我试图绘制最终结果。我已将我的响应变量转换为
expon=^0.2
获得描述效果的回归线。 我想保持直线并在y轴上设置指数刻度,而不是用曲线说明效果。我是否会忽略帮助页面中的某个部分?
到目前为止的相关输入是:
expon = 0.2 ## best fit for assumptions AND correlation
tempMin = 0
dmaxDS_Summer_noSO4$response = dmaxDS_Summer_noSO4$TempChange^expon
plot(response[ind & !indClosed&!indLarge]~Sunshine[ind &!indClosed&!indLarge],
exp="y",data=dmaxDS_Summer_noSO4,pch=21,col="red",
main='All Small Sites',ylab="Temperature Change^0.2",xlab="Sunshine Hrs")
abline(a[1]a[5]lwd=3,col="red")
任何指针都会受到赞赏。
答案 0 :(得分:0)
您可以使用log
函数的plot
自变量。请参见下面的代码:
# Data Simulation
expon <- 0.2 ## best fit for assumptions AND correlation
tempMin <- 0
dmaxDS_Summer_noSO4 <- data.frame(TempChange = 0.2 ^ 0:100 , Sunshine = 0:100 %/% 12)
dmaxDS_Summer_noSO4$response = dmaxDS_Summer_noSO4$TempChange ^ expon
ind = seq_along(nrow(dmaxDS_Summer_noSO4))
indClosed = FALSE
indLarge = FALSE
# Plotting
plot(response[ind & !indClosed & !indLarge] ~ Sunshine [ind & !indClosed & !indLarge],
data = dmaxDS_Summer_noSO4,
pch = 21,
col = "red",
main = "All Small Sites",
ylab = "Temperature Change ^ 0.2",
xlab = "Sunshine Hrs",
log = "y")