I posted earlier today about an error I was getting with using the predict
function. I was able to get that corrected, and thought I was on the right path.
I have a number of observations (actuals) and I have a few data points that I want to extrapolate or predict. I used lm
to create a model, then I tried to use predict
with the actual value that will serve as the predictor input.
This code is all repeated from my previous post, but here it is:
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, data=df)
> model
Call:
lm(formula = df$Total ~ df$Coupon)
Coefficients:
(Intercept) df$Coupon
107286259 1349
Predict code (based on previous help):
(These are the predictor values I want to use to get the predicted value)
Quarter = c("Jun 11", "Sep 11", "Dec 11")
Total = c(79037022, 83100656, 104299800)
Coupon = data.frame(Quarter, Total)
Coupon$estimate <- predict(model, newdate = Coupon$Total)
Now, when I run that, I get this error message:
Error in `$<-.data.frame`(`*tmp*`, "estimate", value = c(60980.3823396919, :
replacement has 21 rows, data has 3
My original data frame that I used to build the model had 21 observations in it. I am now trying to predict 3 values based on the model.
I either don't truly understand this function, or have an error in my code.
Help would be appreciated.
Thanks
instead of newdata you are using newdate in your predict code, verify once. and just use
Coupon$estimate <- predict(model, Coupon)
It will work.First, you want to use
not
model <-lm(df$Total ~ df$Coupon, data=df)
.Second, by saying
lm(Total ~ Coupon)
, you are fitting a model that usesTotal
as the response variable, withCoupon
as the predictor. That is, your model is of the formTotal = a + b*Coupon
, witha
andb
the coefficients to be estimated. Note that the response goes on the left side of the~
, and the predictor(s) on the right.Because of this, when you ask R to give you predicted values for the model, you have to provide a set of new predictor values, ie new values of
Coupon
, notTotal
.Third, judging by your specification of
newdata
, it looks like you're actually after a model to fitCoupon
as a function ofTotal
, not the other way around. To do this:Thanks Hong, that was exactly the problem I was running into. The error you get suggests that the number of rows is wrong, but the problem is actually that the model has been trained using a command that ends up with the wrong names for parameters.
This is really a critical detail that is entirely non-obvious for lm and so on. Some of the tutorial make reference to doing lines like
lm(olive$Area@olive$Palmitic)
- ending up with variable names of olive$Area NOT Area, so creating an entry usinganewdata<-data.frame(Palmitic=2)
can't then be used. If you uselm(Area@Palmitic,data=olive)
then the variable names are right and prediction works.The real problem is that the error message does not indicate the problem at all:
To avoid error, an important point about the new dataset is the name of independent variable. It must be the same as reported in the model. Another way is to nest the two function without creating a new dataset