Getting Warning: " 'newdata' had 1 row but variables found have 32 rows" on predict.lm

This is a problem of using different names between your data and your newdata and not a problem between using vectors or dataframes.

When you fit a model with the lm function and then use predict to make predictions, predict tries to find the same names on your newdata. In your first case name x conflicts with mtcars$wt and hence you get the warning.

See here an illustration of what I say:

This is what you did and didn't get an error:

a <- mtcars$mpg
x <- mtcars$wt

#here you use x as a name
fitCar <- lm(a ~ x) 
#here you use x again as a name in newdata.
predict(fitCar, data.frame(x = mean(x)), interval = "confidence") 

       fit      lwr      upr
1 20.09062 18.99098 21.19027

See that in this case you fit your model using the name x and also predict using the name x in your newdata. This way you get no warnings and it is what you expect.

Let's see what happens when I change the name to something else when I fit the model:

a <- mtcars$mpg
#name it b this time
b <- mtcars$wt 

fitCar <- lm(a ~ b) 
#here I am using name x as previously
predict(fitCar, data.frame(x = mean(x)), interval = "confidence") 

         fit       lwr      upr
1  23.282611 21.988668 24.57655
2  21.919770 20.752751 23.08679
3  24.885952 23.383008 26.38890
4  20.102650 19.003004 21.20230
5  18.900144 17.771469 20.02882
Warning message:
'newdata' had 1 row but variables found have 32 rows 

The only thing I did now was to change the name x when fitting the model to b and then predict using the name x in the newdata. As you can see I got the same error as in your question.

Hope this is clear now!

In the formula for lm function do not refer to the variables using the datasetname$variablename pattern. Instead use variablename + variablename ...This will not throw the warning: 'newdata' had nrow(test) row but variables found have nrow(train) rows.

A way around this without making names is to use the following:

fitCar<-lm(mpg ~ wt, mtcars) #here you use x as a name
predict(fitCar,data.frame(wt=mean(mtcars$wt)), interval="confidence")