Question

r中的条件级联。

key   key1 col1 col2 col3 
cust1 k1   NA    NA   34  
cust2 k2   12    NA   24  
Cust3 k3   1     23   34

。

cols <- names(df[3:ncol(df)])

df[, col4 := do.call(paste,c(lapply(cols, function(x) paste(x, get(x), sep=":")),sep="&&"))]

在这种情况下，我正在获取列名和列值，但是我只想为非null值做同样的事情相应的输出

key   key1 col1 col2 col3 col4
cust1 k1   NA    NA   34  col1:NA&&col2:NA&&col3:34
cust2 k2   12    NA   24  col1:12&&col2:NA&&col3:24
Cust3 k3   1     23   34  col1:1&&col2:23&&col3:34

所需的输出

key   key1 col1 col2 col3 col4
cust1 k1   NA    NA   34  col3:34
cust2 k2   12    NA   24  col1:12&&col13:24
Cust3 k3   1     23   34  col1:1&&col2:23&&col3:34

Answer 1

遵循@Ronak Sha的回答策略，这是与您在问题中使用的data.table语法的集成：

a[, 
  col4 := apply(.SD, 
                1, 
                function(x) paste(names(x)[!is.na(x)], 
                                  x[!is.na(x)], 
                                  sep = ":", 
                                  collapse = "&&")), 
  .SDcols = 3:5]

a

     key key1 col1 col2 col3                     col4
1: cust1   k1   NA   NA   34                  col3:34
2: cust2   k2   12   NA   24         col1:12&&col3:24
3: Cust3   k3    1   23   34 col1:1&&col2:23&&col3:34

带有数据：

a <- fread("key   key1 col1 col2 col3 
cust1 k1   NA    NA   34  
cust2 k2   12    NA   24  
Cust3 k3   1     23   34 ")

Answer 2

我们可以逐行使用apply，删除NA值，然后将sep参数添加为:，并通过collapse &&

cols <- grep("col", names(df))
df$col4 <- apply(df[cols], 1, function(x) {
   x <- x[!is.na(x)]
   paste(names(x), x, sep = ":", collapse = "&&")
})

df
#    key key1 col1 col2 col3                     col4
#1 cust1   k1   NA   NA   34                  col3:34
#2 cust2   k2   12   NA   24         col1:12&&col3:24
#3 Cust3   k3    1   23   34 col1:1&&col2:23&&col3:34

数据

df <- "Cust3"), class = "factor"), key1 = structure(1:3, .Label = c("k1", 
"k2", "k3"), class = "factor"), col1 = c(NA, 12L, 1L), col2 = c(NA, 
NA, 23L), col3 = c(34L, 24L, 34L)), class = "data.frame", row.names = (NA, -3L))

如何执行条件级联

2 个答案:

带有数据：