问题
> reg.len <- lm(chao1.ave ~ lg.std.len, b.div) # b.div is my data frame imported from a CSV file
> reg.len
Call:
lm(formula = chao1.ave ~ lg.std.len, data = b.div)
Coefficients:
(Intercept) lg.std.len
282.4 -115.7
> newx <- seq(0.6, 1.4, 0.01)
> prd.len <- predict(reg.len, newdata=data.frame(x=newx), interval="confidence", level=0.90, type="response")
Error in eval(expr, envir, enclos) : object 'lg.std.len' not found
I've tried doing the lm like this: lm(b.div$chao1.ave ~ b.div$lg.std.len)
, but then, predict()
gives a warnings that the newdata and variables are different lengths. So, I tried the way above, and now predict()
gives an error saying it doesn't recognize the object. How to fix, please?
回答1:
Predict expects newdata to have the same column names (to match the formula in reg.len). You're changing it to "x" in your newdata specification, which isn't part of the formula.
dat <- data.frame(y=rnorm(50),lg.std.len=sample(10:15,50,replace=TRUE))
reg.len <- lm(y ~ lg.std.len,data=dat)
newx <- seq(0.6, 1.4, 0.01)
prd.len <- predict(reg.len, newdata=data.frame(lg.std.len=newx),
interval="confidence", level=0.90, type="response")
The key part is newdata=data.frame(lg.std.len=newx)
来源:https://stackoverflow.com/questions/12923541/r-predict-lm-not-recognizing-an-object