Question

我的数据如下：

> dput(head(updata[,c(1,4,5,6)],10))
structure(list(vehicle = c(2L, 5L, 8L, 9L, 10L, 12L, 13L, 14L, 
18L, 20L), class = c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 2L), 
    velocity = c(42.48, 39.81, 40.63, 47.11, 41.2, 38.92, 35.38, 
    38.62, 38.94, 43.24), lane = c(2L, 4L, 4L, 3L, 1L, 1L, 2L, 
    5L, 5L, 3L)), .Names = c("vehicle", "class", "velocity", 
"lane"), row.names = c(32L, 707L, 1392L, 1772L, 2125L, 2501L, 
2896L, 3262L, 3726L, 3941L), class = "data.frame")

我想通过'class'和'lane'找到'velocity'的累积分布。所以我使用了以下2个函数和ddply：

speedu <- function(x){
  hi <- hist(x)
  speedmph=round(hi$breaks*0.68,1)}
cumdistu <- function(x){
  hi <- hist(x)
  prob=c(0, round(cumsum(hi$counts)/sum(hi$counts),digits=2))}
spdistu <- ddply(updata, c('lane', 'class'), summarise,speed=speedu(velocity),cprob=cumdist(velocity))

在这里，我实际上并不需要2个功能。只有一个函数可以找到速度中断和累积概率，但问题是function不给出2个输出，只给出输出中的最后一个变量。我在函数中尝试list(speedmph, prob)，但输出包含一行中的所有速度。使用上面的配置，我可以得到以下输出，但想知道是否有一种简单的方法来实现相同的目的：

> dput(spdistu)
structure(list(lane = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 
5L), class = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 
1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L), 
    speed = c(6.8, 13.6, 20.4, 27.2, 34, 40.8, 0, 3.4, 6.8, 10.2, 
    13.6, 17, 20.4, 23.8, 27.2, 30.6, 34, 37.4, 27.2, 34, 13.6, 
    20.4, 27.2, 34, 40.8, 3.4, 6.8, 10.2, 13.6, 17, 20.4, 23.8, 
    27.2, 30.6, 34, 37.4, 20.4, 21.8, 23.1, 24.5, 6.8, 10.2, 
    13.6, 17, 20.4, 23.8, 27.2, 30.6, 34, 17.7, 19, 20.4, 21.8, 
    23.1, 24.5, 25.8, 27.2, 28.6, 23.8, 24.5, 25.2, 25.8, 26.5, 
    27.2, 0, 3.4, 6.8, 10.2, 13.6, 17, 20.4, 23.8, 27.2, 30.6, 
    34, 37.4, 13.6, 17, 20.4, 23.8, 27.2, 30.6, 23.1, 24.5, 25.8, 
    27.2, 28.6, 29.9, 0, 3.4, 6.8, 10.2, 13.6, 17, 20.4, 23.8, 
    27.2, 30.6, 34, 37.4, 40.8, 13.6, 17, 20.4, 23.8, 27.2, 30.6
    ), cprob = c(0, 0.14, 0.29, 0.71, 0.86, 1, 0, 0.01, 0.04, 
    0.08, 0.14, 0.22, 0.32, 0.5, 0.74, 0.95, 0.99, 1, 0, 1, 0, 
    0.17, 0.67, 0.83, 1, 0, 0, 0.03, 0.07, 0.16, 0.3, 0.51, 0.8, 
    0.99, 1, 1, 0, 0.5, 0.5, 1, 0, 0.03, 0.05, 0.11, 0.21, 0.49, 
    0.85, 0.99, 1, 0, 0.07, 0.07, 0.27, 0.47, 0.67, 0.73, 0.93, 
    1, 0, 0.25, 0.25, 0.25, 0.75, 1, 0, 0, 0.01, 0.01, 0.06, 
    0.1, 0.17, 0.4, 0.75, 0.96, 1, 1, 0, 0.09, 0.09, 0.52, 0.91, 
    1, 0, 0.29, 0.43, 0.57, 0.57, 1, 0, 0, 0.01, 0.01, 0.02, 
    0.04, 0.06, 0.26, 0.66, 0.95, 1, 1, 1, 0, 0.07, 0.21, 0.43, 
    0.93, 1)), .Names = c("lane", "class", "speed", "cprob"), row.names = c(NA, 
-107L), class = "data.frame")

Answer 1

你在寻找像这样的东西

f <- function(x) {
  hi <- hist(x)
  speedmph=round(hi$breaks*0.68,1)
  prob=c(0, round(cumsum(hi$counts)/sum(hi$counts),digits=2))
  cbind(speedmph, prob)
}

spdistu <- ddply(updata, c('lane', 'class'), summarise, speed=f(velocity))

如何从R中的函数获得2个输出（＆＃39; list＆＃39;不起作用）？

1 个答案: