add a logarithmic regression line to a scatterplot

问题:

In Excel, its pretty easy to fit a logarithmic trend line of a given set of trend line. Just click add trend line and then select "Logarithmic." Switching to R for more power, I am a bit lost as to which function should one use to generate this.

To generate the graph, I used ggplot2 with the following code.

ggplot(data, aes(horizon, success)) + geom_line() + geom_area(alpha=0.3)+
  stat_smooth(method='loess')

But the code does local polynomial regression fitting which is based on averaging out numerous small linear regressions. My question is whether there is a similar log trend line in R that is used in Excel.

Edit: An alternative I am looking for is to get an log equation in form y = (c*ln(x))+b; is there a coef() function to get 'c' and 'b'?

Edit2: Since I have more reputation, I can now post a bit more about what I am struggling to do. Let my data be:

0.599885189,0.588404133,0.577784156,0.567164179,0.556257176,0.545350172,0.535112897,
0.52449292,0.51540375,0.507271336,0.499904325,0.498851894,0.498851894,0.497321087,
0.4964600,0.495885955,0.494068121,0.492154612,0.490145427,0.486892461,0.482395714,
0.477229238,0.471010333

The above data are y-points while the x-points are simply integers from 1:length(y) in increment of 1. In Excel: I can simply plot this and add a logarithmic trend line and the result would look:

With black being the log. In R, how would one do this with the above dataset?

回答1:

You can easily specify alternative smoothing methods (such as lm(), linear least-squares fitting) and an alternative formula

library(ggplot2)
g0 <- ggplot(dat, aes(horizon, success)) + geom_line() + geom_area(alpha=0.3)
g0 + stat_smooth(method="lm",formula=y~log(x),fill="red")

The confidence bands are automatically included: I changed the color to make them visible since they're very narrow. You can use se=FALSE in stat_smooth to turn them off.

The other answer shows you how to get the coefficients:

coef(lm(success~log(horizon),data=dat))

I can imagine you might next want to add the equation to the graph: see Adding Regression Line Equation and R2 on graph

回答2:

I prefer to use base graphics instead of ggplot2:

#some data with a linear model
x <- 1:20
set.seed(1)
y <- 3*log(x)+5+rnorm(20)

#plot data
plot(y~x)

#fit log model
fit <- lm(y~log(x))
#look at result and statistics
summary(fit)
#extract coefficients only
coef(fit)

#plot fit with confidence band
matlines(x=seq(from=1,to=20,length.out=1000),
         y=predict(fit,newdata=list(x=seq(from=1,to=20,length.out=1000)),
                   interval="confidence"))

#some data with a non-linear model
set.seed(1)
y <- log(0.1*x)+rnorm(20,sd=0.1)

#plot data
plot(y~x)

#fit log model
fit <- nls(y~log(a*x),start=list(a=0.2))
#look at result and statistics
summary(fit)

#plot fit
lines(seq(from=1,to=20,length.out=1000),
      predict(fit,newdata=list(x=seq(from=1,to=20,length.out=1000))))

回答3:

I'm pretty sure a simple +scale_y_log10() would get you what you wanted. GGPlot stats are calculated after transformations, so the loess() would then be calculated on the log transformed data.

回答4:

I've just written a blog post here that describes how to match Excel's logarithmic curve fitting exactly. The nub of the approach centers around the lm() function:

# Set x and data.to.fit to the independent and dependent variables
data.to.fit <- c(0.5998,0.5884,0.5777,0.5671,0.5562,0.5453,0.5351,0.524,0.515,0.5072,0.4999,0.4988,0.4988,0.4973,0.49,0.4958,0.4940,0.4921,0.4901,0.4868,0.4823,0.4772,0.4710)
x <- c(seq(1, length(data.to.fit)))
data.set <- data.frame(x, data.to.fit)

# Perform a logarithmic fit to the data set
log.fit <- lm(data.to.fit~log(x), data=data.set)

# Print out the intercept, log(x) parameters, R-squared values, etc.
summary(log.fit)

# Plot the original data set
plot(data.set)

# Add the log.fit line with confidence intervals
matlines(predict(log.fit, data.frame(x=x), interval="confidence"))

Hope that helps.