[我打算通过微基准比较两个基于算法的函数f1,f2之间的时序,该基准适用于rpois模拟数据集,其大小为:[1:7]矢量,由10 ^ seq(1,4,by = 0.5)给出,即:
[1] 10.00000 31.62278 100.00000 316.22777 1000.00000 3162.27766 10000.00000
我也在努力绘制它们,并使用了微基准测试所需的所有信息(即min,lq,平均值,中位数,uq和max-是的,除了expr和neval都需要它们)。我需要通过logp-log规模上的ggplot来实现这一点,它具有单个geom_point()和美观性,每个信息具有不同的颜色,这是我的代码:
library(ggplot2)
library(microbenchmark)
require(dplyr)
library(data.table)
datasetsizes<-c(10^seq(1,4,by=0.5))
f1_min<-integer(length(datasetsizes))
f1_lq<-integer(length(datasetsizes))
f1_mean<-integer(length(datasetsizes))
f1_median<-integer(length(datasetsizes))
f1_uq<-integer(length(datasetsizes))
f1_max<-integer(length(datasetsizes))
f2_min<-integer(length(datasetsizes))
f2_lq<-integer(length(datasetsizes))
f2_mean<-integer(length(datasetsizes))
f2_median<-integer(length(datasetsizes))
f2_uq<-integer(length(datasetsizes))
f2_max<-integer(length(datasetsizes))
for(loopvar in 1:(length(datasetsizes)))
{
s<-summary(microbenchmark(f1(rpois(datasetsizes[loopvar],10), max.segments=3L),f2(rpois(datasetsizes[loopvar],10), maxSegments=3)))
f1_min[loopvar] <- s$min[1]
f2_min[loopvar] <- s$min[2]
f1_lq[loopvar] <- s$lq[1]
f2_lq[loopvar] <- s$lq[2]
f1_mean[loopvar] <- s$mean[1]
f2_mean[loopvar] <- s$mean[2]
f1_median[loopvar] <- s$median[1]
f2_median[loopvar] <- s$median[2]
f1_uq[loopvar] <- s$uq[1]
f2_uq[loopvar] <- s$uq[2]
f1_max[loopvar] <- s$max[1]
f2_max[loopvar] <- s$max[2]
}
algorithm<-data.table(f1_min ,f2_min,
f1_lq, f2_lq,
f1_mean, f2_mean,
f1_median, f2_median,
f1_uq, f2_uq,
f1_max, cdpa_max, datasetsizes)
ggplot(algorithm, aes(x=algorithm,y=datasetsizes)) + geom_point(aes(color=algorithm)) + labs(x="N", y="Runtime") + scale_x_continuous(trans = 'log10') + scale_y_continuous(trans = 'log10')
我在每个步骤中调试我的代码,并使用“算法”的名称将计算值分配给数据表,效果很好。这是计算得出的游程,这些游程作为[1:7] vecs以及最后的数据集大小(以及1:7)一起传递到数据表中:
> algorithm
f1_min f2_min f1_lq f2_lq f1_mean f2_mean f1_median f2_median f1_uq f2_uq f1_max f2_max datasetsizes
1: 86.745000 21.863000 105.080000 23.978000 113.645630 24.898840 113.543500 24.683000 120.243000 25.565500 185.477000 39.141000 10.00000
2: 387.879000 52.893000 451.880000 58.359000 495.963480 66.070390 484.672000 62.061000 518.876500 66.116500 734.149000 110.370000 31.62278
3: 1608.287000 341.335000 1845.951500 382.062000 1963.411800 412.584590 1943.802500 412.739500 2065.103500 443.593500 2611.131000 545.853000 100.00000
4: 5.964166 3.014524 6.863869 3.508541 7.502123 3.847917 7.343956 3.851285 7.849432 4.163704 9.890556 5.096024 316.22777
5: 23.128505 29.687534 25.348581 33.654475 26.860166 37.576444 26.455269 37.080149 28.034113 41.343289 35.305429 51.347386 1000.00000
6: 79.785949 301.548202 88.112824 335.135149 94.248141 370.902821 91.577462 373.456685 98.486816 406.472393 135.355570 463.908240 3162.27766
7: 274.367776 2980.122627 311.613125 3437.044111 337.287131 3829.503738 333.544669 3820.517762 354.347487 4205.737045 546.996092 4746.143252 10000.00000
微基准计算的值符合预期,但ggplot引发此错误:
Don't know how to automatically pick scale for object of type data.table/data.frame. Defaulting to continuous.
Error: Aesthetics must be either length 1 or the same as the data (7): colour, x
无法解决此问题,任何人都可以让我知道可能出了什么问题,并针对该问题更正打印过程吗?
也在旁注中,我必须从计算的基准中分别提取所有值(最小,lq,平均值,中位数,uq,最大),因为我不能将其作为摘要中的数据表,因为它包含expr(表达式)和neval列。我可以使用
消除其中一列algorithm[,!"expr"] or algorithm[,!"neval"]
但是我不能一起消除两个,例如
algorithm[,!"expr",!"neval"] or algorithm[,!("expr","neval")] or algorithm[,!"expr","neval"]
-像这样的所有可能组合都不起作用(抛出“无效参数类型”错误)。
对此的任何可能的解决方法或解决方案以及绘图(主要内容)都将受到高度赞赏!
您的问题主要是由于您引用的是对象中不存在的ggplot公式中的algorithm
列。
根据您的提供,我可以执行以下操作:
algorithm$algorithm <- 1:nrow(algorithm)
ggplot(algorithm, aes(x=algorithm,y=datasetsizes)) + geom_point(aes(color=algorithm)) + labs(x="N", y="Runtime") +
scale_x_continuous(trans = 'log10') + scale_y_continuous(trans = 'log10')
并绘制此罚款:
编辑:让我们清理一下...
根据OP的要求,我已经整理了一下他的代码。
您可以做很多事情来提高代码的可读性,但是在这里我将重点放在实践方面。基本上,如果您知道变量最终会这样,则将变量连接到表中。您可以使用许多技巧来将值分配给正确的位置,您将在下面的代码中看到其中的一些。
library(ggplot2)
library(microbenchmark)
require(dplyr)
library(data.table)
datasetsizes<-c(10^seq(1,4,by=0.5))
l <- length(datasetsizes)
# make a vector with your different conditions
conds <- c('f1', 'f2')
# initalizing a table from the getgo is much cleaner
# than doing everything in separate variables
dat <- data.frame(
datasetsizes = rep(datasetsizes, each = length(conds)), # make replicates for each condition
cond = rep(NA, l*length(conds))
)
dat[, c("min", "lq", "mean", "median", "uq", "max")] <- 0
dat$cond <- factor(dat$cond, levels = conds)
head(dat)
for(i in 1:l){ # for the love of god, don't use something as long as 'loopvar' as an iterative
# I don't have f1 & f2 so I did what I could...
s <- summary(microbenchmark(
"f1" = rpois(datasetsizes[i],10),
"f2" = {length(rpois(datasetsizes[i],10))}))
dat[which(dat$datasetsizes == datasetsizes[i]), # select rows of current ds size
c("cond", "min", "lq", "mean", "median", "uq", "max")] <- s[, !colnames(s)%in%c("neval")]
}
dat <- data.table(dat)
ggplot(dat, aes(x=datasetsizes,y=mean)) +
geom_point(aes(color = cond)) +
geom_line(aes(color = cond)) + # added to see a clear difference btw conds
labs(x="N", y="Runtime") + scale_x_continuous(trans = 'log10') +
scale_y_continuous(trans = 'log10')
给出下面的图。