1

我有数据框

test <- structure(list(
     y2002 = c("freshman","freshman","freshman","sophomore","sophomore","senior"),
     y2003 = c("freshman","junior","junior","sophomore","sophomore","senior"),
     y2004 = c("junior","sophomore","sophomore","senior","senior",NA),
     y2005 = c("senior","senior","senior",NA, NA, NA)), 
              .Names = c("2002","2003","2004","2005"),
              row.names = c(c(1:6)),
              class = "data.frame")
> test
       2002      2003      2004   2005
1  freshman  freshman    junior senior
2  freshman    junior sophomore senior
3  freshman    junior sophomore senior
4 sophomore sophomore    senior   <NA>
5 sophomore sophomore    senior   <NA>
6    senior    senior      <NA>   <NA>

而且我想调整数据以仅获取每一行的各个步骤，如

result <- structure(list(
 y2002 = c("freshman","freshman","freshman","sophomore","sophomore","senior"),
 y2003 = c("junior","junior","junior","senior","senior",NA),
 y2004 = c("senior","sophomore","sophomore",NA,NA,NA),
 y2005 = c(NA,"senior","senior",NA, NA, NA)), 
               .Names = c("1","2","3","4"),
               row.names = c(c(1:6)),
               class = "data.frame")

> result
          1      2         3      4
1  freshman junior    senior   <NA>
2  freshman junior sophomore senior
3  freshman junior sophomore senior
4 sophomore senior      <NA>   <NA>
5 sophomore senior      <NA>   <NA>
6    senior   <NA>      <NA>   <NA>

我知道如果我将每一行视为一个向量，我可以做类似的事情

careerrow <- c(1,2,3,3,4)
pairz <- lapply(careerrow,function(i){c(careerrow[i],careerrow[i+1])})
uniquepairz <- careerrow[sapply(pairz,function(x){x[1]!=x[2]})]

我的困难是将该行应用于我的数据表。我认为 lapply 是要走的路，但到目前为止我无法解决这个问题。

4

2 回答 2

3

如果您的目标是计算每个途径的总数

您可以使用这样的东西（使用data.table是因为它将列表作为 data.table（data.frame-like）对象中的元素处理的好方法。

我正在使用!duplicated(...) 删除重复项，因为这比唯一更有效。

library(data.table)
library(reshape2)
# make the rownames a column 
test$id <- rownames(test)
# put in long format
DT <- as.data.table(melt(test,id='id'))
# get the unique steps and concatenate into a unique identifier for each pathway
DL <- DT[!is.na(value), {.steps <- value[!duplicated(value)]
  stepid <- paste(.steps, sep ='.',collapse = '.')
  list(steps = list(.steps), stepid =stepid)}, by=id]
##    id                            steps                           stepid
## 1:  1           freshman,junior,senior           freshman.junior.senior
## 2:  2 freshman,junior,sophomore,senior freshman.junior.sophomore.senior
## 3:  3 freshman,junior,sophomore,senior freshman.junior.sophomore.senior
## 4:  4                 sophomore,senior                 sophomore.senior
## 5:  5                 sophomore,senior                 sophomore.senior
## 6:  6                           senior                           senior

# count the number per path

DL[, .N, by = stepid]
##                              stepid N
## 1:           freshman.junior.senior 1
## 2: freshman.junior.sophomore.senior 2
## 3:                 sophomore.senior 2
## 4:                           senior 1
于 2012-09-14T04:51:20.840 回答
2

lapply，当传递一个 data.frame 时，对其列进行操作。那是因为 data.frame 是一个列表，其元素是列。代替lapply，您可以使用applywith MARGIN=1：

unique.padded <- function(x) {
   uniq <- unique(x)
   out  <- c(uniq, rep(NA, length(x) - length(uniq)))
}

t(apply(test, 1, unique.padded))

#   [,1]        [,2]     [,3]        [,4]    
# 1 "freshman"  "junior" "senior"    NA      
# 2 "freshman"  "junior" "sophomore" "senior"
# 3 "freshman"  "junior" "sophomore" "senior"
# 4 "sophomore" "senior" NA          NA      
# 5 "sophomore" "senior" NA          NA      
# 6 "senior"    NA       NA          NA

编辑：我看到了您对最终目标的评论。我会做这样的事情：

table(sapply(apply(test, 1, function(x)unique(na.omit(x))),
             paste, collapse = "_"))

#           freshman_junior_senior freshman_junior_sophomore_senior 
#                                1                                2 
#                           senior                 sophomore_senior 
#                                1                                2
于 2012-09-14T03:01:21.090 回答