Graphing Segments in R

Question

I want to plot segmented data in R. That is, say I have data of the form

| Product               | Date   | Origination | Rate | Num | Balance   |
|-----------------------|--------|-------------|------|-----|-----------|
| DEMAND DEPOSITS       | 200505 | 198209      | 0    | 1   | 2586.25   |
| DEMAND DEPOSITS       | 200505 | 198304      | 0    | 1   | 3557.73   |
| DEMAND DEPOSITS       | 200505 | 198308      | 0    | 1   | 14923.72  |
| DEMAND DEPOSITS       | 200505 | 198401      | 0    | 1   | 4431.67   |
| DEMAND DEPOSITS       | 200505 | 198410      | 0    | 1   | 44555.23  |
| MONEY MARKET ACCOUNTS | 200505 | 198209      | 0.25 | 2   | 65710.01  |
| MONEY MARKET ACCOUNTS | 200505 | 198211      | 0.25 | 2   | 41218.41  |
| MONEY MARKET ACCOUNTS | 200505 | 198304      | 0.25 | 1   | 61421.2   |
| MONEY MARKET ACCOUNTS | 200505 | 198402      | 0.25 | 1   | 13620.17  |
| MONEY MARKET ACCOUNTS | 200505 | 198408      | 0.75 | 1   | 281897.74 |
| MONEY MARKET ACCOUNTS | 200505 | 198410      | 0.25 | 1   | 5131.33   |
| NOW ACCOUNTS          | 200505 | 198209      | 0    | 1   | 142744.35 |
| NOW ACCOUNTS          | 200505 | 198303      | 0    | 1   | 12191.6   |
| SAVING ACCOUNTS       | 200505 | 198301      | 0.25 | 1   | 96936.24  |
| SAVING ACCOUNTS       | 200505 | 198302      | 0.25 | 2   | 21764     |
| SAVING ACCOUNTS       | 200505 | 198304      | 0.25 | 1   | 14646.55  |
| SAVING ACCOUNTS       | 200505 | 198305      | 0.25 | 1   | 20909.7   |
| SAVING ACCOUNTS       | 200505 | 198306      | 0.25 | 1   | 66434.56  |
| SAVING ACCOUNTS       | 200505 | 198309      | 0.25 | 1   | 20005.56  |
| SAVING ACCOUNTS       | 200505 | 198404      | 0.25 | 2   | 16766.56  |
| SAVING ACCOUNTS       | 200505 | 198407      | 0.25 | 1   | 47721.97  |

I want to plot on the Y-axis a line per 'Product' type by 'Balance'. On the X-axis, I want to put the 'Origination'. I would ideally also like to set colors to distinguish between the lines. The data is not currently in data.frame form so let me know if I need to change back to that.

I haven't been able to find an informative solution online for this, even though I'm sure there is.

Thanks,

Please make the data [reproducible](http://stackoverflow.com/questions/5963269) and show some code effort, what have you tried and failed. — zx8754, Oct 04 '16 at 15:33

tobiasegli_te · Answer 1 · 2016-10-04T15:49:50.647

As @zx8754 menitioned, you should provide reproducible data. Without having tested the code (because there's no reproducible data), I would suggest the following, assuming that the data is in the data.frame 'data':

all_products <- unique(data$Product)
colors_use <- rainbow(length(all_products))

plot(y = data[data$Product == all_products[1],"Balance"],
    x = data[data$Product == all_products[1],"Origination"],
    type = "l",
    col = colors_use[1],
    ylim = c(min(data$Balance, na.rm = T),max(data$Balance, na.rm = T)),
    xlim = c(min(data$Origination, na.rm = T),max(data$Origination, na.rm = T)))

for(i_product in 2:length(all_products)){
    lines(y = data[data$Product == all_products[i_product],"Balance"],
        x = data[data$Product == all_products[i_product],"Origination"],
        col = colors_use[i_product])
}

This is a nice solution for using many different products. The use of the rainbow functionality was neat. Thanks, I appreciate it. — user5619709, Oct 04 '16 at 17:13

score 1 · Answer 2 · answered Oct 04 '16 at 16:05

I have not enough reputation to comment, so I write it as an answer. To make @tobiasegli_te's answer shorter, the first plot can be plot(Balance~Origination,data=data,type='n') and then make the subsequent lines done for i_product in 1:length(all_products). That way you need not worry about ylim. Here is an example using the Grunfeld data.

z <- read.csv('http://statmath.wu-wien.ac.at/~zeileis/grunfeld/Grunfeld.csv')
plot(invest~year,data=z,type='n')
for (i in unique(as.numeric(z$firm))) lines(invest~year,data=z,
    subset=as.numeric(z$firm)==i, col=i)

Also note that your Origination is not equally spaced. You need to change it to a Date or similar.

score 0 · Accepted Answer · answered Oct 04 '16 at 16:51

I guess you want something like the following:

df <- as.data.frame(df[c('Product', 'Balance', 'Origination')])
head(df)

Product  Balance Origination
1  DEMAND DEPOSITS         2586.25      198209
2  DEMAND DEPOSITS         3557.73      198304
3  DEMAND DEPOSITS        14923.72      198308
4  DEMAND DEPOSITS         4431.67      198401
5  DEMAND DEPOSITS        44555.23      198410
6  MONEY MARKET ACCOUNTS  65710.01      198209

library(ggplot2)
library(scales)
ggplot(df, aes(Origination, Balance, group=Product, col=Product)) + 
         geom_line(lwd=1.2) + scale_y_continuous(labels = comma)

Awesome. This worked and was pretty simple. I ended up having to restart R to get it to work correctly. Thanks. — user5619709, Oct 04 '16 at 17:13

score 0 · Answer 4 · answered Oct 04 '16 at 18:05

I am not sure what you my want, is that what you re looking for?

Assuming you put your data in data.txt, removing the pipes and replacing the spaces in the names by '_'

d = read.table("data.txt", header=T)

prod.col = c("red", "blue", "green", "black" )
prod = unique(d$Product)

par(mai = c(0.8, 1.8, 0.8, 0.8))
plot(1, yaxt = 'n', type = "n", axes = TRUE, xlab = "Origination", ylab = "", xlim = c(min(d$Origination), max(d$Origination)), ylim=c(0, nrow(d)+5) )
axis(2, at=seq(1:nrow(d)), labels=d$Product, las = 2, cex.axis=0.5)
mtext(side=2, line=7, "Products")

for( i in 1:nrow(d) ){
myProd = d$Product[i]
myCol = prod.col[which(prod == myProd)]
myOrig  = d$Origination[i]
segments( x0 = 0, x1 = myOrig, y0 = i, y1 = i, col = myCol, lwd = 5 )
}
legend( "topright", col=prod.col, legend=prod, cex=0.3, lty=c(1,1), bg="white" )

That's a really neat plot. Not quite what I was looking for. It was to construct a time series of the product balances by origination date. But that plotting technique might be useful in the future for sure. Thanks. — user5619709, Oct 04 '16 at 18:19
For your next questions, don't hesitate to draw by hand a sketch of what you want, a picture worths thousand words ;) — Xavier Prudent, Oct 05 '16 at 15:19

Graphing Segments in R

4 Answers4