1

嗨，我有一个看起来像这样的数据集

name1  a b c d  
name2  a c e g i  
name3  t j i m n z

dput输出：

structure(c("name1", "name2", "name3", "a ", "a", "r ", "b", "c", "k ", "c", "e", "l", "d", "t", "o", "e", "j", "m", "", "k", "n"), .Dim = c(3L, 7L), .Dimnames = list(NULL, c("V1", "V2", "V3", "V4", "V5", "V6", "V7")))

我想像这样转换为矩阵

         a b c d e g i j m n t z
name1    1 1 1 1 0 0 0 0 0 0 0 0
name2    1 0 1 0 1 1 1 0 0 0 0 0 
name3    0 0 0 0 0 0 1 1 1 1 1 1

我怎样才能在 R 中做到这一点？

4

3 回答 3

3 
## Assuming this is your starting data
dat <- read.table(text="name1  a b c d  NA NA\nname2  a c e g i NA\nname3  t j i m n z")
rownames(dat) <- dat$V1 
dat$V1 <- NULL

我假设您的数据类似于上面。 

## store the rownames
NM <- rownames(dat)  # or NM <- c("name1", "name2", "name3")

## IMPORTANT. Make sure you have characters, not factors. 
dat <- sapply(dat, as.character)

cols <- sort(unique(as.character(unlist(dat))))

results <- sapply(cols, function(cl) apply(dat, 1, `%in%`, x=cl))
results[] <- as.numeric(results)

rownames(results) <- NM

results

      a b c d e g i j m n t z
name1 1 1 1 1 0 0 0 0 0 0 0 0
name2 1 0 1 0 1 1 1 0 0 0 0 0
name3 0 0 0 0 0 0 1 1 1 1 1 1
于 2013-05-16T20:50:30.137 回答
2

这是一种方法：

qw = function(s) unlist(strsplit(s,'[[:blank:]]+'))
name1 <- qw("a b c d")
name2 <- qw("a c e g i")
name3 <- qw("t j i m n z")

rows <- qw("name1 name2 name3")
cols <- sort(unique(c(name1,name2,name3)))

nr <- length(rows)
nc <- length(cols)

outmat <- matrix(0,nr,nc,dimnames=list(rows,cols))

for (i in rows){
    outmat[i,get(i)] <- 1
}

#       a b c d e g i j m n t z
# name1 1 1 1 1 0 0 0 0 0 0 0 0
# name2 1 0 1 0 1 1 1 0 0 0 0 0
# name3 0 0 0 0 0 0 1 1 1 1 1 1

该功能qw并不是真正必需的，但在您的示例中更易于阅读。

于 2013-05-16T20:54:57.727 回答
1

更新：更快的选择

通过使用矩阵索引，您将获得最佳速度。这是一个示例（带有注释，因此您可以看到发生了什么）。

## Assuming this is your starting data
dat <- read.table(text="name1  a b c d  NA NA\nname2  a c e g i NA\nname3  t j i m n z")
rownames(dat) <- dat$V1 
dat$V1 <- NULL

## Convert the data.frame into a single character vector
A <- unlist(lapply(dat, as.character), use.names = FALSE)

## Identify the unique levels
levs <- sort(unique(na.omit(A)))

## Get the index position for the Row/Column combination
##   that needs to be recoded as "1"
Rows <- rep(sequence(nrow(dat)), ncol(dat))
Cols <- match(A, levs)

## Create an empty matrix
m <- matrix(0, nrow = nrow(dat), ncol = length(levs),
            dimnames = list(rownames(dat), levs))

## Use matrix indexing to replalce the relevant values with 1
m[cbind(Rows, Cols)] <- 1L
m
#       a b c d e g i j m n t z
# name1 1 1 1 1 0 0 0 0 0 0 0 0
# name2 1 0 1 0 1 1 1 0 0 0 0 0
# name3 0 0 0 0 0 0 1 1 1 1 1 1

基准

在创建30000data.table行版本的初始data.frame. 结果如下：

dat2 <- dat ## A backup
dat <- do.call(rbind, replicate(10000, dat, simplify = FALSE))
dim(dat)
# [1] 30000     6

library(microbenchmark)
microbenchmark(AM(), AMDT(), RS(), times = 10)
# Unit: milliseconds
#    expr        min         lq     median        uq       max neval
#    AM()   44.30915   56.21873   57.95815   86.1518  265.3053    10
#  AMDT()  231.71928  245.64236  291.19601  376.8983  515.8216    10
#    RS() 4414.01127 4698.47293 4731.72877 5484.6185 5726.8092    10

矩阵索引显然获胜，但鉴于data.table完成工作的语法简洁，我更喜欢这种方法！@Arun 将 Hadley 的作品与“reshape2”一起移植到了data.table!!!

原始答案

这是一个“data.table”替代方案。它至少需要 1.8.11 版的“data.table”。

加载所需的包

library(data.table)
library(reshape2)
packageVersion("data.table")
# [1] ‘1.8.11’

melt和cast你的data.table

DT <- data.table(dat, keep.rownames=TRUE)
dcast.data.table(melt(DT, id.vars="rn"), rn ~ value)
# Aggregate function missing, defaulting to 'length'
#       rn NA a b c d e g i j m n t z
# 1: name1  0 1 1 1 1 0 0 0 0 0 0 0 0
# 2: name2  0 1 0 1 0 1 1 1 0 0 0 0 0
# 3: name3  0 0 0 0 0 0 0 1 1 1 1 1 1
于 2013-12-19T18:34:32.440 回答