如何在r中平滑ecdf图
nov*_*her 2 r ggplot2 cdf ecdf
我有一个df与5变量,
头(DF,15)
junc N1.ir N2.ir W1.ir W2.ir W3.ir
1 pos$chr1:3197398 0.000000 0.000000 0.000000 0.000000 0.000000
2 pos$chr1:3207049 0.000000 0.000000 0.000000 0.000000 0.000000
3 pos$chr1:3411982 0.000000 0.000000 0.000000 0.000000 0.000000
4 pos$chr1:4342162 0.000000 0.000000 0.000000 0.000000 0.000000
5 pos$chr1:4342918 0.000000 0.000000 0.000000 0.000000 0.000000
6 pos$chr1:4767729 -4.369234 -5.123382 -4.738768 -4.643856 -5.034646
7 pos$chr1:4772814 -3.841302 -3.891419 -4.025029 -3.643856 -3.184425
8 pos$chr1:4798063 -5.038919 -4.847997 -5.497187 -4.035624 -7.543032
9 pos$chr1:4798567 -4.735325 -5.096862 -3.882643 -3.227069 -4.983808
10 pos$chr1:4818730 -8.366322 -7.118941 -8.280771 -6.629357 -6.876517
11 pos$chr1:4820396 -5.514573 -6.330917 -5.898853 -4.700440 -5.830075
12 pos$chr1:4822462 -5.580662 -6.914883 -5.562242 -5.380822 -5.703211
13 pos$chr1:4827155 -4.333273 -4.600904 -4.133399 -4.012824 -3.708345
14 pos$chr1:4829569 -4.287866 -3.874469 -3.977280 -4.209453 -4.490326
15 pos$chr1:4857613 -6.902074 -6.074141 -6.116864 -3.989946 -6.474259
使用后几行 melt
> head(ir.m)
junc variable value
1 pos$chr1:3197398 N1.ir 0.000000
2 pos$chr1:3207049 N1.ir 0.000000
3 pos$chr1:3411982 N1.ir 0.000000
4 pos$chr1:4342162 N1.ir 0.000000
5 pos$chr1:4342918 N1.ir 0.000000
6 pos$chr1:4767729 N1.ir -4.369234
并总结
> summary(ir)
junc N1.ir N2.ir W1.ir
neg$chr1:100030088: 1 Min. :-11.962 Min. :-12.141 Min. :-11.817
neg$chr1:100039873: 1 1st Qu.: -4.379 1st Qu.: -4.217 1st Qu.: -4.158
neg$chr1:10023338 : 1 Median : -2.807 Median : -2.663 Median : -2.585
neg$chr1:10024088 : 1 Mean : -2.556 Mean : -2.434 Mean : -2.362
neg$chr1:10025009 : 1 3rd Qu.: 0.000 3rd Qu.: 0.000 3rd Qu.: 0.000
neg$chr1:10027750 : 1 Max. : 17.708 Max. : 16.162 Max. : 16.210
(Other) :113310
W2.ir W3.ir
Min. :-12.194 Min. :-11.880
1st Qu.: -3.078 1st Qu.: -4.087
Median : -1.000 Median : -2.711
Mean : -1.577 Mean : -2.370
3rd Qu.: 0.000 3rd Qu.: 0.000
Max. : 17.562 Max. : 16.711
我试图使用ggplot和绘制累积概率stat_ecdf,
使用此代码
ggplot(ir.m, aes(x=value)) + stat_ecdf(aes(group=variable,colour = variable))
情节看起来像这样,
如何获得平滑的曲线？我是否需要执行更多统计操作才能获得该操作？
更新的代码
ir.d = as.data.frame(ir.m)
denss = split(ir.d, ir.d$variable) %>%
map_df(function(dw) {
denss = density(dw$value, from=min(ir.d$value) - 0.05*diff(range(ir.d$value)),
to=max(ir.d$value) + 0.05*diff(range(ir.d$value)))
data.frame(x=denss$x, y=denss$y, cd=cumsum(denss$y)/sum(denss$y), group=dw$variable[1])
head(denss)
})
summary(denss)
> summary(denss)
x y cd group
Min. :-13.689 Min. :0.0000000 Min. :0.00000 N1.ir:512
1st Qu.: -5.466 1st Qu.:0.0000046 1st Qu.:0.07061 N2.ir:512
Median : 2.757 Median :0.0002487 Median :0.99552 W1.ir :512
Mean : 2.757 Mean :0.0303942 Mean :0.65315 W2.ir :512
3rd Qu.: 10.980 3rd Qu.:0.0148074 3rd Qu.:0.99997 W3.ir :512
Max. : 19.203 Max. :0.9440592 Max. :1.00000
情节
ggplot() +
stat_ecdf(data=ir.d, aes(x, colour=variable), alpha=0.8) +
geom_line(data=denss, aes(x, cd, colour=group)) +
theme_classic()
ecdf完全遵循数据,没有任何平滑.但是,您可以通过从数据生成核密度估计(基本上是平滑的直方图)并从中创建"ecdf"来创建平滑的累积密度.这是假数据的一个例子:
首先,我们使用该density函数生成核密度估计.默认情况下,这给出了512 x值网格上的密度估计值.然后,我们使用,作为"数据",用于计算ECDF,这仅仅是密度的累积总和(或者,对于任何给定的点一个沿x轴,在ECDF的值一个是核密度下的面积从-Inf到a)的曲线(即积分).
我已将代码保存到下面的函数中,以便您可以看到更改adjust密度函数的参数如何更改平滑的ecdf.较小的值会adjust减少平滑量,从而产生更紧密地跟随数据的密度估计.您可以在下面的图中看到该设置adj=0.1导致平滑的ecdf的平滑度降低,使其更接近原始ecdf中的步骤.
library(ggplot2)
smooth_ecd = function(adj = 1) {
# Fake data
set.seed(2)
dat = data.frame(x=rnorm(15))
# Extend range of density estimate beyond data
e = 0.3 * diff(range(dat$x))
# Kernel density estimate of fake data
dens = density(dat$x, adjust=adj, from=min(dat$x)-e, to=max(dat$x) +e)
dens = data.frame(x=dens$x, y=dens$y)
# Plot kernel density (blue), ecdf (red) and smoothed ecdf (black)
ggplot(dat, aes(x)) +
geom_density(adjust=adj, colour="blue", alpha=0.7) +
geom_line(data=dens, aes(x=x, y=cumsum(y)/sum(y)), size=0.7, colour='grey30') +
stat_ecdf(colour="red", size=0.6, alpha=0.6) +
theme_classic() +
labs(title=paste0("adj=",adj))
}
smooth_ecd(adj=1)
smooth_ecd(adj=0.3)
smooth_ecd(adj=0.1)
以下是按组执行此操作的一些代码:
library(tidyverse)
# Fake data with two groups
set.seed(2)
dat = data.frame(x=c(rnorm(15, 0, 1), rnorm(20, 0.2, 0.8)),
group=rep(LETTERS[1:2], c(15,20)))
# Split the data by group and calculate the smoothed cumulative density for each group
dens = split(dat, dat$group) %>%
map_df(function(d) {
dens = density(d$x, adjust=0.1, from=min(dat$x) - 0.05*diff(range(dat$x)),
to=max(dat$x) + 0.05*diff(range(dat$x)))
data.frame(x=dens$x, y=dens$y, cd=cumsum(dens$y)/sum(dens$y), group=d$group[1])
})
现在我们可以绘制每个平滑的累积密度.在下面的图中,我已经包含了stat_ecdf对原始数据的调用以进行比较.
ggplot() +
stat_ecdf(data=dat, aes(x, colour=group), alpha=0.8, lty="11") +
geom_line(data=dens, aes(x, cd, colour=group)) +
theme_classic()
更新:使用您的数据样本,这是我得到的.我不知道你是如何得到那个长核苷酸串作为你的图中的x值,因为这样的变量不会出现在你发布的数据中的任何地方.
# Melt data
dat = gather(df, variable, value, -junc)
# Split the data by group and calculate the smoothed cumulative density for each group
dens = split(dat, dat$variable) %>%
map_df(function(d) {
dens = density(d$value, adjust=0.1, from=min(dat$value) - 0.05*diff(range(dat$value)),
to=max(dat$value) + 0.05*diff(range(dat$value)))
data.frame(x=dens$x, y=dens$y, cd=cumsum(dens$y)/sum(dens$y), group=d$variable[1])
})
ggplot() +
stat_ecdf(data=dat, aes(value, colour=variable), alpha=0.8, lty="11") +
geom_line(data=dens, aes(x, cd, colour=group)) +
theme_classic()
