是否有一个R函数将向量应用于df中的列，并创建新列？

Question

我正在使用tidyverse对房价数据进行一些分析。我有房价随着时间的推移，我想创建新的列，通过应用一个向量的首付，以输出一个新的栏目为每个首付百分比。

set.seed(100)
dates <- seq.Date(from = as.Date('2010-01-01'), to = as.Date('2010-12-01'), by = 'months')
prices <- rnorm(12, mean = 100000, sd = 5000)
 
home_vals <- data.frame(dates, prices)
 
out <- home_vals %>%
  mutate(
    DownPayment_20 = 0.2 * prices
    ,DownPayment_10 = 0.1 * prices
    ,DownPayment_5 = 0.05 * prices
  )

head(out)
       dates    prices DownPayment_20 DownPayment_10 DownPayment_5
1 2010-01-01  97489.04       19497.81       9748.904      4874.452
2 2010-02-01 100657.66       20131.53      10065.766      5032.883
3 2010-03-01  99605.41       19921.08       9960.541      4980.271
4 2010-04-01 104433.92       20886.78      10443.392      5221.696
5 2010-05-01 100584.86       20116.97      10058.486      5029.243
6 2010-06-01 101593.15       20318.63      10159.315      5079.658

我想通过传递像down_payments <- c(0.2, 0.1, 0.05)这样的向量，对任何数量的首付款输入都能做到这一点，但是，如果不手动修改每一列，我就无法找到一种简单的方法来做到这一点。此外，我可能想包括其他向量的输入，并使用首付向量和新向量在某些计算，并有一个新的列为该数额。

谢谢!

Answer 1

在Base中使用outer函数或简单的%o%

cbind(home_vals, home_vals$prices %o% setNames(downs, paste0('DownPayment_', downs)))

        dates    prices DownPayment_0.2 DownPayment_0.1 DownPayment_0.05
1  2010-01-01  97489.04        19497.81        9748.904         4874.452
2  2010-02-01 100657.66        20131.53       10065.766         5032.883
3  2010-03-01  99605.41        19921.08        9960.541         4980.271
4  2010-04-01 104433.92        20886.78       10443.392         5221.696
5  2010-05-01 100584.86        20116.97       10058.486         5029.243
6  2010-06-01 101593.15        20318.63       10159.315         5079.658
7  2010-07-01  97091.05        19418.21        9709.105         4854.552
8  2010-08-01 103572.66        20714.53       10357.266         5178.633
9  2010-09-01  95873.70        19174.74        9587.370         4793.685
10 2010-10-01  98200.69        19640.14        9820.069         4910.034
11 2010-11-01 100449.43        20089.89       10044.943         5022.472
12 2010-12-01 100481.37        20096.27       10048.137         5024.069

Answer 2

用data.table

library(data.table)

setDT(home_vals)
home_vals[,(paste0("DownPayment_",down_payments*100)):=lapply(down_payments,\(x) x*prices)][]

         dates    prices DownPayment_20 DownPayment_10 DownPayment_5
        <Date>     <num>          <num>          <num>         <num>
 1: 2010-01-01  97489.04       19497.81       9748.904      4874.452
 2: 2010-02-01 100657.66       20131.53      10065.766      5032.883
 3: 2010-03-01  99605.41       19921.08       9960.541      4980.271
 4: 2010-04-01 104433.92       20886.78      10443.392      5221.696
 5: 2010-05-01 100584.86       20116.97      10058.486      5029.243
 6: 2010-06-01 101593.15       20318.63      10159.315      5079.658
 7: 2010-07-01  97091.05       19418.21       9709.105      4854.552
 8: 2010-08-01 103572.66       20714.53      10357.266      5178.633
 9: 2010-09-01  95873.70       19174.74       9587.370      4793.685
10: 2010-10-01  98200.69       19640.14       9820.069      4910.034
11: 2010-11-01 100449.43       20089.89      10044.943      5022.472
12: 2010-12-01 100481.37       20096.27      10048.137      5024.069