将二进制列转换为R中的1行

我对R很陌生，到目前为止只有基本技能，即使我检查了

melt（）

和

gather（）

之类的函数，它们对我还是不起作用。 我想做的是转换这些数据（考虑到所有关于拥有住房/租赁和无家可归的选项只有

1

和

0

，并且你不能拥有超过

1

（你不能同时拥有租赁和无家可归者）

例如

我希望这些数据如下所示：

Passenger ID /// Housing       /// Age /// Gender
1                Has own house     21      Male
2                Renting           24      Female

当谈到预测时，请你告诉我，上述方法（含二元因素）在速度方面是否更有效，还是将所有因素都纳入一列将是更好的解决方案？

试试这个

library(tidyverse)
# importing your data
df <- read_table("Passenger_ID    Has_Own_House   Renting   Homeless   Age Gender
1                  1                 0           0     21      Male
2                  0                 1           0     24     Female")

输出为：

# A tibble: 2 x 4
#   Passenger_ID   Age Gender Housing      
#          <int> <int> <chr>  <chr>        
# 1            1    21 Male   Has_Own_House
# 2            2    24 Female Renting   

#一个tible:2 x 4
#乘客身份证年龄性别住房
#                      
#男性拥有自己的房子
#2 24女性租赁

在带ifelse的基本R中：

# Load Data
dat <- structure(list(Passenger_ID = 1:2, Has_Own_House = c(1L, 0L),
         Renting = 0:1, Homeless = c(0L, 0L), Age = c(21L, 24L), Gender = structure(c(2L,
         1L), .Label = c("Female", "Male"), class = "factor")), .Names = c("Passenger_ID",
         "Has_Own_House", "Renting", "Homeless", "Age", "Gender"), class = "data.frame", row.names = c(NA,
         -2L))

# Assign new column "Housing" based on testing nested ifelse statements:

dat2 <- within(dat, Housing <- ifelse(Has_Own_House==1, "Has_Own_House",
                           ifelse(Renting==1, "Renting",
                           ifelse(Homeless==1, "Homeless", NA))))

# Remove extra columns
dat2$Has_Own_House <- NULL
dat2$Renting <- NULL
dat2$Homeless <- NULL

---

在base R中，您可以通过将返回适当列名的函数应用于数据帧的所有行（

1

argument）来简单地将一个新列分配到一行中（由于

which

），其中值为1：

---

您应该提供一个可复制的示例。请参阅。如果

df

是您的data.frame，您可以制作类似于

df$HOUSING=apply（df[，2:4]，1，function（x）names（df）[2:4][which（x==1）]）的内容。

# A tibble: 2 x 4
#   Passenger_ID   Age Gender Housing      
#          <int> <int> <chr>  <chr>        
# 1            1    21 Male   Has_Own_House
# 2            2    24 Female Renting   

# Load Data
dat <- structure(list(Passenger_ID = 1:2, Has_Own_House = c(1L, 0L),
         Renting = 0:1, Homeless = c(0L, 0L), Age = c(21L, 24L), Gender = structure(c(2L,
         1L), .Label = c("Female", "Male"), class = "factor")), .Names = c("Passenger_ID",
         "Has_Own_House", "Renting", "Homeless", "Age", "Gender"), class = "data.frame", row.names = c(NA,
         -2L))

# Assign new column "Housing" based on testing nested ifelse statements:

dat2 <- within(dat, Housing <- ifelse(Has_Own_House==1, "Has_Own_House",
                           ifelse(Renting==1, "Renting",
                           ifelse(Homeless==1, "Homeless", NA))))

# Remove extra columns
dat2$Has_Own_House <- NULL
dat2$Renting <- NULL
dat2$Homeless <- NULL

>dat2

Passenger_ID Age Gender       Housing
       1  21   Male Has_Own_House
       2  24 Female       Renting

df = data.frame('Passenger ID' = 1:5,
                'Has Own House' = c(1,0,0,1,0),
                'Renting' = c(0,1,0,0,0),
                'Homeless' = c(0,0,1,0,1),
                'Age'=21:25,
                'Gender' = c('Male', 'Female', 'Male', 'Female', 'Male'))


df$HOUSING = apply(df[, 2:4], 1, function(x) names(df)[2:4][which(x==1)])
df
#   Passenger.ID Has.Own.House Renting Homeless Age Gender       HOUSING
# 1            1             1       0        0  21   Male Has.Own.House
# 2            2             0       1        0  22 Female       Renting
# 3            3             0       0        1  23   Male      Homeless
# 4            4             1       0        0  24 Female Has.Own.House
# 5            5             0       0        1  25   Male      Homeless