拆分变量和插入NA在

Question

之间我有一个变量，它看起来像这样：

Var 
[1] 3, 4, 5  2, 4, 5  2, 4  1, 4, 5 

我需要把它拆分成数据帧，看起来像这样：

V1 V2 V3 V4 V5 
NA NA 3 4 5 
NA 2 NA 4 5 
NA 2 NA 4 NA 
1 NA NA 4 5 

抱歉，系统我想不出找到解决我的问题的帖子。有谁知道我该怎么做？ 非常感谢您提前！

编辑：我找到了一个解决方案根据您的答案，并张贴在下面。

编辑2：我使用Ananda的解决方案提高了我的代码效率。

Answer 1

由OP的回答来看， “VAR” 是一个字符串，如：

var <- c("3, 4, 5", "2, 4, 5", "2, 4", "1, 4, 5") 

如果是这样的话，你可以考虑我的 “splitstackshape” 包cSplit_e：

library(splitstackshape) 
cSplit_e(data.frame(var), "var", ",", mode = "value", drop = TRUE) 
# var_1 var_2 var_3 var_4 var_5 
# 1 NA NA  3  4  5 
# 2 NA  2 NA  4  5 
# 3 NA  2 NA  4 NA 
# 4  1 NA NA  4  5 

如果它是list，正如其他答案所假设的那样，您可以使用支持cSplit_e的“splitstackshape”中的（未导出）numMat函数。

var <- list(c(3,4,5), c(2,4,5), c(2,4), c(1,4,5)) 
splitstackshape:::numMat(var, mode = "value") 
#  1 2 3 4 5 
# [1,] NA NA 3 4 5 
# [2,] NA 2 NA 4 5 
# [3,] NA 2 NA 4 NA 
# [4,] 1 NA NA 4 5 

引擎盖下，numMat是一个非常类似的方法，在@ thelatemail的回答中。

---

如果你有-99代表NA和要排除他们，你可以尝试：

var <- c("3, 4, 5", "2, -99, 4, 5", "2, 4", "1, 4, 5, -99") 
splitstackshape:::numMat(
    lapply(strsplit(var, ","), function(x) as.numeric(x)[as.numeric(x) > 0]), 
    mode = "value") 
#  1 2 3 4 5 
# [1,] NA NA 3 4 5 
# [2,] NA 2 NA 4 5 
# [3,] NA 2 NA 4 NA 
# [4,] 1 NA NA 4 5 

Answer 2

如果我们假设您var是这似乎工作清单：

var <- list(c(3,4,5),c(2,4,5),c(2,4),c(1,4,5)) 

#define function find_num to essentially create 
#5 new functions (called closures) inside the for-loop below 
find_num <- function(x) { 
    num <- function(mylist) { 
    sapply(mylist, function(i) if(x %in% i) return(x) else return(NA)) 
    } 
} 

#initiate list 
new_list <- list() 
#find_num is initiated with 5 different values essentially (in each iteration) 
#creating 5 new functions (closures) each for the number we want 
for (i in 1:5){ 
    myfunc <- find_num(i) 
    #this creates the list we want. Each element is a column 
    new_list[[length(new_list)+1]] <- myfunc(var) 
} 

#combine the columns into a new matrix 
new_list <- do.call(cbind, new_list) 

输出：

> new_list 
    [,1] [,2] [,3] [,4] [,5] 
[1,] NA NA 3 4 5 
[2,] NA 2 NA 4 5 
[3,] NA 2 NA 4 NA 
[4,] 1 NA NA 4 5 

Answer 3

使用矩阵索引：

Var <- list(c(3,4,5),c(2,4,5),c(2,4),c(1,4,5)) 
unVar <- unlist(Var) 
out <- matrix(NA, nrow=length(Var), ncol=max(unVar)) 

out[cbind(rep(seq_along(Var),sapply(Var,length)),unVar)] <- unVar 
# and if you're using the new version of R, you can simplify a little: 
out[cbind(rep(seq_along(Var),lengths(Var)),unVar)] <- unVar 

#  [,1] [,2] [,3] [,4] [,5] 
#[1,] NA NA 3 4 5 
#[2,] NA 2 NA 4 5 
#[3,] NA 2 NA 4 NA 
#[4,] 1 NA NA 4 5 

Answer 4

如果无功是只是一个矢量然后我会做以下几点：

Var = c(3,4,5,2,4,5,2,4,1,4,5) 
RowIdx = c(rep(1,3),rep(2,3),rep(3,2),rep(4,3)) 
DF = matrix(NA,nrow=4,ncol=5) 

for (idx in 1:length(Var)){ 
    DF[RowIdx[idx],Var[idx]] = Var[idx] 
} 

当然，如果你有，你可能想找到一种方法来生成更自动化的方式行索引更多数据

Answer 5

Var <- list(c(3, 4, 5), c(2, 4, 5), c(2, 4), c(1, 4, 5)) 
M <- matrix(NA, nrow=length(Var), ncol=max(sapply(Var,max))) 
for(L in seq(Var)) { M [ cbind(rep(L, length(Var[[L]])), Var[[L]]) ] <- Var[[L]]} 
M 
    [,1] [,2] [,3] [,4] [,5] 
[1,] NA NA 3 4 5 
[2,] NA 2 NA 4 5 
[3,] NA 2 NA 4 NA 
[4,] 1 NA NA 4 5 

个人我的投票建议是thelatemail的版本，这是基本同构对此。

Answer 6

我设法根据您的回答找到解决方案！我的最终解决方案如下所示：

# I had the additional problem that my variable was a factor, therefore I had to transform it first. 
df <- data.frame(Var) 
Var <- lapply(strsplit(as.character(df$Var), ", "), "[") 
for(i in 1:length(Var)){ 
    Var[[i]] <- as.numeric(Var[[i]]) 
} 

# Then I created a matrix based on thelatemails and BondedDusts approach. 
M <- matrix(NA, nrow=length(Var), ncol=max(sapply(Var,max))) 

# Additionally, I had the problem that there were some lines with a single -99, which indicates a missing value for the complete line. I had some problems with this negative value. For this reason, I assigned NA's first. 
for(i in 1:length(Var)){ 
    Var[[i]][Var[[i]] == -99] <- NA 
} 

# Final assignment like suggested by BonedDust. 
for(L in seq(Var)) { M [ cbind(rep(L, length(Var[[L]])), Var[[L]]) ] <- Var[[L]]} 
M 

我不确定这是否是最快的解决方案，但现在一切正常！非常感谢您的快速和广泛的答案！