I am trying to add a linear regression to data plotted with ggplot; however, due to the nature of my data I need to plot it such that the responding variable in the linear regression is the x-axis, not the y. Is there a way to change the way the regression is done (I tried changing "formula = y~x" to "formula = x~y" but no luck) by maybe specifying alternate mapping from the mapping specified by the plot? Or is there an easy way to invert the plot after I have added the regression? Thanks! Any help is appreciated.
One straightforward way (which you suggested) would be to make the plot with y and x reversed, and then "inverting" the final plot. I used heavily right skewed "noise" so the example really makes it clear what is being fit.
library(tidyverse)
set.seed(42)
foo <- data_frame(x = 1:100, y = 2 + 0.5*x + 3*rchisq(100, 3))
foo %>%
ggplot(aes(x=y, y=x)) + geom_point() + stat_smooth(method = "lm") + coord_flip()
Related
I have a model which has been created like this
cube_model <- lm(y ~ x + I(x^2) + I(x^3), data = d.r.data)
I have been using ggplot methods like geom_point to plot datapoints and geom_smooth to plot the regression line. Now the question i am trying to solve is to plot fitted data vs observed .. How would i do that? I think i am just unfamiliar with R so not sure what to use here.
--
EDIT
I ended up doing this
predicted <- predict(cube_model)
ggplot() + geom_point(aes(x, y)) + geom_line(aes(x, predicted))
Is this correct approach?
What you need to do is use the predict function to generate the fitted values. You can then add them back to your data.
d.r.data$fit <- predict(cube_model)
If you want to plot the predicted values vs the actual values, you can use something like the following.
library(ggplot2)
ggplot(d.r.data) +
geom_point(aes(x = fit, y = y))
I would like to change the color of coefficient lines based on whether the point estimate is negative or positive in a ggplot2 coefficient plot in R. For example:
require(coefplot)
set.seed(123)
dat <- data.frame(x = rnorm(100), z = rnorm(100))
mod1 <- lm(y1 ~ x + z, data = dat)
coefplot.lm(mod1)
Which produces the following plot:
In this plot, I would like to change the "x" variable to red when plotted. Any ideas? Thanks.
I think, you cannot do this with a plot produced by coefplot.lm. The package coefplot uses ggplot2 as the plotting system, which is good itself, but does not allow to play with colors as easily as you would like. To achieve the desired colors, you need to have a variable in your dataset that would color-code the values; you need to specify color = color-code in aes() function within the layer that draws the dots with CE. Apparently, this is impossible to do with the output of coefplot.lm function. Maybe, you can change the colors using ggplot2 ggplot_build() function. I would say, it's easier to write your own function for this task.
I've done this once to plot odds. If you want, you may use my code. Feel free to change it. The idea is the same as in coefplot. First, we extract coefficients from a model object and prepare the data set for plotting; second, actually plot.
The code for extracting coefficients and data set preparation
df_plot_odds <- function(x){
tmp<-data.frame(cbind(exp(coef(x)), exp(confint.default(x))))
odds<-tmp[-1,]
names(odds)<-c('OR', 'lower', 'upper')
odds$vars<-row.names(odds)
odds$col<-odds$OR>1
odds$col[odds$col==TRUE] <-'blue'
odds$col[odds$col==FALSE] <-'red'
odds$pvalue <- summary(x)$coef[-1, "Pr(>|t|)"]
return(odds)
}
Plot the output of the extract function
plot_odds <- function(df_plot_odds, xlab="Odds Ratio", ylab="", asp=1){
require(ggplot2)
p <- ggplot(df_plot_odds, aes(x=vars, y=OR, ymin=lower, ymax=upper),asp=asp) +
geom_errorbar(aes(color=col),width=0.1) +
geom_point(aes(color=col),size=3)+
geom_hline(yintercept = 1, linetype=2) +
scale_color_manual('Effect', labels=c('Positive','Negative'),
values=c('blue','red'))+
coord_flip() +
theme_bw() +
theme(legend.position="none",aspect.ratio = asp)+
ylab(xlab) +
xlab(ylab) #switch because of the coord_flip() above
return(p)
}
Plotting your example
set.seed(123)
dat <- data.frame(x = rnorm(100),y = rnorm(100), z = rnorm(100))
mod1 <- lm(y ~ x + z, data = dat)
df <- df_plot_odds(mod1)
plot <- plot_odds(df)
plot
Which yields
Note that I chose theme_wb() as the default. Output is a ggplot2object. So, you may change it quite a lot.
I'm trying to display some frequencies convolved with a Gaussian kernel in ggplot2. I tried smoothing the lines with:
+ stat_smooth(se = F,method = "lm", formula = y ~ poly(x, 24))
Without success.
I read an article suggesting the frequencies should be convolved with a Gaussian kernel. Which ggplot2's stat_density function (http://docs.ggplot2.org/current/stat_density.html) seem to be able to produce.
However, I can't seem to be able to replace my geometry with stat_density. I there anything wrong with my code?
require(reshape2)
library(ggplot2)
library(RColorBrewer)
fileName = "/1.csv" # downloadable there: https://www.dropbox.com/s/l5j7ckmm5s9lo8j/1.csv?dl=0
mydata = read.csv(fileName,sep=",", header=TRUE)
dataM = melt(mydata,c("bins"))
myPalette <- colorRampPalette(rev(brewer.pal(11, "Spectral")))
ggplot(data=dataM,
aes(x=bins, y=value, colour=variable)) +
geom_line() + scale_x_continuous(limits = c(0, 2))
This code produces the following plot:
I'm looking at smoothing the lines a little bit, so they look more like this:
(from http://journal.frontiersin.org/Journal/10.3389/fncom.2013.00189/full)
Since my comments solved your problem, I'll convert them to an answer:
The density function takes individual measurements and calculates a kernel density distribution by convolution (gaussian is the default kernel). For example, plot(density(rnorm(1000))). You can control the smoothness with the bw (bandwidth) parameter. For example, plot(density(rnorm(1000), bw=0.01)).
But your data frame is already a density distribution (analogous to the output of the density function). To generate a smoother density estimate, you need to start with the underlying data and run density on it, adjusting bw to get the smoothness where you want it.
If you don't have access to the underlying data, you can smooth out your existing density distributions as follows:
ggplot(data=dataM, aes(x=bins, y=value, colour=variable)) +
geom_smooth(se=FALSE, span=0.3) +
scale_x_continuous(limits = c(0, 2)).
Play around with the span parameter to get the smoothness you want.
I have a data set with some points in it and want to fit a line on it. I tried it with the loess function. Unfortunately I get very strange results. See the plot bellow. I expect a line that goes more through the points and over the whole plot. How can I achieve that?
How to reproduce it:
Download the dataset from https://www.dropbox.com/s/ud32tbptyvjsnp4/data.R?dl=1 (only two kb) and use this code:
load(url('https://www.dropbox.com/s/ud32tbptyvjsnp4/data.R?dl=1'))
lw1 = loess(y ~ x,data=data)
plot(y ~ x, data=data,pch=19,cex=0.1)
lines(data$y,lw1$fitted,col="blue",lwd=3)
Any help is greatly appreciated. Thanks!
You've plotted fitted values against y instead of against x. Also, you will need to order the x values before plotting a line. Try this:
lw1 <- loess(y ~ x,data=data)
plot(y ~ x, data=data,pch=19,cex=0.1)
j <- order(data$x)
lines(data$x[j],lw1$fitted[j],col="red",lwd=3)
Unfortunately the data are not available anymore, but an easier way how to fit a non-parametric line (Locally Weighted Scatterplot Smoothing or just a LOESS if you want) is to use following code:
scatter.smooth(y ~ x, span = 2/3, degree = 2)
Note that you can play with parameters span and degree to get arbitrary smoothness.
May be is to late, but you have options with ggplot (and dplyr). First if you want only plot a loess line over points, you can try:
library(ggplot2)
load(url("https://www.dropbox.com/s/ud32tbptyvjsnp4/data.R?dl=1"))
ggplot(data, aes(x, y)) +
geom_point() +
geom_smooth(method = "loess", se = FALSE)
Other way, is by predict() function using a loess fit. For instance I used dplyr functions to add predictions to new column called "loess":
library(dplyr)
data %>%
mutate(loess = predict(loess(y ~ x, data = data))) %>%
ggplot(aes(x, y)) +
geom_point(color = "grey50") +
geom_line(aes(y = loess))
Update: Added line of code to load the example data provided
Update2: Correction on geom_smoot() function name acoording #phi comment
I'm trying to plot a scatter-plot with two layers. The reason is I want to represent the size of the points by its number of answers. Then I need to have a smooth-curve layed over it. So I use two datasets to achieve this.
The problem is, when I lay the second layer with the smoother using the original dataset, then the smoother is shifted by one point on the x-scale to the left.
Does anyone know, how to correct this in the R code? Is there maybe something wrong in it?
I thought about to add 1 to the x variable, but I don't want to have to go this far.
library(ggplot2)
q.tab <- xtabs(~x + y, mydata)
q.df <- as.data.frame(q.tab)
pointsize <- q.df$Freq
qplot(x, y, data=q.df) + geom_point(aes(size=as.factor(pointsize)))
+ geom_smooth(data=mydata, method="loess", span=1))
With ggplot2 , when you think in terms of layer it is better to use ggplot function and not qplot.
I generate your data (sample function is very convenient to generate data)
mydata$x <- sample(1:10,100,replace=TRUE)
mydata$y <- sample(1:10,100,replace=TRUE)
q.tab <- xtabs(~x + y, mydata)
q.df <- as.data.frame(q.tab)
ggplot version:
library(ggplot2)
ggplot(data=mydata,aes(x,y,size=Freq)) +
geom_point() +
geom_smooth( method="loess", span=1)
qplot version:
qplot(data=mydata,x=x,y=y,size=Freq,geom='point')+
geom_smooth( method="loess", span=1)