How to deal with error “Error in eval(predvars, data, env) : numeric 'envir' arg not of length one” in R?

This error occurs when we do not pass the argument for independent variable as a data frame. The predict function will predict the value of the dependent variable for the provided values of the independent variable and we can also use the values of the independent variable using which the model is created.

Example

Consider the below data frame −

set.seed(1)
x <-rnorm(20)
y <-runif(20,5,10)
df <-data.frame(x,y)
df

Output

      x       y
1 -0.62645381 9.104731
2 0.18364332 8.235301
3 -0.83562861 8.914664
4 1.59528080 7.765182
5 0.32950777 7.648598
6 -0.82046838 8.946781
7 0.48742905 5.116656
8 0.73832471 7.386150
9 0.57578135 8.661569
10 -0.30538839 8.463658
11 1.51178117 7.388098
12 0.38984324 9.306047
13 -0.62124058 7.190486
14 -2.21469989 6.223986
15 1.12493092 5.353395
16 -0.04493361 5.497331
17 -0.01619026 6.581359
18 0.94383621 7.593171
19 0.82122120 8.310025
20 0.59390132 7.034151

Creating the linear model −

M <-lm(y~x,data=df)

Formula for prediction that results in error −

predict(M,newdata=df$x,interval="confidence")
Error in eval(predvars, data, env) :
numeric 'envir' arg not of length one

Formula for prediction that does not result in error −

predict(M,newdata=data.frame(df$x),interval="confidence")

Output

      fit    lwr       upr
1 7.642084 6.814446 8.469722
2 7.536960 6.927195 8.146725
3 7.669228 6.738695 8.599762
4 7.353775 6.214584 8.492966
5 7.518031 6.900897 8.135166
6 7.667261 6.744547 8.589975
7 7.497538 6.854767 8.140310
8 7.464980 6.749018 8.180943
9 7.486073 6.821666 8.150480
10 7.600420 6.902430 8.298410
11 7.364611 6.273305 8.455917
12 7.510202 6.885355 8.135048
13 7.641408 6.816180 8.466635
14 7.848187 6.091378 9.604995
15 7.414811 6.530792 8.298831
16 7.566622 6.935903 8.197340
17 7.562892 6.936919 8.188865
18 7.438312 6.639516 8.237107
19 7.454223 6.706932 8.201514
20 7.483722 6.814287 8.153156

We can simply use the Model object as well, if we want to predict the dependent variable for the independent variable

Example

predict(M)

Output

1 2 3 4 5 6 7 8
7.642084 7.536960 7.669228 7.353775 7.518031 7.667261 7.497538 7.464980
9 10 11 12 13 14 15 16
7.486073 7.600420 7.364611 7.510202 7.641408 7.848187 7.414811 7.566622
17 18 19 20
7.562892 7.438312 7.454223 7.483722

Example

predict(M,interval="confidence")

Output

   fit       lwr       upr
1 7.642084 6.814446 8.469722
2 7.536960 6.927195 8.146725
3 7.669228 6.738695 8.599762
4 7.353775 6.214584 8.492966
5 7.518031 6.900897 8.135166
6 7.667261 6.744547 8.589975
7 7.497538 6.854767 8.140310
8 7.464980 6.749018 8.180943
9 7.486073 6.821666 8.150480
10 7.600420 6.902430 8.298410
11 7.364611 6.273305 8.455917
12 7.510202 6.885355 8.135048
13 7.641408 6.816180 8.466635
14 7.848187 6.091378 9.604995
15 7.414811 6.530792 8.298831
16 7.566622 6.935903 8.197340
17 7.562892 6.936919 8.188865
18 7.438312 6.639516 8.237107
19 7.454223 6.706932 8.201514
20 7.483722 6.814287 8.153156

Nizamuddin Siddiqui

Published on 21-Aug-2020 11:35:23

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