sapply()
,使用 ==
检查是否存在并使用 as.integer()
将其强制转换为二进制。cbind(df[1:2], sapply(levels(df$c), function(x) as.integer(x == df$c)), df[4])
# a b Pink Red Rose d
# 1 1 1 0 0 1 2
# 2 2 1 1 0 0 3
# 3 3 2 0 1 0 4
但是,由于您有一百万行数据,您可能希望选择 data.table。
library(data.table)
setDT(df)[, c(levels(df$c), "c") :=
c(lapply(levels(c), function(x) as.integer(x == c)), .(NULL))]
提供
df
# a b d Pink Red Rose
# 1: 1 1 2 0 0 1
# 2: 2 1 3 1 0 0
# 3: 3 2 4 0 1 0
如果需要,您可以使用setcolorder(df, c(1, 2, 4:6, 3))
重置列的顺序。
[.data.table
(setDT(df), , :=
(levels(c), lapply(levels(c), :
LHS of := isn't column names ('character') or positions ('integer' or 'numeric')) - prasanth您可以通过重新塑形来完成此操作:
library(dplyr)
library(tidyr)
df %>%
mutate(value = 1,
c = paste0("Is", c)) %>%
spread(c, value, fill = 0)
c
没有重复实例的情况。 - timothy.s.lau为了完整性,基于此解决方案(https://dev59.com/D5Hea4cB1Zd3GeqPq5GN#33990970),这里提供了使用最新的tidyverse包进行更新。
library(tidyverse)
df %>%
mutate(value = 1,
c = paste0("Is", c)) %>%
pivot_wider(names_from = c,
values_from = value,
values_fill = 0)
expand_factor <- function(df,variable){
variable = as.name(variable)
paste0('~ ',variable,' -1',collapse = '') %>%
as.formula ->formulae
current.na.action <- options('na.action')
options(na.action='na.pass')
expanded<-model.matrix(data=df,object = formulae)
options(na.action=current.na.action)
colnames(expanded) <-gsub(replacement = 'is_',x = colnames(expanded),pattern=variable)
expanded %>%
tbl_df %>%
mutate_each(funs(as.integer)) ->expanded
return(bind_cols(df,expanded))
}
library(dplyr)
df <-data_frame(x = iris$Species,y = iris$Petal.Width)
df <- rbind(data_frame(x=NA,y = NA),df)
df %>%
expand_factor('x')
> df %>%
+ expand_factor('x')
# A tibble: 151 <U+00D7> 5
x y is_setosa is_versicolor is_virginica
<chr> <dbl> <int> <int> <int>
1 <NA> NA NA NA NA
2 setosa 0.2 1 0 0
3 setosa 0.2 1 0 0
4 setosa 0.2 1 0 0
5 setosa 0.2 1 0 0
6 setosa 0.2 1 0 0
7 setosa 0.4 1 0 0
8 setosa 0.3 1 0 0
9 setosa 0.2 1 0 0
10 setosa 0.2 1 0 0
# ... with 141 more rows
dummy <- function(df) {
NUM <- function(dataframe)dataframe[,sapply(dataframe,is.numeric)]
FAC <- function(dataframe)dataframe[,sapply(dataframe,is.factor)]
require(ade4)
if (is.null(ncol(NUM(df)))) {
DF <- data.frame(NUM(df), acm.disjonctif(FAC(df)))
names(DF)[1] <- colnames(df)[which(sapply(df, is.numeric))]
} else {
DF <- data.frame(NUM(df), acm.disjonctif(FAC(df)))
}
return(DF)
}
model.matrix
。类似这样的代码model.matrix(~ 0 + c, df)
可以实现大部分你所需求的功能。然后,只需要将其与其他列绑定即可。 - A5C1D2H2I1M1N2O1R2T1