R 从布尔表中保留具有现有值的列表
具有如下布尔帧：R 从布尔表中保留具有现有值的列表,r,R,具有如下布尔帧： dftest <- data.frame(id = c("A1","A2","A3","A4","A5"), stock1= c(1,0,0,1,0), stock2 = c(1,1,0,1,0), stock3 = c(0,1,0,0,0)) id stock1 stock2 stock3 A1 1 1 0 A2 0 1 1 A3 0 0 0 A4 1 1
dftest <- data.frame(id = c("A1","A2","A3","A4","A5"), stock1= c(1,0,0,1,0), stock2 = c(1,1,0,1,0), stock3 = c(0,1,0,0,0))
id stock1 stock2 stock3
A1 1 1 0
A2 0 1 1
A3 0 0 0
A4 1 1 0
A5 0 0 0
stock1 = c("A1,","A4")
stock2 = c("A1","A2","A4")
stock3 = c("A2")
一种具有
lappylapply(names(dftest)[-1], function(x) dftest[dftest[x] == 1, 'id'])
#[[1]]
#[1] A1 A4
#5 Levels: A1 A2 A3 ... A5
#[[2]]
#[1] A1 A2 A4
#5 Levels: A1 A2 A3 ... A5
#[[3]]
#[1] A2
#5 Levels: A1 A2 A3 ... A5
您的id列当前为factor类型。如果您不想看到输出中的级别，只需将其转换为字符，即
dftest$id一个具有lapply
和简单子集的解决方案：
lapply(names(dftest)[-1], function(x) dftest[dftest[x] == 1, 'id'])
#[[1]]
#[1] A1 A4
#5 Levels: A1 A2 A3 ... A5
#[[2]]
#[1] A1 A2 A4
#5 Levels: A1 A2 A3 ... A5
#[[3]]
#[1] A2
#5 Levels: A1 A2 A3 ... A5
您的id列当前为factor类型。如果您不想看到输出中的级别，只需将其转换为字符，即dftest$id另一个选项可以是使用mapply
，将1/0数据转换为逻辑数据，并将每列中相应的id
s子集，以获得id
s列表
mapply(`[`, dftest["id"], data.frame(dftest[-1] == 1), USE.NAMES = FALSE)
#[[1]]
#[1] "A1" "A4"
#[[2]]
#[1] "A1" "A2" "A4"
#[[3]]
#[1] "A2"
数据
dftest <- data.frame(id = c("A1","A2","A3","A4","A5"),
stock1= c(1,0,0,1,0),
stock2 = c(1,1,0,1,0),
stock3 = c(0,1,0,0,0), stringsAsFactors = FALSE)
dftest另一个选项可以是使用mapply
，将1/0数据转换为逻辑数据，并从每列中子集相应的id
s，以获得id
s的列表
mapply(`[`, dftest["id"], data.frame(dftest[-1] == 1), USE.NAMES = FALSE)
#[[1]]
#[1] "A1" "A4"
#[[2]]
#[1] "A1" "A2" "A4"
#[[3]]
#[1] "A2"
数据
dftest <- data.frame(id = c("A1","A2","A3","A4","A5"),
stock1= c(1,0,0,1,0),
stock2 = c(1,1,0,1,0),
stock3 = c(0,1,0,0,0), stringsAsFactors = FALSE)
dftest玩弄数据。表
：
library(data.table)
data.table::melt(dftest, id.vars = "id")[as.logical(value), list(list(id)), by = variable][[2]]
[[1]]
[1] A1 A4
Levels: A1 A2 A3 A4 A5
[[2]]
[1] A1 A2 A4
Levels: A1 A2 A3 A4 A5
[[3]]
[1] A2
Levels: A1 A2 A3 A4 A5
玩弄数据。表
：
library(data.table)
data.table::melt(dftest, id.vars = "id")[as.logical(value), list(list(id)), by = variable][[2]]
[[1]]
[1] A1 A4
Levels: A1 A2 A3 A4 A5
[[2]]
[1] A1 A2 A4
Levels: A1 A2 A3 A4 A5
[[3]]
[1] A2
Levels: A1 A2 A3 A4 A5
这里有一个带有tidyverse
library(tidyverse)
gather(dftest, key, val, -id) %>%
filter(val == 1) %>%
group_split(key) %>%
map(pluck, 'id')
#[[1]]
#[1] A1 A4
#Levels: A1 A2 A3 A4 A5
#[[2]]
#[1] A1 A2 A4
#Levels: A1 A2 A3 A4 A5
#[[3]]
#[1] A2
#Levels: A1 A2 A3 A4 A5
或者使用map
map(dftest[-1], ~ as.character(dftest$id)[!! .x])
#$stock1
#[1] "A1" "A4"
#$stock2
#[1] "A1" "A2" "A4"
#$stock3
#[1] "A2"
或者使用map
和discard
map(dftest[-1], ~ discard(dftest$id, !.x))
或者使用split
with(subset(cbind(dftest['id'], stack(dftest, -id)),
values == 1), split(as.character(id), ind))
#$stock1
#[1] "A1" "A4"
#$stock2
#[1] "A1" "A2" "A4"
#$stock3
#[1] "A2"
或者通过复制“id”列和名称来使用split
split(dftest[col(dftest[-1])][c(!!dftest[-1])],
names(dftest)[-1][col(dftest[-1])][c(!!dftest[-1])])
这里有一个带有tidyverse
library(tidyverse)
gather(dftest, key, val, -id) %>%
filter(val == 1) %>%
group_split(key) %>%
map(pluck, 'id')
#[[1]]
#[1] A1 A4
#Levels: A1 A2 A3 A4 A5
#[[2]]
#[1] A1 A2 A4
#Levels: A1 A2 A3 A4 A5
#[[3]]
#[1] A2
#Levels: A1 A2 A3 A4 A5
或者使用map
map(dftest[-1], ~ as.character(dftest$id)[!! .x])
#$stock1
#[1] "A1" "A4"
#$stock2
#[1] "A1" "A2" "A4"
#$stock3
#[1] "A2"
或者使用map
和discard
map(dftest[-1], ~ discard(dftest$id, !.x))
或者使用split
with(subset(cbind(dftest['id'], stack(dftest, -id)),
values == 1), split(as.character(id), ind))
#$stock1
#[1] "A1" "A4"
#$stock2
#[1] "A1" "A2" "A4"
#$stock3
#[1] "A2"
或者通过复制“id”列和名称来使用split
split(dftest[col(dftest[-1])][c(!!dftest[-1])],
names(dftest)[-1][col(dftest[-1])][c(!!dftest[-1])])
或者lappy（dftest[-1]，FUN=function（x）dftest$id[as.logical（x）]）
或者lappy（dftest[-1]，FUN=function（x）dftest$id[as.logical（x）]