拆分不同的長度值並綁定到列

Question

我有一個相當大的（大約10萬個觀測值）數據集，類似於：

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

我想將“值”列拆分為"," ，對於遺漏的單元格","使用NA ：

ID v1 v2 v3 v4 v5 v6 v7
1  1  2  3  NA NA NA NA
2  4  NA NA NA NA NA NA
3  NA NA NA NA NA NA NA
4  4  1  6  5  1  1  6
5  0  0  NA NA NA NA NA
...

最好的嘗試是strsplit + rbind ：

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

但是rbind函數只是回收所有“短”行而不是設置“NA”。 發現了類似的問題

非常感謝，Leo

Answer 1

我建議查看我的cSplit功能或手動解決問題。

cSplit方法很簡單：

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

手動接近問題看起來像：

## 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)

^^這幾乎是在cSplit中發生的cSplit ，但是該函數有一些其他選項和一些基本的錯誤檢查等等，這可能會使它比純手動方法（或為解決您的特定問題而編寫的函數）慢一點）。

這兩種方法都比“data.table”+“reshape2”方法更快。 此外，由於每行都是單獨處理的，即使您有重復的ID值，也不應該有任何問題 - 您的輸出應該與輸入具有相同的行數。

---

基准

我已經在更多行和數據上做了基准測試，這些測試會產生“更廣泛”的結果（因為在你對David的答案的評論中暗示了這一點）。

以下是示例數據：

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)

以下是要測試的功能：

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)
}

結果如下：

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

注：結果fun1a和fun1b不會是相同的fun2和fun3因為重復的ID。

Answer 2

這是一個data.table結合reshape2方法（應該非常有效）

library(data.table) # Loading `data.table` package
data2 <- setDT(data)[, list(Values = unlist(strsplit(Values, ","))), by = ID] # splitting the values by `,` for each `ID`
data2[, Var := paste0("v", seq_len(.N)), by = ID] # Adding the `Var` variable

library(reshape2) # Loading `reshape2` package
dcast.data.table(data2, ID ~ Var, fill = NA_character_, value.var = "Values") # decasting

#    ID v1 v2 v3 v4 v5 v6 v7
# 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