I'm trying to predict a value in R using the predict() function, by passing along variables into the model.
I am getting the following error:
Error in eval(predvars, data, env) :
numeric 'envir' arg not of length one
Here is my data frame, name df:
df <- read.table(text = '
Quarter Coupon Total
1 "Dec 06" 25027.072 132450574
2 "Dec 07" 76386.820 194154767
3 "Dec 08" 79622.147 221571135
4 "Dec 09" 74114.416 205880072
5 "Dec 10" 70993.058 188666980
6 "Jun 06" 12048.162 139137919
7 "Jun 07" 46889.369 165276325
8 "Jun 08" 84732.537 207074374
9 "Jun 09" 83240.084 221945162
10 "Jun 10" 81970.143 236954249
11 "Mar 06" 3451.248 116811392
12 "Mar 07" 34201.197 155190418
13 "Mar 08" 73232.900 212492488
14 "Mar 09" 70644.948 203663201
15 "Mar 10" 72314.945 203427892
16 "Mar 11" 88708.663 214061240
17 "Sep 06" 15027.252 121285335
18 "Sep 07" 60228.793 195428991
19 "Sep 08" 85507.062 257651399
20 "Sep 09" 77763.365 215048147
21 "Sep 10" 62259.691 168862119', header=TRUE)
str(df)
'data.frame': 21 obs. of 3 variables:
$ Quarter : Factor w/ 24 levels "Dec 06","Dec 07",..: 1 2 3 4 5 7 8 9 10 11 ...
$ Coupon: num 25027 76387 79622 74114 70993 ...
$ Total: num 132450574 194154767 221571135 205880072 188666980 ...
Code:
model <- lm(df$Total ~ df$Coupon)
> model
Call:
lm(formula = df$Total ~ df$Coupon)
Coefficients:
(Intercept) df$Coupon
107286259 1349
Now, when I run predict, I get the error I showed above.
> predict(model, df$Total, interval="confidence")
Error in eval(predvars, data, env) :
numeric 'envir' arg not of length one
Any idea where I am going wrong?
Thanks
There are several problems here:
The newdata argument of predict() needs a predictor variable. You should thus pass it values for Coupon, instead of Total, which is the response variable in your model.
The predictor variable needs to be passed in as a named column in a data frame, so that
predict() knows what the numbers its been handed represent. (The need for this becomes clear when you consider more complicated models, having more than one predictor variable).
For this to work, your original call should pass df in through the data argument, rather than using it directly in your formula. (This way, the name of the column in newdata will be able to match the name on the RHS of the formula).
With those changes incorporated, this will work:
model <- lm(Total ~ Coupon, data=df)
new <- data.frame(Coupon = df$Coupon)
predict(model, newdata = new, interval="confidence")
Source: Stackoverflow.com