Plot table objects with ggplot? - r

I've got this data:
No Yes
Female 411 130
Male 435 124
which was created using the standard table command. Now with plot I can plot this as such:
plot(table(df$gender, df$fraud))
and it then outputs a 2x2 bar chart.
So my question is, how can I do this with ggplot2? Is there any way with out transforming the table-object to a data frame? I would do that, but it becomes a mess and you then need to rename column and row headers and it just becomes a mess for what is really a quite simple thing?


Something such as
ggplot(as.data.frame(table(df)), aes(x=gender, y = Freq, fill=fraud)) +
geom_bar(stat="identity")
gets a similar chart with a minimum amount of relabelling.


ggplot2 works with data frame, so, you have to convert table into a frame. Here is a sample code:
myTable <- table(df$gender, df$fraud)
myFrame <- as.data.frame(table(myTable))
Now, you can use myFrame in ggplot2:
ggplot(myFrame, aes(x=gender))+
geom_bar(y = Freq)
see Coerce to a Data Frame for more information.


For the record, janitor::tabyl() outputs contingency tables that are data.frames. As such, they are more convenient for a workflow based on tidyverse tools.
For example:
# Data
df <- data.frame(gender = c(rep("female", times = 411),
rep("female", times = 130),
rep("male", times = 435),
rep("male", times = 124)),
fraud = c(rep("no", times = 411),
rep("yes", times = 130),
rep("no", times = 435),
rep("yes", times = 124)))
# Plotting the tabulation with tidyverse tools
df |>
janitor::tabyl(gender, fraud) |>
tidyr::gather(key = fraud, value = how_many, no:yes) |>
ggplot2::ggplot(aes(y = how_many, x = gender, fill = fraud)) +
geom_col()
Note: The sequence of piped results with janitor and tidyr has the benefit of being more transparent, but it essentially replicates the same result achieved with as.data.frame(table(df)).

Related

How to efficiently draw lots of graphs in R from data in a wide format?

I'm trying to draw 18 graphs using R and the ggplot2 package. My data look like this:
v1 v2 v3 ... v18 subject group
534 543 512 ... 410 1 (6.5, 18]
437 576 465 ... 420 2 (0, 6.5]
466 487 492 ... 501 3 (18, 55]
And I need to create a "faceted" histogram showing distributions for all of the groups in one frame (i. e. to conveniently present all of the subgroups' distributions) like this:
I came up with this code for a single plot:
ggplot(data = df, aes (x = v1)) + geom_histogram (boundary = 500) + facet_wrap(~Group, nrow = 2)
But since there are 18 variables (v1, v2,...), I'm looking for a way to write an efficient function/loop/command that would draw all the 18 graphs without me having to copy/paste and change the variable name 18 times. Like this:
ggplot(data = df, aes (x = **v1**)) + geom_histogram (boundary = 500) + facet_wrap(~Group, nrow = 2)
ggplot(data = df, aes (x = **v2**)) + geom_histogram (boundary = 500) + facet_wrap(~Group, nrow = 2)
ggplot(data = df, aes (x = **v3**)) + geom_histogram (boundary = 500) + facet_wrap(~Group, nrow = 2)
I know that the solution probably lies in looping and it seems like a useful skill to have, so I'm also using this opportunity to learn this right.
Thank you, any help is appreciated! (And thanks to all the suggestions so far!)
This is where I've gotten so far with the kind help of the user below:
for (v in c(v1,v2)) {
pdf("plots.pdf")
histograms <- ggplot(data = data, aes (x = v)) + geom_histogram (boundary = 500) + facet_wrap(~Group, nrow = 2)
print(histograms)
}
dev.off()

EDIT A significantly revised answer is provided having clarified the needs.
The problem presents several common issues, each of which are addressed in other posts. However, perhaps this suggestion allows for a one-stop solution to these common issues.
My first suggestion is to reformat the data into a "long" format. There are many resources describing this and packages to help. Many users embrace the "tidyverse" set of tools and I'll leave that to others. I'll demonstrate a simple approach using base functions. I don't recommend the reshape() function in the stats package. I find it to be useful for repeated measures with time as one of the variables but find it rather complicated for other data.
A large fake data set will be generated in the "wide" format with demographic data (id, sex, weight, age, group) and 18 variables named "v01", "v02", ..., "v18" as random integers between 400 and 500.
# Set random number generator and number of "individuals" in fake data
set.seed(1234) # to ensure reproducibility
N <- 936 # number of "individuals" in the fake data
# Create typical fake demographic data and divide the age into 4 groups
id <- factor(sample(1e4:9e4, N, replace = FALSE))
age <- rpois(N, 36)
sex <- sample(c("F","M"), N, replace = TRUE)
weight <- 16 * log(age)
group <- cut(age, breaks = c(12, 32, 36, 40, 62))
Generate 18 fake values for each individual for the wide format and then create the fake "wide" data.frame.
# 18 variable measurements for wide format
V <- replicate(18, sample(400:600, N, replace = TRUE), simplify = FALSE)
names(V) <- sprintf("v%02d", 1:18)
# Add a little variation to the fake data
adj <- sample(1:6, 18, replace = TRUE)
V <- Map("/", V, adj) # divide each value by the number in 'adj'
V <- lapply(V, round, 1) # simplify
# Create data.frame with variable data in wide format
vars <- as.data.frame(V)
names(vars)
# Assemble demographic and variable data into a typical "wide" data set
wide <- data.frame(id, sex, weight, age, group, vars)
names(wide)
head(wide)
In the "wide" format, each row corresponds to a unique individual with demographic information and 18 values for 18 variables. This is going to be changed into the "long" format with each value represented by a row. The new "long" data frame will have two new variables for the data (values) and a factor indicating the group from which the data came (ind). Typically they get renamed but I will simply work with the default names here.
As noted above, the simple base function stack() will be used to stack the variables into a single vector. In contrast to cbind(), the data.frame() function will replicate values only as long as they are an even multiple of each other. The following code takes advantage of this property to build the "long" data.frame.
# Identify those variables to be stacked (they all start with 'v')
sel <- grepl("^v", names(wide))
long <- data.frame(wide[!sel], stack(wide[sel]))
head(long)
My second suggestion is to use one of the "apply" functions to create a list of ggplot objects. By storing the plots in this variable, you have the option of plotting them with different formats without running the plotting code each time.
The code creates a plot for each of the 18 different variables, which are identified by the new variable ind. I changed boundary = 500 to a bins = 10 since I don't know what your actual data looks like. I also added a "caption" to each plot identifying the original variable.
library(ggplot2) # to use ggplot...
plotList <- lapply(levels(long$ind), function(i)
ggplot(data = subset(long, ind == i), aes(x = values))
+ geom_histogram(bins = 10)
+ facet_wrap(~ group, nrow = 2)
+ labs(caption = paste("Variable", i)))
names(plotList) <- levels(long$ind) # name the list elements for convenience
Now to examine each of the 18 plots (this may not work in RStudio):
opar <- par(ask = TRUE)
plotList # This is the same as print(plotList)
par(opar) # turn off the 'ask' option
To save the plots to file, the advice of Imo is good. But it would be wise to take control of the size and nature of the file output. I suggest you look at the help files for pdf() and dev.print(). The last part of this answer shows one possibility with the pdf() function using a for loop to generate single page plots.
for (v in levels(long$ind)) {
fname <- paste(v, "pdf", sep = ".")
fname <- file.path("~", fname) # change this to specify a directory
pdf(fname, width = 6.5, height = 7, paper = "letter")
print(plotList[[v]])
dev.off()
}
And just to add another possible approach, here's a solution with lattice showing 6 groups of variables per plot. (Personally, I'm a fan of this simpler approach.)
library(lattice)
idx <- split(levels(long$ind), gl(3, 6, 18))
opar <- par(ask = TRUE)
for (i in idx)
plot(histogram(~values | group + ind, data = long,
subset = ind %in% i, as.table = TRUE))
par(opar)

same y axis variable , scatter-plot and long format

Let's say two different raters are evaluating behavioral problems. They use the same scale (from 0 to 50) and the child being evaluated is the same for both raters. In social sciences, this method is common and there are some useful statistics, such as correlation coefficient and Intra-Class Correlation.
In addition, one graph that comes to my mind is the scatter-plot, and in the x-axys I'll plot the behavioral problems scores considering the first rater and in the y-axis, I'll do the same for the second rater.
gplot2 creates amazing plots, however, some simple routines and action become really difficult to do.
Please see the code below and the r base plot. I would like to know if ggplot can create this plot as well.
Thanks much
set.seed(123)
ds <- data.frame(behavior_problems = rnorm(100,30,2), evaluator=sample(1:2))
plot(ds$behavior_problems[ds$evaluator == '1'] ,
y = ds$behavior_problems[ds$evaluator == '2'])
== I had to edit to make clear why a scatter-plot would be informative==

I think the key problem here is the way in which you have set up the data frame. It is not clear that each individual gets a pair of scores, one from each evaluator. So the first thing to do is add an ID for each individual: 50 IDs in your example data.
library(tidyverse)
ds %>%
mutate(id = rep(1:50, each = 2)
Now we can use tidyr::spread to create a column for each evaluator. But numbers for column names are not a great idea, so we'll rename them to e1 and e2.
ds %>%
mutate(id = rep(1:50, each = 2)) %>%
spread(evaluator, behavior_problems) %>%
rename(e1 = `1`, e2 = `2`)
Now we have column names that can be supplied to ggplot:
ds %>%
mutate(id = rep(1:50, each = 2)) %>%
spread(evaluator, behavior_problems) %>%
rename(e1 = `1`, e2 = `2`) %>%
ggplot(aes(e1, e2)) +
geom_point()
If this seems like a "long way around", it's because ggplot2 works better with "long" data (before the spread) than "wide" (after the spread). To illustrate, here's another way to visualize the difference in scores by individual, which I think works quite well:
ds %>%
mutate(id = rep(1:50, each = 2),
evaluator = factor(evaluator)) %>%
ggplot(aes(id, behavior_problems)) +
geom_point(aes(color = evaluator)) +
geom_line(aes(group = id))

Specify data to be graphed by one parameter

I want to be able to graph some points based on if one parameter is true in a data frame. I want to only use the position coordinates if in the "Time" column is equal to 2. How do I write this in R?

In the future it'd helpful to dput some data. The packages dplyr and ggplot2 are two popular packages you might want to look into.
library(dplyr)
library(ggplot2)
Create something that your data might look like:
df <- data.frame(x = runif(10), y = runif(10), Time = sample(1:2, 10, replace = T))
Filter then plot:
df_tmp <- df %>% filter(Time == 2)
ggplot(df_tmp, aes(x,y)) + geom_point()

Subset of data included in more than one ggplot facet

I have a population and a sample of that population. I've made a few plots comparing them using ggplot2 and its faceting option, but it occurred to me that having the sample in its own facet will distort the population plots (however slightly). Is there a way to facet the plots so that all records are in the population plot, and just the sampled records in the second plot?

Matt,
If I understood your question properly - you want to have a faceted plot where one panel contains all of your data, and the subsequent facets contain only a subset of that first plot?
There's probably a cleaner way to do this, but you can create a new data.frame object with the appropriate faceting variable that corresponds to each subset. Consider:
library(ggplot2)
df <- data.frame(x = rnorm(100), y = rnorm(100), sub = sample(letters[1:5], 100, TRUE))
df2 <- rbind(
cbind(df, faceter = "Whole Sample")
, cbind(df[df$sub == "a" ,], faceter = "Subset A")
#other subsets go here...
)
qplot(x,y, data = df2) + facet_wrap(~ faceter)
Let me know if I've misunderstood your question.
-Chase

Setting up a CSV file for R to display histograms

Greetings,
Basically, I have two vectors of data (let's call it experimental and baseline). I want to use the lattice library and histogram functions of R to plot the two histograms side-by-side, just as seen at the end of this page.
I have my data in a CSV file like this:
Label1,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
Label2,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
Label3,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
Label4,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
Label5,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
Label6,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
Each row should be a new pair of histograms. Columns 1-9 represents the data for the experiment (left-side histogram). Columns 10-18 represents the baseline data (right-side histogram).
Can anyone help me on this? Thanks.

Your data is poorly formatted for faceting with lattice. You can restructure it using reshape.
read.csv(textConnection("Label1,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
Label2,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
Label3,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
Label4,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
Label5,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18
Label6,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18"), header = F)->data
colnames(data)[1] <- "ID"
colnames(data)[2:10] <- paste("exp",1:9, sep = "_")
colnames(data)[11:19] <- paste("base", 1:9, sep = "_")
library(reshape)
data.m <- melt(data, id = "ID")
data.m <- cbind(data.m, colsplit(data.m$variable, "_", names = c("Source","Measure")))
data.m is now in the format you really want your data to be in for almost everything. I don't know if each of the 9 measurements from the experiment and the baseline are meaningful or can be meaningfully compared so I kept them distinct.
Now, you can use lattice properly.
histogram(~value | Source + ID, data = data.m)
If the measurements are meaningfully compared (that is, data[,2] and data[,11] are somehow the "same"), you could recast the data to directly compare experiment to baseline
data.comp <- cast(data.m, ID + Measure ~ Source)
## I know ggplot2 better
library(ggplot2)
qplot(base, exp, data = data.comp)+
geom_abline()+
expand_limits(x = 0, y = 0)

Something like this should work:
library(lattice)
data <- matrix(1:18, ncol=18, nrow=3, byrow=T)
for (i in 1:nrow(data))
{
tmp <- cbind(data[i,], rep(1:2, each=9))
print(histogram(~tmp[,1]|tmp[,2]), split=c(1,i,1,nrow(data)), more=T)
}
Note: this will work only for few rows of data... for larger datasets you may want to think of slightly different layout (change the split parameter in histogram)

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