R blueprints / floorplan - r

I'm working on trying to represent an office building in R. Later, I'll need to represent multiple floors, but for now I need to start with one floor. There are clusters of cubes all in a regular structure. There are four small cubes for junior staff (4x4), and two larger cubes for a senior engineer and a manager (4x6). Once these are mapped out, I need to be able to show if they are occupied or free for new hires -- by color (like red for occupied, green for available). These are all laid out the same way, with the big ones on one end. For example,
+----+--+--+
| S |J1|J2|
+----+--+--+
<-hallway-->
+----+--+--+
| M |J3|J4|
+----+--+--+
I first thought I could use ggplot and just scatter plot everybody out, but I can't figure out how to capture the different size cubes with geom_point. I spent some time looking at maps, but it seems like I can't really take advantage of the regular structure of my floorplan -- maybe that really is the way to go and I take advantage of my regular structure in building out a map? Does R have a concept I should Google for this kind of structure?
In the end, I'll get a long data file, with the type of cubicle, the x and y coordinates of the cluster, and a "R" or "G" (4 columns).

You could also write a low-level graphic function; it's sometimes easier to tune than removing more and more components from a complex plot,
library(grid)
library(gridExtra)
floorGrob <- function(S = c(TRUE, FALSE), J = c(TRUE, FALSE, TRUE, TRUE),
draw=TRUE, newpage=is.null(vp), vp=NULL){
m <- rbind(c(1,3,4), # S1 J1 J2
c(7,7,7), # hall
c(2,5,6)) # S2 J3 J4
fills <- c(c("#FBB4AE","#CCEBC5")[c(S, J)+1], "grey90")
cellGrob <- function(f) rectGrob(gp=gpar(fill=f, col="white", lwd=2))
grobs <- mapply(cellGrob, f=fills, SIMPLIFY = FALSE)
g <- arrangeGrob(grobs = grobs, layout_matrix = m, vp = vp, as.table = FALSE,
heights = unit(c(4/14, 1/14, 4/14), "null"),
widths = unit(c(6/14, 4/14, 4/14), "null"), respect=TRUE)
if(draw) {
if(newpage) grid.newpage()
grid.draw(g)
}
invisible(g)
}
floorGrob()

How about?
df <- expand.grid(x = 0:5, y = 0:5)
df$color <- factor(sample(c("green", "red"), 36, replace = T))
head(df)
# x y color
# 1 0 0 green
# 2 1 0 green
# 3 2 0 green
# 4 3 0 red
# 5 4 0 green
# 6 5 0 red
library(ggplot2)
ggplot(df, aes(x, y, fill = color)) +
geom_tile() +
scale_fill_manual(name = "Is it open?",
values = c("lightgreen", "#FF3333"),
labels = c("open", "not open"))

Related

Depict supply network structures

I would like to plot a supply network structure. I have tried to use igraph, but until now did not come up with a reasonable result. An example would look like this:
library(igraph)
d <- read.table(text = "V1 V2 weight
s1 p1 88
s3 p1 100
s2 p2 100
s3 p2 43
p1 c1 21
p1 c2 79
p1 c3 88
p2 c1 22
p2 c2 121
", stringsAsFactors = F, header = T)
g <- graph_from_data_frame(d, directed = T)
plot(g, layout=layout.fruchterman.reingold,
edge.width=E(g)$weight/20,
vertex.shape = "none", vertex.label.font = 2,
vertex.label.cex=1.1, edge.color="gray70")
Which gives:
The problem is that the network has an additional structure. A resonable - among others - result would show the "s"-nodes (for suppliers) should be in the left third, the "p"-nodes (plants) should be in the middle and the c-nodes (customers) on the right hand side. Is this even doable with igraph (and how)? Is there another package that could do this?
Yes, this is doable with igraph. One way to make your own layout. A simple way to do this is to place all "s" nodes at x=1, "p" nodes at x=2 and "c" nodes at x=3. Each distinct node of each type (s,p,c) should get a unique y value so that they do not overlap. Using your example graph:
LO = matrix(0, nrow=vcount(g), ncol=2)
LO[grep("s", V(g)$name), 1] = 1
LO[grep("p", V(g)$name), 1] = 2
LO[grep("c", V(g)$name), 1] = 3
LO[,2] = ave(rep(1, vcount(g)), LO[,1], FUN = seq_along)
plot(g, layout=LO, edge.width=E(g)$weight/20,
vertex.shape = "none", vertex.label.font = 2,
vertex.label.cex=1.1, edge.color="gray70")
Also, following up on the comment of #Henrik, you can use layout_with_sugiyama. You still need to define the (s,p,c)-layers. Also, sugiyama arranges the layers vertically. You need to swap the x and y coordinates to get a horizontal layout.
Layers = rep(0,vcount(g))
Layers[grep("s", V(g)$name)] = 3
Layers[grep("p", V(g)$name)] = 2
Layers[grep("c", V(g)$name)] = 1
LO2 = layout_with_sugiyama(g, layers=Layers)$layout
LO2 = LO2[,2:1]
plot(g, layout=LO2, edge.width=E(g)$weight/20,
vertex.shape = "none", vertex.label.font = 2,
vertex.label.cex=1.1, edge.color="gray70")

Mapping slope of an area and returning percent above and below a threshold in R

I am trying to figure our the proportion of an area that has a slope of 0, +/- 5 degrees. Another way of saying it is anything above 5 degrees and below 5 degrees are bad. I am trying to find the actual number, and a graphic.
To achieve this I turned to R and using the Raster package.
Let's use a generic country, in this case, the Philippines
{list.of.packages <- c("sp","raster","rasterVis","maptools","rgeos")
new.packages <- list.of.packages[!(list.of.packages %in% installed.packages()[,"Package"])]
if(length(new.packages)) install.packages(new.packages)}
library(sp) # classes for spatial data
library(raster) # grids, rasters
library(rasterVis) # raster visualisation
library(maptools)
library(rgeos)
Now let's get the altitude information and plot the slopes.
elevation <- getData("alt", country = "PHL")
x <- terrain(elevation, opt = c("slope", "aspect"), unit = "degrees")
plot(x$slope)
Not very helpful due to the scale, so let's simply look at the Island of Palawan
e <- drawExtent(show=TRUE) #to crop out Palawan (it's the long skinny island that is roughly midway on the left and is oriented between 2 and 8 O'clock)
gewataSub <- crop(x,e)
plot(gewataSub, 1)## Now visualize the new cropped object
A little bit better to visualize. I get a sense of the magnitude of the slopes and that with a 5 degree restriction, I am mostly confined to the coast. But I need a little bit more for analysis.
I would like Results to be something to be in two parts:
1. " 35 % (made up) of the selected area has a slope exceeding +/- 5 degrees" or " 65 % of the selected area is within +/- 5 degrees". (with the code to get it)
2. A picture where everything within +/- 5 degrees is one color, call it good or green, and everything else is in another color, call it bad or red.
Thanks
There are no negative slopes, so I assume you want those that are less than 5 degrees
library(raster)
elevation <- getData('alt', country='CHE')
x <- terrain(elevation, opt='slope', unit='degrees')
z <- x <= 5
Now you can count cells with freq
f <- freq(z)
If you have a planar coordinate reference system (that is, with units in meters or similar) you can do
f <- cbind(f, area=f[,2] * prod(res(z)))
to get areas. But for lon/lat data, you would need to correct for different sized cells and do
a <- area(z)
zonal(a, z, fun=sum)
And there are different ways to plot, but the most basic one
plot(z)
You can use reclassify from the raster package to achieve that. The function assigns each cell value that lies within a defined interval a certain value. For example, you can assign cell values within interval (0,5] to value 0 and cell values within the interval (5, maxSlope] to value 1.
library(raster)
library(rasterVis)
elevation <- getData("alt", country = "PHL")
x <- terrain(elevation, opt = c("slope", "aspect"), unit = "degrees")
plot(x$slope)
e <- drawExtent(show = TRUE)
gewataSub <- crop(x, e)
plot(gewataSub$slope, 1)
m <- c(0, 5, 0, 5, maxValue(gewataSub$slope), 1)
rclmat <- matrix(m, ncol = 3, byrow = TRUE)
rc <- reclassify(gewataSub$slope, rclmat)
levelplot(
rc,
margin = F,
col.regions = c("wheat", "gray"),
colorkey = list(at = c(0, 1, 2), labels = list(at = c(0.5, 1.5), labels = c("<= 5", "> 5")))
)
After the reclassification you can calculate the percentages:
length(rc[rc == 0]) / (length(rc[rc == 0]) + length(rc[rc == 1])) # <= 5 degrees
[1] 0.6628788
length(rc[rc == 1]) / (length(rc[rc == 0]) + length(rc[rc == 1])) # > 5 degrees
[1] 0.3371212

specify colorRamp endpoints

Can I specify endpoints to colorRamp so that a value maps consistently to a single color, regardless of the range of other data?
I'm trying to create an interactive correlation plot in plotly. Here's some sample data.
set.seed(1)
m <- 4
cm <- matrix(runif(m**2,-1,1),
nrow=m, ncol=m,
dimnames=list(letters[1:m],letters[1:m]))
diag(cm) <- 1
cm
# a b c d
# a 1.0000000 -0.5966361 0.2582281 0.3740457
# b -0.2557522 1.0000000 -0.8764275 -0.2317926
# c 0.1457067 0.8893505 1.0000000 0.5396828
# d 0.8164156 0.3215956 -0.6468865 1.0000000
I'm basically trying to create an interactive version of this:
library(corrplot)
corrplot(cm,method='shade')
Here's the (kind of hacky) interactive correlation plot I created.
div_colors <- c('dark red','white','navy blue')
grid_labels <- matrix(paste0('Cor(',
do.call(paste,c(expand.grid(rownames(cm),colnames(cm)), sep=', ') ),
'): ',
t(round(cm,2))
),
ncol=m,byrow=TRUE)
library(plotly)
plot_ly(x = colnames(cm),
y = rownames(cm),
z = cm,
colors = colorRamp(div_colors),
type='heatmap',
hoverinfo='text',
text = grid_labels
) %>% layout(yaxis=list(autorange='reversed'))
My problem is that without forcing the colorRamp endpoints to c(-1,1), the white color doesn't match correlation of 0, and the dark red maps to the minimum observed, rather than -1.
As #rawr mentioned in a comment, the solution is to set zmin and zmax, as in:
plot_ly(x = colnames(cm),
y = rownames(cm),
z = cm,
zmin=-1, # <============
zmax=1, # <============
colors = colorRamp(div_colors),
type='heatmap',
hoverinfo='text',
text = grid_labels
) %>% layout(yaxis=list(autorange='reversed'))
Which produces the desired result. (The legend bar is shorter, presumably due to a change in default sizes in a newer version of plotly.)

Add a line to coplot {graphics}, classic approaches don't work

I found coplot {graphics} very useful for my plots. However, I would like to include there not only one line, but add there one another. For basic graphic I just need to add = TRUE to add another line, or tu use plot(..) and lines(..). For {lattice} I can save my plots as objects
a<-xyplot(..)
b<-xyplot(..)
and display it simply by a + as.layer(b). No one of these approaches works for coplot(), apparently because creating objects as a<-coplot() doesn't produce trellis graphic but NULL object.
Please, any help how to add data line in coplot()? I really like its graphic so I wish to keep it. Thank you !!
my exemle data are here: http://ulozto.cz/xPfS1uRH/repr-exemple-csv
My code:
sub.tab<-read.csv("repr_exemple.csv", , header = T, sep = "")
attach(sub.tab)
cells.f<-factor(cells, levels=c(2, 25, 100, 250, 500), # unique(cells.in.cluster)???
labels=c("size2", "size25", "size100", "size250", "size500"))
perc.f<-factor(perc, levels=c(5, 10), # unique(cells.in.cluster)???
labels=c("perc5", "perc10"))
# how to put these plots together?
a<- coplot(max_dist ~ time |cells.f + perc.f, data = sub.tab,
xlab = "ticks", type = "l", col = "black", lwd = 1)
b<- coplot(mean_dist ~ time |cells.f * perc.f, data = sub.tab,
xlab = "ticks", type = "l", col = "grey", lwd = 1)
a + as.layer(b) # this doesn't work
Please, how to merge these two plots (grey and black lines)? I couldn't figure it out... Thank you !
Linking to sample data isn't really as helpful. Here's a randomly created sample data set
set.seed(15)
dd <- do.call("rbind",
do.call("Map", c(list(function(a,b) {
cbind.data.frame(a,b, x=1:5,
y1=cumsum(rpois(5,7)),
y2=cumsum(rpois(5,9)))
}),
expand.grid(a=letters[1:5], b=letters[20:22])))
)
head(dd)
# a b x y1 y2
# 1 a t 1 8 16
# 2 a t 2 13 28
# 3 a t 3 25 35
# 4 a t 4 33 45
# 5 a t 5 39 57
# 6 b t 1 4 12
I will note the coplot is a base graphics function, not Lattice. But it does have a panel= parameter. And you can have the coplot() take care of subsetting your data for you (well, calculating the indexes at least). But, like other base graphics functions, plotting different groups isn't exactly trivial. You can do it in this case with
coplot(y~x|a+b,
# make a fake y col to cover range of all y1 and y2 values
cbind(dd, y=seq(min(dd$y1, dd$y2), max(dd$y1, dd$y2), length.out=nrow(dd))),
#request subscripts to be sent to panel function
subscripts=TRUE,
panel=function(x,y,subscripts, ...) {
# draw group 1
lines(x, dd$y1[subscripts])
# draw group 2
lines(x, dd$y2[subscripts], col="red")
})
This gives

Print frequencies (as numbers) in plot

In R, I would like to insert frequencies (as numbers) in a plot:
my code to create the plot:
par(mar=c(4.5,4.5,9.5,4), xpd=TRUE)
plot(factor(ArtMehrspr)~Mehrspr_Vielf, data=datProjektMehr, col=terrain.colors(4),
bty='L', main="Vielfalt nutzen")
legend("topright", inset=c(0,-.225), title="Art der Mehrsprachigkeit", levels(factor(datProjektMehr$ArtMehrspr)),
fill=terrain.colors(4), horiz=TRUE)
par(mar=c(5,4,4,2)+0.1)
In the plot, 2 columns of my dataframe are depicted: ArtMehrspr and Mehrspr_Vielf.
Now what I would like to know is, how many "Kombi" are in category "1", how many "Paral" are in category "1" and so on, and then to print this number in the plot, so that in every box of the plot, I can see the corresponding number of observations. R must know these numbers, otherwise it could not vary the height of the different boxes according to the number of observations. So it cannot be that hard to get these numbers into the plot, can it?
With the command table(), I can get these numbers, but I would have to have 5 table()-commands to get all the numbers. Example for category = 1:
> table(subset(datProjektMehr, Mehrspr_Vielf=="1")$ArtMehrspr)
einspr Kombi Paral Versc Wechs
0 1 9 2 1
Apparently, you can achieve what I am looking for by adding the command labels = TRUE. But it does not work:
par(mar=c(4.5,4.5,9.5,4), xpd=TRUE, labels = TRUE)
plot(factor(ArtMehrspr)~Mehrspr_Vielf, data=datProjektMehr, col=terrain.colors(4),
bty='L', main="Vielfalt nutzen")
legend("topright", inset=c(0,-.225), title="Art der Mehrsprachigkeit", levels(factor(datProjektMehr$ArtMehrspr)),
fill=terrain.colors(4), horiz=TRUE)
par(mar=c(5,4,4,2)+0.1)
R gives me the following warning message:
Warning message:
In par(mar = c(4.5, 4.5, 9.5, 4), xpd = TRUE, labels = TRUE) :
"labels" is not a graphical parameter
Is this not the right command? Does anyone know how to do this?
First of all, the warning informs that there is not a labels argument you can use inside par.
Regarding the plotting of the table output, I'm not aware if there is an easy way of doing this, but I managed a pretty UNreliable and, maybe, inefficient code. In my machine, though, it works every time I run it.
The concept I had in mind is to text all values from your table inside the plot. To do so, coordinates in xx' and yy' had to be estimated. I prefer the term "estimated" instead of "calculated" because I didn't find a way to compute absolute values for the coordinates, due to the fact that the plot method was plot.factor.
So:
#random data. DF = datProjektMehr, artmehr = ArtMehrspr, mehrviel = Mehrspr_Vielf
DF <- data.frame(artmehr = sample(letters[1:4], 20, T), mehrviel = as.factor(sample(1:5, 20, T)))
#your code of plotting
par(mar = c(4.5,4.5,9.5,4), xpd = TRUE)
plot(factor(artmehr) ~ mehrviel, data = DF, col = terrain.colors(4),
bty = 'L', main = "Vielfalt nutzen")
legend("topright", inset=c(0,-.225), title="Art der Mehrsprachigkeit", levels(factor(DF$artmehr)),
fill=terrain.colors(4), horiz=TRUE)
#no need to "table()" many times
tab = table(DF$artmehr, DF$mehrviel)
#maximum value of x axis (at least in my machine)
#I found -through trial and error- that for a factor of n levels, x.max = 1 + (n-1)*0.02
x.max = 1 + (length(levels(DF$mehrviel)) - 1) * 0.02
#coordinates of "mehrviel" (as I named it)
mehrviel.coords = ((cumsum(apply(tab, 2, sum)) / sum(tab)) * x.max) - ((apply(tab, 2, sum) / sum(tab)) / 2)
#coordinates of "artmehr" (as I named it)
artmehr.coords <- apply(tab, 2, function(x) { cumsum(x / sum(x)) })
artmehr.coords <- apply(artmehr.coords, 2, function(x) { x - c(x[1]/2, diff(x)/2) })
#"text" the values in your table
#don't plot "0"s
for(i in 1:ncol(artmehr.coords))
{
text(x = mehrviel.coords[i], y = artmehr.coords[,i], labels = ifelse(tab[,i] != 0, tab[,i], ""), cex = 2)
}
The values of table:
tab
1 2 3 4 5
a 1 1 0 1 0
b 0 0 2 1 2
c 1 1 2 1 0
d 2 0 0 3 2
The plot:
EDIT: 1) "Tidied" the answer. 2) Aadded an extra level to the factor ploted in xx' axis to match your data exactly. 3)texted the frequencies in the middle of each box.

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