I am trying to plot a point histogram (a histogram that shows the values with a point instead of bars) that is log-scaled. The result should look like this:
MWE:
Lets simulate some Data:
set.seed(123)
d <- data.frame(x = rnorm(1000))
To get the point histogram I need to calculate the histogram data (hdata) first
hdata <- hist(d$x, plot = FALSE)
tmp <- data.frame(mids = hdata$mids,
density = hdata$density,
counts = hdata$counts)
which we can plot like this
p <- ggplot(tmp, aes(x = mids, y = density)) + geom_point() +
stat_function(fun = dnorm, col = "red")
p
to get this graph:
In theory we should be able to apply the log scales (and set the y-limits to be above 0) and we should have a similar picture to the target graph.
However, if I apply it I get the following graph:
p + scale_y_log10(limits = c(0.001, 10))
The stat_function clearly shows non-scaled values instead of producing a figure closer to the solid line in the first picture.
Any ideas?
Bonus
Are there any ways to graph the histogram with dots without using the hist(..., plot = FALSE) function?
EDIT Workaround
One possible solution is to calculate the dnorm-data outside of ggplot and then insert it as a line. For example
tmp2 <- data.frame(mids = seq(from = min(tmp$mids), to = max(tmp$mids),
by = (max(tmp$mids) - min(tmp$mids))/10000))
tmp2$dnorm <- dnorm(tmp2$mids)
# Plot it
ggplot() +
geom_point(data = tmp, aes(x = mids, y = density)) +
geom_line(data = tmp2, aes(x = mids, y = dnorm), col = "red") +
scale_y_log10()
This returns a graph like the following. This is basically the graph, but it doesn't resolve the stat_function issue.
library(ggplot2)
set.seed(123)
d <- data.frame(x = rnorm(1000))
ggplot(d, aes(x)) +
stat_bin(geom = "point",
aes(y = ..density..),
#same breaks as function hist's default:
breaks = pretty(range(d$x), n = nclass.Sturges(d$x), min.n = 1),
position = "identity") +
stat_function(fun = dnorm, col = "red") +
scale_y_log10(limits = c(0.001, 10))
Another possible solution that I found while revisiting this issue is to apply the log10 to the stat_function-call.
library(ggplot2)
set.seed(123)
d <- data.frame(x = rnorm(1000))
hdata <- hist(d$x, plot = FALSE)
tmp <- data.frame(mids = hdata$mids,
density = hdata$density,
counts = hdata$counts)
ggplot(tmp, aes(x = mids, y = density)) + geom_point() +
stat_function(fun = function(x) log10(dnorm(x)), col = "red") +
scale_y_log10()
Created on 2018-07-25 by the reprex package (v0.2.0).
Related
I have a data.frame that has counts for several groups:
set.seed(1)
df <- data.frame(group = sample(c("a","b"),200,replace = T),
n = round(runif(200,1,2)))
df$n <- as.integer(df$n)
And I'm trying to display a histogram of df$n, facetted by the group using ggplot2's geom_histogram:
library(ggplot2)
ggplot(data = df, aes(x = n)) + geom_histogram() + facet_grid(~group) + theme_minimal()
Any idea how to get ggplot2 to label the x-axis ticks with the integers the histogram is summarizing rather than the numeric values it is currently showing?
You could tweak this by the binwidth argument of geom_histogram:
library(ggplot2)
ggplot(data = df, aes(x = n)) +
geom_histogram(binwidth = 0.5) +
facet_grid(~group) +
theme_minimal()
Another example:
set.seed(1)
df <- data.frame(group = sample(c("a","b"),200,replace = T),
n = round(runif(200,1,5)))
library(ggplot2)
ggplot(data = df, aes(x = n)) +
geom_histogram(binwidth = 0.5) +
facet_grid(~group) +
theme_minimal()
You can manually specify the breaks with scale_x_continuous(breaks = seq(1, 2)). Alternatively, you can set the breaks and labels separately as well.
I am trying to make a histogram of density values and overlay that with the curve of a density function (not the density estimate).
Using a simple standard normal example, here is some data:
x <- rnorm(1000)
I can do:
q <- qplot( x, geom="histogram")
q + stat_function( fun = dnorm )
but this gives the scale of the histogram in frequencies and not densities. with ..density.. I can get the proper scale on the histogram:
q <- qplot( x,..density.., geom="histogram")
q
But now this gives an error:
q + stat_function( fun = dnorm )
Is there something I am not seeing?
Another question, is there a way to plot the curve of a function, like curve(), but then not as layer?
Here you go!
# create some data to work with
x = rnorm(1000);
# overlay histogram, empirical density and normal density
p0 = qplot(x, geom = 'blank') +
geom_line(aes(y = ..density.., colour = 'Empirical'), stat = 'density') +
stat_function(fun = dnorm, aes(colour = 'Normal')) +
geom_histogram(aes(y = ..density..), alpha = 0.4) +
scale_colour_manual(name = 'Density', values = c('red', 'blue')) +
theme(legend.position = c(0.85, 0.85))
print(p0)
A more bare-bones alternative to Ramnath's answer, passing the observed mean and standard deviation, and using ggplot instead of qplot:
df <- data.frame(x = rnorm(1000, 2, 2))
# overlay histogram and normal density
ggplot(df, aes(x)) +
geom_histogram(aes(y = after_stat(density))) +
stat_function(
fun = dnorm,
args = list(mean = mean(df$x), sd = sd(df$x)),
lwd = 2,
col = 'red'
)
What about using geom_density() from ggplot2? Like so:
df <- data.frame(x = rnorm(1000, 2, 2))
ggplot(df, aes(x)) +
geom_histogram(aes(y=..density..)) + # scale histogram y
geom_density(col = "red")
This also works for multimodal distributions, for example:
df <- data.frame(x = c(rnorm(1000, 2, 2), rnorm(1000, 12, 2), rnorm(500, -8, 2)))
ggplot(df, aes(x)) +
geom_histogram(aes(y=..density..)) + # scale histogram y
geom_density(col = "red")
I'm trying for iris data set. You should be able to see graph you need in these simple code:
ker_graph <- ggplot(iris, aes(x = Sepal.Length)) +
geom_histogram(aes(y = ..density..),
colour = 1, fill = "white") +
geom_density(lwd = 1.2,
linetype = 2,
colour = 2)
I am trying to make a histogram of density values and overlay that with the curve of a density function (not the density estimate).
Using a simple standard normal example, here is some data:
x <- rnorm(1000)
I can do:
q <- qplot( x, geom="histogram")
q + stat_function( fun = dnorm )
but this gives the scale of the histogram in frequencies and not densities. with ..density.. I can get the proper scale on the histogram:
q <- qplot( x,..density.., geom="histogram")
q
But now this gives an error:
q + stat_function( fun = dnorm )
Is there something I am not seeing?
Another question, is there a way to plot the curve of a function, like curve(), but then not as layer?
Here you go!
# create some data to work with
x = rnorm(1000);
# overlay histogram, empirical density and normal density
p0 = qplot(x, geom = 'blank') +
geom_line(aes(y = ..density.., colour = 'Empirical'), stat = 'density') +
stat_function(fun = dnorm, aes(colour = 'Normal')) +
geom_histogram(aes(y = ..density..), alpha = 0.4) +
scale_colour_manual(name = 'Density', values = c('red', 'blue')) +
theme(legend.position = c(0.85, 0.85))
print(p0)
A more bare-bones alternative to Ramnath's answer, passing the observed mean and standard deviation, and using ggplot instead of qplot:
df <- data.frame(x = rnorm(1000, 2, 2))
# overlay histogram and normal density
ggplot(df, aes(x)) +
geom_histogram(aes(y = after_stat(density))) +
stat_function(
fun = dnorm,
args = list(mean = mean(df$x), sd = sd(df$x)),
lwd = 2,
col = 'red'
)
What about using geom_density() from ggplot2? Like so:
df <- data.frame(x = rnorm(1000, 2, 2))
ggplot(df, aes(x)) +
geom_histogram(aes(y=..density..)) + # scale histogram y
geom_density(col = "red")
This also works for multimodal distributions, for example:
df <- data.frame(x = c(rnorm(1000, 2, 2), rnorm(1000, 12, 2), rnorm(500, -8, 2)))
ggplot(df, aes(x)) +
geom_histogram(aes(y=..density..)) + # scale histogram y
geom_density(col = "red")
I'm trying for iris data set. You should be able to see graph you need in these simple code:
ker_graph <- ggplot(iris, aes(x = Sepal.Length)) +
geom_histogram(aes(y = ..density..),
colour = 1, fill = "white") +
geom_density(lwd = 1.2,
linetype = 2,
colour = 2)
I've plotted a confusion matrix (predicting 5 outcomes) in R using ggplot and scales for geom_text labeling.
The way geom_text(aes(label = percent(Freq/sum(Freq))) is written in code, it's showing Frequency of each box divided by sum of all observations, but what I want to do is get Frequency of each box divided by sum Frequency for each Reference.
In other words, instead of A,A = 15.8%,
it should be A,A = 15.8%/(0.0%+0.0%+0.0%+0.0%+15.8%%) = 100.0%
library(ggplot2)
library(scales)
valid_actual <- as.factor(c("A","B","B","C","C","C","E","E","D","D","A","A","A","E","E","D","D","C","B"))
valid_pred <- as.factor(c("A","B","C","C","E","C","E","E","D","B","A","B","A","E","D","E","D","C","B"))
cfm <- confusionMatrix(valid_actual, valid_pred)
ggplotConfusionMatrix <- function(m){
mytitle <- paste("Accuracy", percent_format()(m$overall[1]),
"Kappa", percent_format()(m$overall[2]))
p <-
ggplot(data = as.data.frame(m$table) ,
aes(x = Reference, y = Prediction)) +
geom_tile(aes(fill = log(Freq)), colour = "white") +
scale_fill_gradient(low = "white", high = "green") +
geom_text(aes(x = Reference, y = Prediction, label = percent(Freq/sum(Freq)))) +
theme(legend.position = "none") +
ggtitle(mytitle)
return(p)
}
ggplotConfusionMatrix(cfm)
The problem is that, as far as I know, ggplot is not able to do group calculation. See this recent post for similar question.
To solve your problem you should take advantage of the dplyrpackage.
This should work
library(ggplot2)
library(scales)
library(caret)
library(dplyr)
valid_actual <- as.factor(c("A","B","B","C","C","C","E","E","D","D","A","A","A","E","E","D","D","C","B"))
valid_pred <- as.factor(c("A","B","C","C","E","C","E","E","D","B","A","B","A","E","D","E","D","C","B"))
cfm <- confusionMatrix(valid_actual, valid_pred)
ggplotConfusionMatrix <- function(m){
mytitle <- paste("Accuracy", percent_format()(m$overall[1]),
"Kappa", percent_format()(m$overall[2]))
data_c <- mutate(group_by(as.data.frame(m$table), Reference ), percentage =
percent(Freq/sum(Freq)))
p <-
ggplot(data = data_c,
aes(x = Reference, y = Prediction)) +
geom_tile(aes(fill = log(Freq)), colour = "white") +
scale_fill_gradient(low = "white", high = "green") +
geom_text(aes(x = Reference, y = Prediction, label = percentage)) +
theme(legend.position = "none") +
ggtitle(mytitle)
return(p)
}
ggplotConfusionMatrix(cfm)
And the result:
I am trying to make a histogram of density values and overlay that with the curve of a density function (not the density estimate).
Using a simple standard normal example, here is some data:
x <- rnorm(1000)
I can do:
q <- qplot( x, geom="histogram")
q + stat_function( fun = dnorm )
but this gives the scale of the histogram in frequencies and not densities. with ..density.. I can get the proper scale on the histogram:
q <- qplot( x,..density.., geom="histogram")
q
But now this gives an error:
q + stat_function( fun = dnorm )
Is there something I am not seeing?
Another question, is there a way to plot the curve of a function, like curve(), but then not as layer?
Here you go!
# create some data to work with
x = rnorm(1000);
# overlay histogram, empirical density and normal density
p0 = qplot(x, geom = 'blank') +
geom_line(aes(y = ..density.., colour = 'Empirical'), stat = 'density') +
stat_function(fun = dnorm, aes(colour = 'Normal')) +
geom_histogram(aes(y = ..density..), alpha = 0.4) +
scale_colour_manual(name = 'Density', values = c('red', 'blue')) +
theme(legend.position = c(0.85, 0.85))
print(p0)
A more bare-bones alternative to Ramnath's answer, passing the observed mean and standard deviation, and using ggplot instead of qplot:
df <- data.frame(x = rnorm(1000, 2, 2))
# overlay histogram and normal density
ggplot(df, aes(x)) +
geom_histogram(aes(y = after_stat(density))) +
stat_function(
fun = dnorm,
args = list(mean = mean(df$x), sd = sd(df$x)),
lwd = 2,
col = 'red'
)
What about using geom_density() from ggplot2? Like so:
df <- data.frame(x = rnorm(1000, 2, 2))
ggplot(df, aes(x)) +
geom_histogram(aes(y=..density..)) + # scale histogram y
geom_density(col = "red")
This also works for multimodal distributions, for example:
df <- data.frame(x = c(rnorm(1000, 2, 2), rnorm(1000, 12, 2), rnorm(500, -8, 2)))
ggplot(df, aes(x)) +
geom_histogram(aes(y=..density..)) + # scale histogram y
geom_density(col = "red")
I'm trying for iris data set. You should be able to see graph you need in these simple code:
ker_graph <- ggplot(iris, aes(x = Sepal.Length)) +
geom_histogram(aes(y = ..density..),
colour = 1, fill = "white") +
geom_density(lwd = 1.2,
linetype = 2,
colour = 2)