2

I would like to tabulate by row within a data frame. I can obtain adequate results using table within apply in the following example:

df.1 <- read.table(text = '
  state  county  city  year1  year2  year3  year4  year5
      1       2     4      0      0      0      1      2
      2       5     3     10     20     10     NA     10
      2       7     1    200    200     NA     NA    200
      3       1     1     NA     NA     NA     NA     NA
', na.strings = "NA", header=TRUE)

tdf <- t(df.1)
apply(tdf[4:nrow(tdf),1:nrow(df.1)], 2, function(x) {table(x, useNA = "ifany")})

Here are the results:

[[1]]
x
0 1 2 
3 1 1 

[[2]]
x
  10   20 <NA> 
   3    1    1 

[[3]]
x
 200 <NA> 
   3    2 

[[4]]
x
<NA> 
   5

However, in the following example each row consists of a single value.

df.2 <- read.table(text = '
  state  county  city  year1  year2  year3  year4  year5
      1       2     4      0      0      0      0      0
      2       5     3      1      1      1      1      1
      2       7     1      2      2      2      2      2
      3       1     1     NA     NA     NA     NA     NA
', na.strings = "NA", header=TRUE)

tdf.2 <- t(df.2)
apply(tdf.2[4:nrow(tdf.2),1:nrow(df.2)], 2, function(x) {table(x, useNA = "ifany")})

The output I obtain is:

# [1] 5 5 5 5

As such, I cannot tell from this output that the first 5 is for 0, the second 5 is for 1, the third 5 is for 2 and the last 5 is for NA. Is there a way I can have R return the value represented by each 5 in the second example?

4

4 回答 4

6

You can use lapply to systematically output a list. You would have to loop over the row indices:

sub.df <- as.matrix(df.2[grepl("year", names(df.2))])
lapply(seq_len(nrow(sub.df)), 
       function(i)table(sub.df[i, ], useNA = "ifany"))
于 2013-05-29T23:16:55.667 回答
4

通过包装来保护结果list：

apply(tdf.2[4:nrow(tdf.2),1:nrow(df.2)], 2, 
              function(x) {list(table(x, useNA = "ifany")) })
于 2013-05-29T23:58:43.770 回答
3

这是一个table解决方案：

table(
    rep(rownames(df.1),5),
    unlist(df.1[,4:8]),
    useNA="ifany")

这给

    0 1 2 10 20 200 <NA>
  1 3 1 1  0  0   0    0
  2 0 0 0  3  1   0    1
  3 0 0 0  0  0   3    2
  4 0 0 0  0  0   0    5

...和你的df.2：

    0 1 2 <NA>
  1 5 0 0    0
  2 0 5 0    0
  3 0 0 5    0
  4 0 0 0    5

好吧，这是一个解决方案，除非您出于某种原因真的喜欢拥有一个表格列表。

于 2013-05-30T00:11:19.627 回答
2

I think the problem is stated in applys help:

... If n equals 1, apply returns a vector if MARGIN has length 1 and an array of dimension dim(X)[MARGIN] otherwise ...

基础 R 的 apply 系列返回值的不一致是我完全转向plyrs **ply 函数的原因。所以这可以按需要工作：

library(plyr)
alply( df.2[ 4:8 ], 1, function(x) table( unlist(x), useNA = "ifany" ) )
于 2013-05-29T23:33:16.487 回答