R函数仅打印向量的最后一个值

问题描述：

感谢您对此功能的帮助，该功能应该:

Thank you for your help with this function, which should :

输入一个特定值
将值插入函数
采用向量中生成的一组其他值
为向量的每个元素计算一个值
返回包含矢量和计算值的数据框.

这是我尝试过的:

rate<-function(Y2) {
   ran<-seq(0.001,1,0.001)
   for(i in ran) {
     calculated<-as.vector(Y2/(1+i)+Y2/(1+i)^2+Y2/(1+i)^3+Y2/(1+i)^4)
     tableau<-data.frame(ran,calculated)
   }
   return(tableau)
}

使用res<-rate(500)进行测试时，仅最后一个值返回1000次:

When testing with res<-rate(500), only the last value is returned 1000 times:

...

ran calculated
1   0.001   468.75
2   0.002   468.75
3   0.003   468.75
...

996  0.996     468.75
997  0.997     468.75
998  0.998     468.75
999  0.999     468.75
1000 1.000     468.75

我的循环怎么了?

答

每次循环时，您要将as.vector(...)计算的输出分配给相同的变量.然后，您将在每次循环时构建一个名为tableau的data.frame.您只返回最后一次迭代.如果要保存每次迭代，则需要索引一些内容:

You're assigning the output of your as.vector(...) calculation to the same variable each time you loop. Then you're building a data.frame, named tableau each time you loop. You're only returning the last iteration. If you want to save each iteration, you'll need to index into something:

res[n] <- as.vector(...)

或更高级的R-ish版本，请使用Apply系列之一(特别是lapply)，并且完全没有循环:

Or the more R-ish version, use one of the apply family (specifically lapply) and no loop at all:

rate <- function(Y2) {
    ran <- seq(0.001, 1, 0.001)
    result <- lapply(ran, 
                     function(i) data.frame(ran = i, 
                                            calculated = as.vector(Y2/(1+i)+Y2/(1+i)^2+Y2/(1+i)^3+Y2/(1+i)^4)))

    return (do.call(rbind, result))
}

话虽如此，没有理由要循环或套用功能.使用R是矢量化的事实:

With that said, there is no reason for a loop or an apply function. Use the fact that R is vectorized:

ran <- seq(0.001, 1, 0.001)
Y2 <- 500

calculated <- as.vector(Y2/(1+ran)+Y2/(1+ran)^2+Y2/(1+ran)^3+Y2/(1+ran)^4)

result <- data.frame(ran, calculated)

all.equal(result, rate(500))
# [1] TRUE

R函数仅打印向量的最后一个值

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