是否有R函数将矢量应用于DF中的列并创建新列？

发布于 2025-02-12 13:39:42 字数 1120 浏览 1 评论 0原文

我正在使用tidyverse对房价数据集进行一些分析。随着时间的流逝，我有房价，我想通过应用下降付款的向量来为每个首付百分比输出新列来创建新的专栏。

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

我希望通过仅通过down_payments＆lt; - c（0.2，0.1，0.05）的向量来执行此操作，但我无法弄清楚一个直接执行此操作的方法，而无需手动突变每一列。此外，我可能想在某些计算中包括其他带有输入的向量，并同时使用新的付款向量和新向量，并为该金额提供新的列。

谢谢！

原文

I am using tidyverse to do some analysis to a dataset of home prices. I have home prices over time, and I would like to create new columns by applying a vector of down payments to output a new column for each down payment percentage.

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

I'd like to be able to do this for any number of down payment inputs, by just passing a vector like down_payments <- c(0.2, 0.1, 0.05), but I cannot figure out a straightforward way to do this without manually mutating every column. Additionally, I might want to include other vectors with inputs, and use both the down payment vector and the new vector in some calculation, and have a new column for that amount.

Thanks!

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听风念你 2025-02-19 13:39:42

在基本中，r使用外部函数或简单地％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

in Base R use the outer function or simply %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

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不…忘初心 2025-02-19 13:39:42

怎么样：

library(dplyr)
library(purrr)
library(tidyr)
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)

downs <- c(.2, .1, .05)

map(downs, function(x){
  home_vals %>% 
    mutate(!!sym(paste0("DownPayment_", round(x*100))) := prices*x)
}) %>% 
  do.call(coalesce, .)
#>         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
#> 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

^由

How about this:

library(dplyr)
library(purrr)
library(tidyr)
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)

downs <- c(.2, .1, .05)

map(downs, function(x){
  home_vals %>% 
    mutate(!!sym(paste0("DownPayment_", round(x*100))) := prices*x)
}) %>% 
  do.call(coalesce, .)
#>         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
#> 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

^{Created on 2022-06-30 by the reprex package (v2.0.1)}

回复收藏 0 原文

飘过的浮云 2025-02-19 13:39:42

使用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

With 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

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