(data$Values), split = "," , fixed = FALSE)
                        ))

---

df这里有一个
数据表
与
重塑2
方法相结合（应该非常有效）

library（data.table）#加载`data.table`包
数据2我建议手动查看或处理问题

cSplit
方法就是：

cSplit(data, "Values", ",")
#    ID Values_1 Values_2 Values_3 Values_4 Values_5 Values_6 Values_7
# 1:  1        1        2        3       NA       NA       NA       NA
# 2:  2        4       NA       NA       NA       NA       NA       NA
# 3:  3                NA       NA       NA       NA       NA       NA
# 4:  4        4        1        6        5        1        1        6
# 5:  5        0        0       NA       NA       NA       NA       NA

手动处理问题的方式如下所示：

## Split up the values
Split <- strsplit(data$Values, ",", fixed = TRUE)
## How long is each list element?
Ncol <- vapply(Split, length, 1L)
## Create an empty character matrix to store the results
M <- matrix(NA_character_, nrow = nrow(data),
            ncol = max(Ncol), 
            dimnames = list(NULL, paste0("V", sequence(max(Ncol)))))
## Use matrix indexing to figure out where to put the results
M[cbind(rep(1:nrow(data), Ncol), 
        sequence(Ncol))] <- unlist(Split, use.names = FALSE)
## Bind the values back together, here as a "data.table" (faster)
data.table(ID = data$ID, M)

以下是要测试的功能：

fun1a <- function(inDT) {
  data2 <- DT[, list(Values = unlist(
    strsplit(Values, ","))), by = ID]
  data2[, Var := paste0("v", seq_len(.N)), by = ID] 
  dcast.data.table(data2, ID ~ Var, 
                   fill = NA_character_, 
                   value.var = "Values")
}

fun1b <- function(inDT) {
  data2 <- DT[, list(Values = unlist(
    strsplit(Values, ",", fixed = TRUE), 
    use.names = FALSE)), by = ID]
  data2[, Var := paste0("v", seq_len(.N)), by = ID] 
  dcast.data.table(data2, ID ~ Var, 
                   fill = NA_character_, 
                   value.var = "Values")
}

fun2 <- function(inDT) {
  cSplit(DT, "Values", ",")
}

fun3 <- function(inDF) {
  Split <- strsplit(inDF$Values, ",", fixed = TRUE)
  Ncol <- vapply(Split, length, 1L)
  M <- matrix(NA_character_, nrow = nrow(inDF),
              ncol = max(Ncol), 
              dimnames = list(NULL, paste0("V", sequence(max(Ncol)))))
  M[cbind(rep(1:nrow(inDF), Ncol), 
          sequence(Ncol))] <- unlist(Split, use.names = FALSE)
  data.table(ID = inDF$ID, M)
}

注意：
fun1a
和
fun1b
的结果与
fun2
和
fun3
的结果不一样，因为ID重复。
很抱歉出现这种混乱。当然它应该有7列。谢谢，@David Arenburg，它工作得很好，但是如果ID不是唯一的呢？
ID
是唯一的。我不明白你的问题。我准确地达到了您想要的输出在我的真实数据中有一些ID相等的字符串。在发送问题之前，我想我有点匆忙。那么？这应该仍然有效。我在代码中使用了
by=ID
，因此它会起作用。我同意
cSplit
会更快，但我认为它不会仅通过这一行就达到所需的输出。关于
vapply
+
序列
，我不确定它们是否会像
循环的
那样快。你有在大数据集上做基准吗？@DavidArenburg，你会错过什么？只是列名？还是我误解了这个问题？我会带着一些基准回来的。@DavidArenburg，当我的基准运行时，我期望的不是速度，而是矩阵索引的东西……好的，我把它拿回来，关于
cSplit
。看起来这正是他所需要的。我曾想过使用它，但懒得找到要点Id。你真的需要使用@Arun将其添加到
数据中。table
packagecSplit工作得很好！我甚至不需要添加具有唯一ID的额外列。还有一个更好的列名，比DT+Reformate2更好。阿南达马托，非常感谢你对大卫伦堡的帮助。我真的很感激。R太棒了+1针对具有预期结果的可重复性问题，并演示您已尝试的内容。
## Split up the values
Split <- strsplit(data$Values, ",", fixed = TRUE)
## How long is each list element?
Ncol <- vapply(Split, length, 1L)
## Create an empty character matrix to store the results
M <- matrix(NA_character_, nrow = nrow(data),
            ncol = max(Ncol), 
            dimnames = list(NULL, paste0("V", sequence(max(Ncol)))))
## Use matrix indexing to figure out where to put the results
M[cbind(rep(1:nrow(data), Ncol), 
        sequence(Ncol))] <- unlist(Split, use.names = FALSE)
## Bind the values back together, here as a "data.table" (faster)
data.table(ID = data$ID, M)

set.seed(1)
a <- sample(0:100, 100000, TRUE)
Values <- vapply(a, function(x) 
  paste(sample(0:100, x, TRUE), collapse = ","), character(1L))
Values[sample(length(Values), length(Values) * .15)] <- ""
ID <- c(1:80000, 1:20000)
data <- data.frame(ID, Values, stringsAsFactors = FALSE)
DT <- as.data.table(data)

fun1a <- function(inDT) {
  data2 <- DT[, list(Values = unlist(
    strsplit(Values, ","))), by = ID]
  data2[, Var := paste0("v", seq_len(.N)), by = ID] 
  dcast.data.table(data2, ID ~ Var, 
                   fill = NA_character_, 
                   value.var = "Values")
}

fun1b <- function(inDT) {
  data2 <- DT[, list(Values = unlist(
    strsplit(Values, ",", fixed = TRUE), 
    use.names = FALSE)), by = ID]
  data2[, Var := paste0("v", seq_len(.N)), by = ID] 
  dcast.data.table(data2, ID ~ Var, 
                   fill = NA_character_, 
                   value.var = "Values")
}

fun2 <- function(inDT) {
  cSplit(DT, "Values", ",")
}

fun3 <- function(inDF) {
  Split <- strsplit(inDF$Values, ",", fixed = TRUE)
  Ncol <- vapply(Split, length, 1L)
  M <- matrix(NA_character_, nrow = nrow(inDF),
              ncol = max(Ncol), 
              dimnames = list(NULL, paste0("V", sequence(max(Ncol)))))
  M[cbind(rep(1:nrow(inDF), Ncol), 
          sequence(Ncol))] <- unlist(Split, use.names = FALSE)
  data.table(ID = inDF$ID, M)
}

library(microbenchmark)
microbenchmark(fun2(DT), fun3(data), times = 20)
# Unit: seconds
#        expr      min       lq   median       uq      max neval
#    fun2(DT) 4.810942 5.173103 5.498279 5.622279 6.003339    20
#  fun3(data) 3.847228 3.929311 4.058728 4.160082 4.664568    20

## Didn't want to microbenchmark here...
system.time(fun1a(DT))
#    user  system elapsed 
#   16.92    0.50   17.59
system.time(fun1b(DT))  # fixed = TRUE & use.names = FALSE
#    user  system elapsed 
#   11.54    0.42   12.01