我更喜欢使用基本图形，而不是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))))

您可以轻松指定替代平滑方法（如lm()、线性最小二乘拟合）和替代公式

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")

置信带自动包括：我改变了颜色，使他们可见，因为他们是非常狭窄的。你可以使用se=FALSE在stat_smooth关闭他们。

另一个答案告诉你如何得到系数：

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

我可以想象你接下来可能想把这个方程加到这个图上：参见Adding Regression Line Equation and R2 on graph

我确信一个简单的+scale_y_log10（）就能得到你想要的结果。GGPlot统计数据是在转换后计算的，所以loess（）是在对数转换后的数据上计算的。

我刚刚编写了一个blog post here，它描述了如何精确地匹配Excel的对数曲线拟合，该方法的核心是lm()函数：

# 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"))

希望能有所帮助。