如何在不考虑Na值的情况下返回多个列，并按R中的其他列名称进行分组？

Question

mexico <- c(1,2,5,1,NA,1)
argentina <- c(2,2,2,2,NA,2)
italy<- c(NA,10,10,10,NA,10)
spain <- c(NA,NA,11,11,11,11)
england <- c(5,NA,10,NA,NA,12)
germany <- c(1,NA,NA,NA,NA,10)

Data_Risk = data.frame( Mexico, Argentina, Italy, Spain, England, Germany)

Data_Risk 

给

 mexico     argentina italy spain england germany

1      1         2    NA    NA       5       1
2      2         2    10    NA      NA      NA
3      5         2    10    11      10      NA
4      1         2    10    11      NA      NA
5     NA        NA    NA    11      NA      NA
6      1         2    10    11      12      10

在这种情况下，我不需要考虑NA情况，因此我尝试了这个

Data_Risk <- as.data.table(Data_Risk)
my_c <- !apply(Data_Risk, 1, is.na)[,1]
my_L <- Data_Risk[1]
as.data.frame(my_L)[my_c]

结果：

  Mexico Argentina England Germany
1      1         2       5       1

在这种情况下，我不仅需要它考虑一行，而且还要考虑所有这些行。 此外，每行的组需要放在新列中而不考虑值，因此最终表必须如下所示：

var1           var2          var3       var4     var5    var6
mexico    argentina       england    germany     null    null
mexico    argentina         italy       null     null    null 
mexico    argentina         italy      spain  england    null
mexico    argentina         italy      spain     null    null
spain      null             null       null      null    null
mexico    argentina         italy      spain england  germany

Answer 1

一种选择是查看which(!is.na(Data_Risk), arr.ind = T)并将其扩展为宽形式，用col替换order(col)变量，并添加colnm列以用作spread-to-long（dcast）过程中的value.var。

library(data.table)
library(magrittr)

nms <- as.data.table(which(!is.na(Data_Risk), arr.ind = T))

nms[, .(colnm = names(Data_Risk)[col], col = paste0('var', order(col)))
    , by = row] %>% 
  dcast(row ~ col, value.var = 'colnm')

#    row   var1      var2    var3    var4    var5    var6
# 1:   1 mexico argentina england germany    <NA>    <NA>
# 2:   2 mexico argentina   italy    <NA>    <NA>    <NA>
# 3:   3 mexico argentina   italy   spain england    <NA>
# 4:   4 mexico argentina   italy   spain    <NA>    <NA>
# 5:   5  spain      <NA>    <NA>    <NA>    <NA>    <NA>
# 6:   6 mexico argentina   italy   spain england germany

等效的dplyr代码：

library(dplyr)

nms <- as.data.frame(which(!is.na(Data_Risk), arr.ind = T))

nms %>% 
  group_by(row) %>% 
  mutate(colnm = names(Data_Risk)[col],
         col = paste0('var', order(col))) %>% 
  spread(col, value = colnm) %>% 
  ungroup

Answer 2

有一些问题需要什么，但如果你想要的是在每一行中分别用下面的非NA替换每个NA，下面给出了该形式的矩阵：

library(zoo)
t(apply(Data_Risk, 1, na.locf0, fromLast = TRUE))

赠送：

     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    2    5    5    5    1
[2,]    2    2   10   NA   NA   NA
[3,]    5    2   10   11   10   NA
[4,]    1    2   10   11   NA   NA
[5,]   11   11   11   11   NA   NA
[6,]    1    2   10   11   12   10

或者如果你想要的是将每行中的NA移动到最后：

t(apply(Data_Risk, 1, function(x) c(na.omit(x), rep(NA, sum(is.na(x))))))

赠送：

     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    2    5    1   NA   NA
[2,]    2    2   10   NA   NA   NA
[3,]    5    2   10   11   10   NA
[4,]    1    2   10   11   NA   NA
[5,]   11   NA   NA   NA   NA   NA
[6,]    1    2   10   11   12   10

或等效地：

t(apply(Data_Risk, 1, function(x) "length<-"(na.omit(x), length(x))))

Answer 3

我们可以逐行使用apply，找出非NA索引，用列名替换它们，并用NA附加其余的索引。

t(apply(Data_Risk, 1, function(x) {
    inds <- which(!is.na(x))
   c(names(Data_Risk)[inds], rep(NA,ncol(Data_Risk) - length(inds)))
}))

#        [,1]         [,2]     [,3]      [,4]      [,5]      [,6]     
#[1,] "mexico" "argentina" "england" "germany" NA        NA       
#[2,] "mexico" "argentina" "italy"   NA        NA        NA       
#[3,] "mexico" "argentina" "italy"   "spain"   "england" NA       
#[4,] "mexico" "argentina" "italy"   "spain"   NA        NA       
#[5,] "spain"  NA          NA        NA        NA        NA       
#[6,] "mexico" "argentina" "italy"   "spain"   "england" "germany"

如果您希望最终输出为数据框，请将apply包装在data.frame()中。