从函数向data.frame添加列

I am in the process of building a function for drug efficacy that depends both on age of administration and years since administration. The idea is to figure out the efficacy of a drug at the time you get it and then the decrease in efficacy as time goes on; for example, get drug at age 50, and then follow each year and see how efficacy changes. A modified version of the function is below.

我正在建立药物功效的功能，这取决于给药年龄和给药后的年数。这个想法是在你得到药物时弄清楚药物的疗效，然后随着时间的推移降低药效;例如，在50岁时获得药物，然后每年跟随，看看疗效如何变化。该功能的修改版本如下。

What I would like to have at the end of this is a data.frame for drug efficacy with each column being the age at which the drug was administered (e.g., 50, 51, 52, ... , 85) and each row being the efficacy at each year after dosage (e.g., 1, 2, ... , 10).

我想在最后得到的是药物功效的数据框架，每列是药物的给药年龄（例如，50,51,52，...，85），每一行都是剂量后每年的疗效（例如，1,2，......，10）。

I have tried a variety of things but have gotten basically nowhere. This code below is the closest I have got, but is still way off. If need be, the function can be restructured to two parts if that's easier.

我尝试了各种各样的东西，但基本上没有。下面这段代码是我最接近的代码，但仍然没有。如果需要，可以将功能重构为两个部分，如果这更容易的话。

Also if there is an easy way to name the columns that would be great, but that's the least of my concerns right now.

此外，如果有一种简单的方法可以将列命名为伟大，但这是我现在最不关心的问题。

Any help would be greatly appreciated. Thank you

任何帮助将不胜感激。谢谢

b2 <- 0.28
a1 <- 0.02
a2 <- 0.04
e  <- exp(1)

p_tot   <- 19132 + 18827 + 14505 
p1   <- 19132 
p2   <- 18827 
p3   <- 14505 

years_i <- seq(1, 10, by = 1)
age_j   <- seq(50, 85, by = 1)

VEij <- function(age_j, years_i){
  b1j <-   ((a1 - a2*age_j)*p_tot + b2*(p1 + 2*p2 + 3*p3)) / p_tot
  Efficacy <- b1j - b2*years_i
  return(Efficacy)
}

EffDF <- data.frame(years_i)

for (x in age_j){
  for (y in years_i){
    EffDF <- cbind(EffDF, VEij(x, y))
  }
}

1 个解决方案

#1

I am guessing this is what you want.

我猜这就是你想要的。

EffDF <- data.frame(sapply(as.list(age_j), VEij, years_i)
names(EffDF) <- age_j

Truncated output:

截断输出：

> EffDF
          50        51        52        53        54        55        56        57
1  -1.724694 -1.764694 -1.804694 -1.844694 -1.884694 -1.924694 -1.964694 -2.004694
2  -2.004694 -2.044694 -2.084694 -2.124694 -2.164694 -2.204694 -2.244694 -2.284694
3  -2.284694 -2.324694 -2.364694 -2.404694 -2.444694 -2.484694 -2.524694 -2.564694
4  -2.564694 -2.604694 -2.644694 -2.684694 -2.724694 -2.764694 -2.804694 -2.844694
5  -2.844694 -2.884694 -2.924694 -2.964694 -3.004694 -3.044694 -3.084694 -3.124694
6  -3.124694 -3.164694 -3.204694 -3.244694 -3.284694 -3.324694 -3.364694 -3.404694
7  -3.404694 -3.444694 -3.484694 -3.524694 -3.564694 -3.604694 -3.644694 -3.684694
8  -3.684694 -3.724694 -3.764694 -3.804694 -3.844694 -3.884694 -3.924694 -3.964694
9  -3.964694 -4.004694 -4.044694 -4.084694 -4.124694 -4.164694 -4.204694 -4.244694
10 -4.244694 -4.284694 -4.324694 -4.364694 -4.404694 -4.444694 -4.484694 -4.524694

#1