使用ggplot2对分组数据进行多重stat_function [英] Multiple stat_function on grouped data with ggplot2

问题描述

我正在研究一个多参数超时观察的数据集。数据如下：

  test< -data.frame（t = c（0,0,0,0,0， 0,0,03,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.33,0.33， 0.67,0.67,0.67,0.67,0.67,0.67,0.67,0.67,1,1,1,1,1,1,1,1,1,1,1,3,3,1,3,3,1,1,3,1,1,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,3,3， 1.33,1.33,1.33,1.33,1.33,1.33,1.33,1.33,1.33,1.33,1.33,1.67,1.67,1.67,1.67,1.67， 2，2，2，2，2，2，2，2，2，2，2，2，2，2，2，2，2，2，4，4，4，4，4，4，6， 6，6,6，6,6，6,8，8,8，8,8，8,8，8,8，8,8，8,8，8,8，8,8，8,8，... 8,8,10,10,10,10,10,10,10,10,10,10,10），int = c（76.44609375,94.6619686800895,112.148907103825,75.1003097802036,74.1037037037037,76.7526662128432,74.0734830988873,87.9052100068855,81.0525931336742 ，92.1907873244038,84.0708929788684,88.8232221775814,98.1323678006063,115.175322139789,91.2653104925053,76.3661620658949,152.637799717913,107.054702135 631，83.4693197755961，91.658991910392，81.3991787335206，106.153762268266，100.919789842382，67.2119436084271，137.558914728682，89.1182608695652，156.10352233677，108.180911207183，87.9794680354643，77.7501400560224，80.7675382653061，95.6662793399954，92.5649630541872，88.3301402668491，84.3891875746714，76.4318673395818，111.413893510815，82.4753828420879，119.099190283401，192.539417212559，208.49203187251，106.919937512205， 105.370936371214，180.028767711464，130.29369773608，170.193357597816，172.703180212014，178.061569518042，182.097607918614，227.066976984743，153.856101031661，432.991580916745，299.143735224586，144.118156808803，396.36644895153，334.538796516231，350.186359610275，200.781101530882，279.866079790223，122.542700519331，235.199555308505，204.924140655867，229.181848967152，225.542753383955，468.308974987739，269.306058221873，229.969282013323， 255.553846153846,621.021220159151,255.017211703959,396.658265826583,273.300663227708,2232.449965010497,303.343894 502483，276.952483801296，327.419805194805，241.136864249474，457.961489497136，498.901714285714，280.9558101473，322.089588377724，386.754533152909，364.356809338521，340.416035518412，428.482916666667，668.447197400487，387.671341748481，471.049545829893，255.8802020688，361.979536152797，192.224629418472，284.088954468803，170.763997760358，237.869065100343，365.08237271854，294.266488413547，718.279750479846，211.599427030671， 294.045375597047，207.099267015707，194.209973045822，251.306358381503，190.786794766966，400.396083385976，183.133240482823，130.442107867392，167.231452991453，345.110896351776，299.304645622394，192.078204692282，121.273544841369，153.996295438759，97.6034616378197，362.80049522462，130.498551774077，106.031656035908，117.682936668011，90.1247837370242，140.855475040258，169.050049067713，244.290241606527，120.603356419819，173.413333333333， 125.896389002872,206.543873212215,186.668320340184,85.0988108720272,106.57849117175,102.867232728676,216.23295 7110609，86.6538461538462，149.459777852575，212.498573059361，93.3816390633923，105.567730417318，120.095470383275，137.205696941396，141.156985871272，90.578857338351，84.8457760314342，127.092660685395，136.859870967742，188.406440382942，86.0879705400982））
类（测试）
  

我设法使用绘制密度对于每个时间点：

  ggplot（test，aes（int，group = as.factor（t），color = t））+ geom_density（）
  

但我想做同样的图，但我想绘制一个对数密度的对数正态拟合。而我知道如何绘制对数正态分布使用 fitdistr 并将参数传递给 stat_function whit

 <$ c 
 $ （test，t == 0，select ='int'）$ int，lognormal）
 ggplot（data = subset（test， t = 0，select ='int'），aes（x = int））+ stat_function（fun = dlnorm，args = list（mean = fit $ estimate [1]，sd = fit $ estimate [2]））
'/ code> 

但我怎么能做到这一点对所有的 T 与线的颜色由 T 的值给定为它可以提供在 ARGS 列表的功能？

解决方案

我想到了另一种天真的解决方案：预测每个dlnorm（）的值。

  ##根据t 
 tt < -  split（test，test $ t）分解数据
 
＃ ＃对每个数据集适合对数正态分布
 fits<  -  lapply（tt，function（x）fitdistr（x $ int，lognormal））
 
 ##预测值
 （x）dlnorm（x = 1：max（test $ int），
 mean = x $ estimate [1]，sd = x $ estimate [2]））
 
 ##将所有东西都包装成一个data.frame ggplot可以处理
 plot.data<  -  data.frame（y = unlist（fitted），int = 1：max（test $ int），
t = rep（unique（test $ t），
 each = length（unlist（fitted））/ length（unique（test $ t））））
 
 ##地块
 ggplot（test，aes（int，group = as.factor（t），color = t））+ 
 #geom_density（）+ 
 geom_line（data = plot.data，aes（y = y），lwd = 1）
  

I am studying a data set with multiple observation of a parameter overtime. the data is like:

test<-data.frame(t = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.33, 0.67, 0.67, 0.67, 0.67, 0.67, 0.67, 0.67, 0.67, 0.67, 0.67, 0.67, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1.33, 1.33, 1.33, 1.33, 1.33, 1.33, 1.33, 1.33, 1.33, 1.33, 1.33, 1.33, 1.33, 1.33, 1.33, 1.33, 1.67, 1.67, 1.67, 1.67, 1.67, 1.67, 1.67, 1.67, 1.67, 1.67, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10), int = c(76.44609375, 94.6619686800895, 112.148907103825, 75.1003097802036, 74.1037037037037, 76.7526662128432, 74.0734830988873, 87.9052100068855, 81.0525931336742, 92.1907873244038, 84.0708929788684, 88.8232221775814, 98.1323678006063, 115.175322139789, 91.2653104925053, 76.3661620658949, 152.637799717913, 107.054702135631, 83.4693197755961, 91.658991910392, 81.3991787335206, 106.153762268266, 100.919789842382, 67.2119436084271, 137.558914728682, 89.1182608695652, 156.10352233677, 108.180911207183, 87.9794680354643, 77.7501400560224, 80.7675382653061, 95.6662793399954, 92.5649630541872, 88.3301402668491, 84.3891875746714, 76.4318673395818, 111.413893510815, 82.4753828420879, 119.099190283401, 192.539417212559, 208.49203187251, 106.919937512205, 105.370936371214, 180.028767711464, 130.29369773608, 170.193357597816, 172.703180212014, 178.061569518042, 182.097607918614, 227.066976984743, 153.856101031661, 432.991580916745, 299.143735224586, 144.118156808803, 396.36644895153, 334.538796516231, 350.186359610275, 200.781101530882, 279.866079790223, 122.542700519331, 235.199555308505, 204.924140655867, 229.181848967152, 225.542753383955, 468.308974987739, 269.306058221873, 229.969282013323, 255.553846153846, 621.021220159151, 255.017211703959, 396.658265826583, 273.300663227708, 232.449965010497, 303.343894502483, 276.952483801296, 327.419805194805, 241.136864249474, 457.961489497136, 498.901714285714, 280.9558101473, 322.089588377724, 386.754533152909, 364.356809338521, 340.416035518412, 428.482916666667, 668.447197400487, 387.671341748481, 471.049545829893, 255.8802020688, 361.979536152797, 192.224629418472, 284.088954468803, 170.763997760358, 237.869065100343, 365.08237271854, 294.266488413547, 718.279750479846, 211.599427030671, 294.045375597047, 207.099267015707, 194.209973045822, 251.306358381503, 190.786794766966, 400.396083385976, 183.133240482823, 130.442107867392, 167.231452991453, 345.110896351776, 299.304645622394, 192.078204692282, 121.273544841369, 153.996295438759, 97.6034616378197, 362.80049522462, 130.498551774077, 106.031656035908, 117.682936668011, 90.1247837370242, 140.855475040258, 169.050049067713, 244.290241606527, 120.603356419819, 173.413333333333, 125.896389002872, 206.543873212215, 186.668320340184, 85.0988108720272, 106.57849117175, 102.867232728676, 216.232957110609, 86.6538461538462, 149.459777852575, 212.498573059361, 93.3816390633923, 105.567730417318, 120.095470383275, 137.205696941396, 141.156985871272, 90.578857338351, 84.8457760314342, 127.092660685395, 136.859870967742, 188.406440382942, 86.0879705400982))
class(test)

I managed to plot the density for each time point using:

ggplot(test, aes(int, group = as.factor(t),colour=t))+ geom_density()

But I would like to do the same graph but instead of the density I would like to plot a log normal fit of the density.

I know how to plot the lognormal fitting on one time point using fitdistr and passing parameter to stat_function whit this code

library(MASS)
fit <- fitdistr(subset(test, t == 0,select='int')$int, "lognormal")
ggplot(data=subset(test, t == 0,select='int'), aes(x=int)) +stat_function(fun = dlnorm,args = list(mean = fit$estimate[1], sd = fit$estimate[2]))

But how can I do it for all t with the colour of the line being given by the value of t is it possible to provide a function in the args list?

解决方案

I thought of another naive solution: Predicting the values of every dlnorm().

## Split up the data according to t
tt     <- split(test, test$t)

## Fit a lognormal to every dataset
fits   <- lapply(tt, function(x) fitdistr(x$int, "lognormal"))

## Predict values
fitted <- lapply(fits, function(x) dlnorm(x = 1:max(test$int),
                               mean = x$estimate[1], sd = x$estimate[2]))

## Wrap everything into a data.frame ggplot can handle
plot.data <- data.frame(y = unlist(fitted), int = 1:max(test$int),
                        t = rep(unique(test$t),
                            each = length(unlist(fitted))/length(unique(test$t))))

## Plot
ggplot(test, aes(int, group = as.factor(t), colour=t)) +
  #geom_density() +
  geom_line(data = plot.data, aes(y = y), lwd = 1)

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