我的数据如下所示：
ID Diagnosis_1 Diagnosis_2 Diagnosis_3 Diagnosis_4
A 1 0 0 0
A 1 0 0 0
A 1 0 0 0
B 0 1 0 0
C 0 0 0 1
C 0 1 0 0
D 0 0 0 1
E 0 0 1 0
E 0 1 0 0
E 0 0 1 0
Diagnosis_1：Diagnosis_4都是二进制，表示诊断的存在（1）或不存在（0）。我想要做的是创建一个如下所示的数据框：
ID Diagnosis
A 1
A 1
A 1
B 2
C 4
C 2
D 4
E 3
E 2
E 3
无论我阅读有关reshape / reshape2 / tidyr的文档多少次，我都无法绕过他们的实现。
我可以使用dplyr的mutate解决我的问题，但这是一个耗时，迂回的方式来实现我的目标。
编辑：编辑的数据更真实地代表我的实际数据框。
答案 0 :(得分：6)
尝试矩阵乘法：
nc <- ncol(DF)
data.frame(ID = DF$ID, Diagnosis = as.matrix(DF[-1]) %*% seq(nc-1))
，并提供：
ID Diagnosis
1 A 1
2 B 2
3 C 2
4 D 4
5 E 3
注意：我们将此用作输入：
Lines <- "ID Diagnosis_1 Diagnosis_2 Diagnosis_3 Diagnosis_4
A 1 0 0 0
B 0 1 0 0
C 0 1 0 0
D 0 0 0 1
E 0 0 1 0"
DF <- read.table(text = Lines, header = TRUE)
答案 1 :(得分：3)
您可以尝试max.col获取每行的列索引。
data.frame(ID=df1$ID, Diagnosis=max.col(df1[-1]))
# ID Diagnosis
#1 A 1
#2 B 2
#3 C 2
#4 D 4
#5 E 3
获取索引的其他选项是
unname(which(t(df1[-1])!=0, arr.ind=TRUE)[,1])
#[1] 1 2 2 4 3
set.seed(25)
m1 <- matrix(sample(0:1, 1e8*5, replace=TRUE, prob=c(0.9, 0.1)), ncol=5)
m2 <- m1[rowSums(m1)==1,]
dim(m2)
#[1] 32812201 5
set.seed(395)
df1 <- data.frame(ID= sample(LETTERS, nrow(m2), replace=TRUE), m2,
stringsAsFactors=FALSE)
colnames(df1)[-1] <- paste('X', 1:5, sep="_")
Grothendieck <- function() {nc <- ncol(df1)
data.frame(ID = df1$ID, Diagnosis =
as.matrix(df1[-1]) %*% seq(nc-1))}
akrun <- function() {data.frame(ID=df1$ID,
Diagnosis=max.col(df1[-1], 'first'))}
ananda <- function() {df1 %>%
gather(var, val, -ID) %>%
separate(var, into = c("var", "value")) %>%
filter(val == 1) %>%
select(ID, value)}
system.time(akrun())
# user system elapsed
# 3.690 0.396 4.085
system.time(Grothendieck())
# user system elapsed
# 3.121 0.459 3.581
在'df1'的较小子集上尝试dplyr解决方案，即。 1e6行
df1 <- df1[1:1e6,]
system.time(ananda())
# user system elapsed
# 6.279 0.177 6.454
使用microbenchmark，
library(microbenchmark)
microbenchmark(akrun(), Grothendieck(), unit='relative', times=20L)
#Unit: relative
# expr min lq mean median uq max neval cld
# akrun() 1.019108 1.252443 1.084306 1.180743 1.16463 0.6928535 20 a
#Grothendieck() 1.000000 1.000000 1.000000 1.000000 1.00000 1.0000000 20 a
df1 <- structure(list(ID = c("A", "B", "C", "D", "E"),
Diagnosis_1 = c(1L,
0L, 0L, 0L, 0L), Diagnosis_2 = c(0L, 1L, 1L, 0L, 0L),
Diagnosis_3 = c(0L,
0L, 0L, 0L, 1L), Diagnosis_4 = c(0L, 0L, 0L, 1L, 0L)),
.Names = c("ID",
"Diagnosis_1", "Diagnosis_2", "Diagnosis_3", "Diagnosis_4"),
class = "data.frame", row.names = c(NA, -5L))
答案 2 :(得分：3)
由于您提及＆＃34; reshape2＆＃34;，＆＃34; tidyr＆＃34;以及相关工具，以下是一些需要考虑的选项：
## Using "tidyr" and "dplyr"
library(dplyr)
library(tidyr)
df1 %>%
gather(var, val, -ID) %>%
separate(var, into = c("var", "value")) %>%
filter(val == 1) %>%
select(ID, value)
# ID value
# 1 A 1
# 2 B 2
# 3 C 2
# 4 E 3
# 5 D 4
## Getting half-way there with "melt" from "reshape2"
library(reshape2)
melt(replace(df1, df1 == 0, NA), id.vars = "ID", na.rm = TRUE)
# ID variable value
# 1 A Diagnosis_1 1
# 7 B Diagnosis_2 1
# 8 C Diagnosis_2 1
# 15 E Diagnosis_3 1
# 19 D Diagnosis_4 1
考虑到您的更新，您只需添加辅助ID：
library(dplyr)
library(tidyr)
mydf %>%
group_by(ID) %>%
mutate(ID2 = row_number()) %>%
gather(var, val, Diagnosis_1:Diagnosis_4) %>%
separate(var, into = c("var", "value")) %>%
filter(val == 1) %>%
arrange(ID, ID2)
# Source: local data frame [10 x 5]
#
# ID ID2 var value val
# 1 A 1 Diagnosis 1 1
# 2 A 2 Diagnosis 1 1
# 3 A 3 Diagnosis 1 1
# 4 B 1 Diagnosis 2 1
# 5 C 1 Diagnosis 4 1
# 6 C 2 Diagnosis 2 1
# 7 D 1 Diagnosis 4 1
# 8 E 1 Diagnosis 3 1
# 9 E 2 Diagnosis 2 1
# 10 E 3 Diagnosis 3 1