Reputation: 153
I want to write the code for plotting my logistic regression model, i.e., the "S"-shape logistic curve. How could that be done please as I have have two independent covariates? I'm attaching my data set, and the code for my model. Thank you in advance.
239 0.72 1
324.6 0.83 1
331.8 0.95 1
334.3 0.83 1
259.7 0.89 1
212.3 0.88 1
204.7 0.65 1
253.86 0.75 1
258.94 0.85 1
329.66 0.95 0
469.68 1.46 0
459.74 1.11 0
293.2 0.64 0
297.88 0.98 0
267.9 0.82 0
374.1 1.29 0
333.62 0.74 0
dat <- read.table("data.txt")
colnames(dat)<-c("press","v","gender")
# logostic regression
dat$gender <- factor(dat$gender)
mylogit<- glm(gender~press+v,data=dat,family="binomial")
summary(mylogit)
######## the code below are irrelevant to making plot, ignore if you want
mylogit$fitted.values
newdat <- data.frame(t(c(300,0.1)))
colnames(newdat)<-c("press","v")
# this is your new dataset, we name it as "newdat"
pred <- predict(mylogit,newdata = newdat,type="response")
pred # the probability of being in class 1 will stored in this object
pred <- predict(mylogit,newdata = dat,type="response")
pred # the probability of being in class 1 will stored in this object
# accuracy
dat$pred <- 0
factor(dat$pred)
dat$pred[which(pred>0.5)] <- 1
table(dat$gender,dat$pred)
Upvotes: 3
Views: 1683
Reputation: 73265
You have 2 continuous, non-categorical variables, so the logistic curve will be a 3D curve. I will offer you two ways for presentation.
persp
function to produce a real 3D smooth curve;v
at a number of values, then produce a number of 2D logistic curve (which you called "S"-shape curve).3D curve
press_grid <- seq(200, 480, by = 5)
v_grid <- seq(0.6, 1.5, by = 0.1)
newdat <- data.frame(press = rep(press_grid, times = length(v_grid)), v = rep(v_grid, each = length(press_grid)))
pred <- predict.glm(mylogit, newdata = newdat, type="response")
z <- matrix(pred, length(press_grid))
persp(press_grid, v_grid, z, xlab = "pressure", ylab = "velocity", zlab = "predicted probability", main = "logistic curve (3D)", theta = 30, phi = 20)
You need to first generate a 2D grid. The newdat
holds this grid, and you can do plot(newdat)
to see this grid. Then prediction will take place on this grid, by calling predict.glm(..., type = "response")
. The result pred
is a vector. To plot it, cast it to a matrix z
, then invoke persp
to make 3D plot. xlab
, ylab
and zlab
are labels for three axis. The parameters theta
and phi
are used to tweak your viewing angles.
In the above, the marginal grid for press
and v
are based on the range of your original data: range(dat$press)
and range(dat$v)
. We don't make prediction beyond this range too far. But even within this range, you only have 17 observations. So you need still be sceptical on the plot.
Here is the curve:
2D curves
This toy function is useful for making a 2D curve, with v
fixed as some level:
curve_2D_fix_v <- function(model, v = 1, press_grid = seq(200, 480, by = 5), add = FALSE, col = "black") {
newdat <- data.frame(press = press_grid, v = v)
pred <- predict.glm(model, newdat, type = "response")
if (add) lines(press_grid, pred, col = col) else {
plot(press_grid, pred, xlab = "pressure", ylab = "predicted probability", type = "l", col = col, main = "logistic curve (2D)")
abline(h = c(0, 0.5, 1), lty = 2, col = col)
}
}
If add = FALSE
, it opens a new plotting window; while it is TRUE
, it plots on previous window (but it is your duty to make sure there is such a window!) The 2D plot gives more information, because you can add a horizontal line at 0, 0.5 and 1.
Let's have a go:
curve_2D_fix_v(mylogit, v = 0.4, add = FALSE, col = "black")
curve_2D_fix_v(mylogit, v = 0.6, add = TRUE, col = "red")
curve_2D_fix_v(mylogit, v = 0.8, add = TRUE, col = "green")
curve_2D_fix_v(mylogit, v = 1, add = TRUE, col = "blue")
curve_2D_fix_v(mylogit, v = 1.2, add = TRUE, col = "cyan")
curve_2D_fix_v(mylogit, v = 0.4, add = TRUE, col = "yellow")
Here is the curve:
Discussion
In both plots, we see that the relationship between gender
(predicted probability) and v
(velocity) is not very strong. In 2D plot, almost all values of v
produce the same curve. On the other hand, press
(pressure) is a strong effect.
Back to your model:
> summary(mylogit)
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 8.08326 4.45463 1.815 0.0696 .
press -0.02575 0.01618 -1.591 0.1115
v -0.15385 4.83824 -0.032 0.9746
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
You can see that v
is not significant at all! While strictly speaking, press
is also not significant at 0.1 level. So this is a very weak model. I suggest you drop variable v
and do the model again, using press
as the only variable.
Upvotes: 11