Reproduced from this code:
library(haven)
library(survey)
library(dplyr)
nhanesDemo <- read_xpt(url("https://wwwn.cdc.gov/Nchs/Nhanes/2015-2016/DEMO_I.XPT"))
# Rename variables into something more readable
nhanesDemo$fpl <- nhanesDemo$INDFMPIR
nhanesDemo$age <- nhanesDemo$RIDAGEYR
nhanesDemo$gender <- nhanesDemo$RIAGENDR
nhanesDemo$persWeight <- nhanesDemo$WTINT2YR
nhanesDemo$psu <- nhanesDemo$SDMVPSU
nhanesDemo$strata <- nhanesDemo$SDMVSTRA
nhanesAnalysis <- nhanesDemo %>%
mutate(LowIncome = case_when(
INDFMIN2 < 40 ~ T,
T ~ F
)) %>%
# Select the necessary columns
select(INDFMIN2, LowIncome, persWeight, psu, strata)
# Set up the design
nhanesDesign <- svydesign(id = ~psu,
strata = ~strata,
weights = ~persWeight,
nest = TRUE,
data = nhanesAnalysis)
svyhist(~log10(INDFMIN2), design=nhanesDesign, main = '')
How do I color the histogram by independent variable, say, LowIncome? I want to have two separate histograms, one for each value of LowIncome. Unfortunately I picked a bad example, but I want them to be see-through in case their values overlap.
If you want to plot a histogram from your model, you can get its data from model.frame (this is what svyhist does under the hood). To get the histogram filled by group, you could use this data frame inside ggplot:
library(ggplot2)
ggplot(model.frame(nhanesDesign), aes(log10(INDFMIN2), fill = LowIncome)) +
geom_histogram(alpha = 0.5, color = "gray60", breaks = 0:20 / 10) +
theme_classic()
Edit
As Thomas Lumley points out, this does not incorporate sampling weights, so if you wanted this you could do:
ggplot(model.frame(nhanesDesign), aes(log10(INDFMIN2), fill = LowIncome)) +
geom_histogram(aes(weight = persWeight), alpha = 0.5,
color = "gray60", breaks = 0:20 / 10) +
theme_classic()
To demonstrate this approach works, we can replicate Thomas's approach in ggplot using the data example from svyhist. To get the uneven bin sizes (if this is desired), we need two histogram layers, though I'm guessing this would not be required for most use-cases.
ggplot(model.frame(dstrat), aes(enroll)) +
geom_histogram(aes(fill = "E", weight = pw, y = after_stat(density)),
data = subset(model.frame(dstrat), stype == "E"),
breaks = 0:35 * 100,
position = "identity", col = "gray50") +
geom_histogram(aes(fill = "Not E", weight = pw, y = after_stat(density)),
data = subset(model.frame(dstrat), stype != "E"),
position = "identity", col = "gray50",
breaks = 0:7 * 500) +
scale_fill_manual(NULL, values = c("#00880020", "#88000020")) +
theme_classic()
You can't just extract the data and use ggplot, because that won't use the weights and so misses the whole point of svyhist. You can use the add=TRUE argument, though. You do need to set the x and y axis ranges correctly to make sure the whole plot is visible
Using the data example from ?svyhist
svyhist(~enroll, subset(dstrat,stype=="E"), col="#00880020",ylim=c(0,0.003),xlim=c(0,3500))
svyhist(~enroll, subset(dstrat,stype!="E"), col="#88000020",add=TRUE)
Related
I'm currently finishing off my Masters project and need to include some graphics for the write-up. Without boring you too much, I have some data which is associated with AR(1) parameters ranging from 0.1 to 0.9 by 0.1 increments. As such I thought of doing a faceted histogram like the one below (worry not about the hideous fruit salad of colours, it will not be used).
I used this code.
ggplot(opt_lens_geom,aes(x=l_1024,fill=factor(rho))) + geom_histogram()+coord_flip()+facet_grid(.~rho,scales = "free_x")
I also would like to draw a trend line for the median values since the AR(1) parameter is continuous. In a later iteration I deleted the padding and made it "look" like it was one graph, but I have had issues with the endpoints matching up since each facet is a separate graphical device. Can anyone give me some advice on how to do this? I am not particularly partial to the faceting so if it is not needed I do away with it.
I will try and upload sample data, but all simulating 100 values for each of the 9 rhos would work just to get it started like:
opt_lens_geom <- data.frame(rho= rep(seq(0.1,0.9,by=0.1),each=100),l_1024=rnorm(900))
You might consider ggridges. I've assumed here that you want a median value for each value of rho.
library(ggplot2)
library(ggridges)
library(dplyr)
set.seed(1001)
opt_lens_geom <- data.frame(rho = rep(seq(0.1, 0.9, by = 0.1), each = 100),
l_1024 = rnorm(900))
opt_lens_geom %>%
mutate(rho_f = factor(rho)) %>%
ggplot(aes(l_1024, rho_f)) +
stat_density_ridges(quantiles = 2, quantile_lines = TRUE)
Result. You can add scale = 1 as a parameter to stat_density_ridges if you don't like the amount of overlap.
Try the following. It uses a pre-computed data frame of the medians.
library(ggplot2)
df <- iris[c(1, 5)]
names(df) <- c("val", "rho")
med <- plyr::ddply(df, "rho", summarise, m = median(val))
ggplot(data = df, aes(x = val, fill = factor(rho))) +
geom_histogram() +
coord_flip() +
geom_vline(data = med, aes(xintercept = m), colour = 'black') +
facet_wrap(~ factor(rho))
You could do a variant on this using geom_violin instead of using histograms, although you wouldn't get labelled counts, just an idea of the relative density. Example with made up data:
df = data.frame(
rho = rep(c(0.1, 0.2, 0.3), each = 50),
val = sample(1:10, 150, replace = TRUE)
)
df$val = df$val + (5 * (df$rho == 0.2)) + (8 * (df$rho == 0.3))
ggplot(df, aes(x = rho, y = val, fill = factor(rho))) +
geom_violin() +
stat_summary(aes(group = 1), colour = "black",
geom = "line", fun.y = "median")
This produces a violin for each value of rho, and joins the medians for each violin.
How could I draw different smooth lines on my plot for data included in separate time periods, but draw them both on the full range of the plot?
In my working example below, even when setting the fullrange argument to TRUE, the smooth lines limit themselves, and I get the missing values warnings (which does make sense as we are setting a new data range locally in each one of the geom_smooth() functions).
# convert time series to data.frame, conserving date info
sb <- data.frame(Seatbelts, date = time(Seatbelts))
# convert from ts to date
library(lubridate)
sb$date <- as_date(date_decimal(as.numeric(sb$date)))
# store seatbelt law date
law <- ymd(19830131)
# plot
library(ggplot2)
ggplot(sb) + aes(x = date, y = front) +
geom_line() +
geom_vline(xintercept = law, colour = "red") +
geom_smooth(data = sb[sb$date < law,],
fullrange = TRUE) +
geom_smooth(data = sb[sb$date > law,],
fullrange = TRUE)
Warning messages:
Warning messages:
1: Removed 10 rows containing missing values (geom_smooth).
2: Removed 71 rows containing missing values (geom_smooth).
(currently using ggplot2 3.1.0 and R 3.5.2)
Edit:
As I thought the issue was the preliminary subsetting of the data, I also tried this cleaner version, to no avail:
# add before/after
sb$relative <- ifelse(sb$date < law, "before", "after")
# plot v.2
ggplot(sb) + aes(x = date, y = front) +
geom_line() +
geom_vline(xintercept = law, colour = "red") +
geom_smooth(aes(colour = relative),
fullrange = TRUE)
The explanation for the behaviour you're seeing has to do with the way the LOESS fit is performed; by default
dates <- seq(as.Date("1960-01-01"), law, by = "1 day")
head(setNames(predict(
loess(front ~ as.numeric(date), data = sb[sb$date < law, ]),
data.frame(date = as.numeric(dates))), dates))
1960-01-01 1960-01-02 1960-01-03 1960-01-04 1960-01-05 1960-01-06
NA NA NA NA NA NA
the behaviour of which is explained in ?predict.loess (bold-face mine)
When the fit was made using ‘surface = "interpolate"’ (the
default), ‘predict.loess’ will not extrapolate - so points outside
an axis-aligned hypercube enclosing the original data will have
missing (‘NA’) predictions and standard errors.
In order to extrapolate to points outside of the range of points used for the LOESS model we can use control = loess.control(surface = "direct") inside loess.
Unfortunately this means that we need to manually perform the two LOESS fits, predict values for two ranges of interest, and plot everything.
Here is what I would do:
Define a convenience function extrapolate.loess that predicts values for dates with a lower/upper confidence interval (based on an alpha level)
library(tidyverse)
library(broom)
extrapolate.loess <- function(data, dates, alpha = 0.95) {
loess(
front ~ as.numeric(date), data = data,
control = loess.control(surface = "direct")) %>%
augment(newdata = data.frame(date = as.numeric(dates))) %>%
transmute(
date = dates,
front = .fitted,
front.l = front - qnorm((1 - alpha) / 2) * .se.fit,
front.h = front + qnorm((1 - alpha) / 2) * .se.fit)
}
We now store LOESS estimates with CI's for the two ranges in a data.frame
dates.left <- seq(as.Date("1960-01-01"), law, by = "1 day")
df.left <- extrapolate.loess(sb[sb$date < law, ], dates.left)
dates.right <- seq(law, as.Date("1990-01-01"), by = "1 day")
df.right <- extrapolate.loess(sb[sb$date > law, ], dates.right)
Now we can plot
ggplot(sb) + aes(x = date, y = front) +
geom_line() +
geom_vline(xintercept = law, colour = "red") +
geom_line(data = df.left, colour = "blue", size = 1) +
geom_ribbon(data = df.left, aes(x = date, ymin = front.l, ymax = front.h), alpha = 0.2) +
geom_line(data = df.right, colour = "blue", size = 1) +
geom_ribbon(data = df.right, aes(x = date, ymin = front.l, ymax = front.h), alpha = 0.2) +
coord_cartesian(ylim = c(400, 1300))
I will not/cannot comment on how useful/meaningful these LOESS extrapolations are.
I have recently came across a problem with ggplot2::geom_density that I am not able to solve. I am trying to visualise a density of some variable and compare it to a constant. To plot the density, I am using the ggplot2::geom_density. The variable for which I am plotting the density, however, happens to be a constant (this time):
df <- data.frame(matrix(1,ncol = 1, nrow = 100))
colnames(df) <- "dummy"
dfV <- data.frame(matrix(5,ncol = 1, nrow = 1))
colnames(dfV) <- "latent"
ggplot() +
geom_density(data = df, aes(x = dummy, colour = 's'),
fill = '#FF6666', alpha = 0.2, position = "identity") +
geom_vline(data = dfV, aes(xintercept = latent, color = 'ls'), size = 2)
This is OK and something I would expect. But, when I shift this distribution to the far right, I get a plot like this:
df <- data.frame(matrix(71,ncol = 1, nrow = 100))
colnames(df) <- "dummy"
dfV <- data.frame(matrix(75,ncol = 1, nrow = 1))
colnames(dfV) <- "latent"
ggplot() +
geom_density(data = df, aes(x = dummy, colour = 's'),
fill = '#FF6666', alpha = 0.2, position = "identity") +
geom_vline(data = dfV, aes(xintercept = latent, color = 'ls'), size = 2)
which probably means that the kernel estimation is still taking 0 as the centre of the distribution (right?).
Is there any way to circumvent this? I would like to see a plot like the one above, only the centre of the kerner density would be in 71 and the vline in 75.
Thanks
Well I am not sure what the code does, but I suspect the geom_density primitive was not designed for a case where the values are all the same, and it is making some assumptions about the distribution that are not what you expect. Here is some code and a plot that sheds some light:
# Generate 10 data sets with 100 constant values from 0 to 90
# and then merge them into a single dataframe
dfs <- list()
for (i in 1:10){
v <- 10*(i-1)
dfs[[i]] <- data.frame(dummy=rep(v,100),facet=v)
}
df <- do.call(rbind,dfs)
# facet plot them
ggplot() +
geom_density(data = df, aes(x = dummy, colour = 's'),
fill = '#FF6666', alpha = 0.5, position = "identity") +
facet_wrap( ~ facet,ncol=5 )
Yielding:
So it is not doing what you thought it was, but it is also probably not doing what you want. You could of course make it "translation-invariant" (almost) by adding some noise like this for example:
set.seed(1234)
noise <- +rnorm(100,0,1e-3)
dfs <- list()
for (i in 1:10){
v <- 10*(i-1)
dfs[[i]] <- data.frame(dummy=rep(v,100)+noise,facet=v)
}
df <- do.call(rbind,dfs)
ggplot() +
geom_density(data = df, aes(x = dummy, colour = 's'),
fill = '#FF6666', alpha = 0.5, position = "identity") +
facet_wrap( ~ facet,ncol=5 )
Yielding:
Note that there is apparently a random component to the geom_density function, and I can't see how to set the seed before each instance, so the estimated density is a bit different each time.
this is my first stack overflow post and I am a relatively new R user, so please go gently!
I have a data frame with three columns, a participant identifier, a condition (factor with 2 levels either Placebo or Experimental), and an outcome score.
set.seed(1)
dat <- data.frame(Condition = c(rep("Placebo",10),rep("Experimental",10)),
Outcome = rnorm(20,15,2),
ID = factor(rep(1:10,2)))
I would like to construct a bar plot with two bars with the mean outcome score for each condition and the standard deviation as an error bar. I would like to then overlay lines connecting points for each participant's score in each condition. So the plot displays the individual response as well as the group mean.If it is also possible I would like to include an axis break.
I don't seem to be able to find any advice in other threads, apologies if I am repeating a question.
Many Thanks.
p.s. I realise that presenting data in this way will not be to everyones tastes. It is for a specific requirement!
This ought to work:
library(ggplot2)
library(dplyr)
dat.summ <- dat %>% group_by(Condition) %>%
summarize(mean.outcome = mean(Outcome),
sd.outcome = sd(Outcome))
ggplot(dat.summ, aes(x = Condition, y = mean.outcome)) +
geom_bar(stat = "identity") +
geom_errorbar(aes(ymin = mean.outcome - sd.outcome,
ymax = mean.outcome + sd.outcome),
color = "dodgerblue", width = 0.3) +
geom_point(data = dat, aes(x = Condition, y = Outcome),
color = "firebrick", size = 1.2) +
geom_line(data = dat, aes(x = Condition, y = Outcome, group = ID),
color = "firebrick", size = 1.2, alpha = 0.5) +
scale_y_continuous(limits = c(0, max(dat$Outcome)))
Some people are better with ggplot's stat functions and arguments than I am and might do it differently. I prefer to just transform my data first.
set.seed(1)
dat <- data.frame(Condition = c(rep("Placebo",10),rep("Experimental",10)),
Outcome = rnorm(20,15,2),
ID = factor(rep(1:10,2)))
dat.w <- reshape(dat, direction = 'wide', idvar = 'ID', timevar = 'Condition')
means <- colMeans(dat.w[, 2:3])
sds <- apply(dat.w[, 2:3], 2, sd)
ci.l <- means - sds
ci.u <- means + sds
ci.width <- .25
bp <- barplot(means, ylim = c(0,20))
segments(bp, ci.l, bp, ci.u)
segments(bp - ci.width, ci.u, bp + ci.width, ci.u)
segments(bp - ci.width, ci.l, bp + ci.width, ci.l)
segments(x0 = bp[1], x1 = bp[2], y0 = dat.w[, 2], y1 = dat.w[, 3], col = 1:10)
points(c(rep(bp[1], 10), rep(bp[2], 10)), dat$Outcome, col = 1:10, pch = 19)
Here is a method using the transfomations inside ggplot2
ggplot(dat) +
stat_summary(aes(x=Condition, y=Outcome, group=Condition), fun.y="mean", geom="bar") +
stat_summary(aes(x=Condition, y=Outcome, group=Condition), fun.data="mean_se", geom="errorbar", col="green", width=.8, size=2) +
geom_line(aes(x=Condition, y=Outcome, group=ID), col="red")
I need to add a legend of the two lines (best fit line and 45 degree line) on TOP of my two plots. Sorry I don't know how to add plots! Please please please help me, I really appreciate it!!!!
Here is an example
type=factor(rep(c("A","B","C"),5))
xvariable=seq(1,15)
yvariable=2*xvariable+rnorm(15,0,2)
newdata=data.frame(type,xvariable,yvariable)
p = ggplot(newdata,aes(x=xvariable,y=yvariable))
p+geom_point(size=3)+ facet_wrap(~ type) +
geom_abline(intercept =0, slope =1,color="red",size=1)+
stat_smooth(method="lm", se=FALSE,size=1)
Here is another approach which uses aesthetic mapping to string constants to identify different groups and create a legend.
First an alternate way to create your test data (and naming it DF instead of newdata)
DF <- data.frame(type = factor(rep(c("A", "B", "C"), 5)),
xvariable = 1:15,
yvariable = 2 * (1:15) + rnorm(15, 0, 2))
Now the ggplot code. Note that for both geom_abline and stat_smooth, the colour is set inside and aes call which means each of the two values used will be mapped to a different color and a guide (legend) will be created for that mapping.
ggplot(DF, aes(x = xvariable, y = yvariable)) +
geom_point(size = 3) +
geom_abline(aes(colour="one-to-one"), intercept =0, slope = 1, size = 1) +
stat_smooth(aes(colour="best fit"), method = "lm", se = FALSE, size = 1) +
facet_wrap(~ type) +
scale_colour_discrete("")
Try this:
# original data
type <- factor(rep(c("A", "B", "C"), 5))
x <- 1:15
y <- 2 * x + rnorm(15, 0, 2)
df <- data.frame(type, x, y)
# create a copy of original data, but set y = x
# this data will be used for the one-to-one line
df2 <- data.frame(type, x, y = x)
# bind original and 'one-to-one data' together
df3 <- rbind.data.frame(df, df2)
# create a grouping variable to separate stat_smoothers based on original and one-to-one data
df3$grp <- as.factor(rep(1:2, each = nrow(df)))
# plot
# use original data for points
# use 'double data' for abline and one-to-one line, set colours by group
ggplot(df, aes(x = x, y = y)) +
geom_point(size = 3) +
facet_wrap(~ type) +
stat_smooth(data = df3, aes(colour = grp), method = "lm", se = FALSE, size = 1) +
scale_colour_manual(values = c("red","blue"),
labels = c("abline", "one-to-one"),
name = "") +
theme(legend.position = "top")
# If you rather want to stack the two keys in the legend you can add:
# guide = guide_legend(direction = "vertical")
#...as argument in scale_colour_manual
Please note that this solution does not extrapolate the one-to-one line outside the range of your data, which seemed to be the case for the original geom_abline.