user2543601 - 1 year ago 123

R Question

I'm trying to make a linear regression using the lm() function in R. When I run it, it appears that R is taking each record and making it an independent variable.

`fit3 <- lm(no_outliers$strat_count ~ no_outliers$CDS_emerg + no_outliers$weighted_volume_avg)`

summary(fit3)

This is a sample of the output (there are 118 independent variables, but I didn't want to put all of it on the post):

`Coefficients:`

Estimate Std. Error t value Pr(>|t|)

(Intercept) -336.1017 1065.5530 -0.315 0.7652

no_outliers$weighted_volume_avg941931862 12.2047 31.2349 0.391 0.7121

no_outliers$weighted_volume_avg949989365.5 4.0453 33.3295 0.121 0.9081

no_outliers$weighted_volume_avg955100055.4 10.4469 31.3577 0.333 0.7525

no_outliers$weighted_volume_avg961033059.1 -17.2295 32.3160 -0.533 0.6168

no_outliers$weighted_volume_avg973785580 85.1891 40.0488 2.127 0.0867 .

no_outliers$weighted_volume_avg976666189.1 48.1133 39.9253 1.205 0.2821

no_outliers$weighted_volume_avg979529996 26.2521 31.2707 0.840 0.4395

no_outliers$weighted_volume_avg985701661.4 35.3185 35.8381 0.986 0.3696

no_outliers$weighted_volume_avg988019226.8 66.4781 31.3502 2.120 0.0875 .

no_outliers$weighted_volume_avg994324495.5 -13.4227 32.9220 -0.408 0.7004

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 29.2 on 5 degrees of freedom

Multiple R-squared: 0.9752, Adjusted R-squared: 0.3894

F-statistic: 1.665 on 118 and 5 DF, p-value: 0.3005

Any help or pointers would be greatly appreciated.

Answer Source

I guess `weighted_volume_avg`

is a factor. Try `class(no_outliers$weighted_volume_avg)`

. If it is should to convert it to numeric first. Try:

```
no_outliers$weighted_volume_avg <- as.numeric(as.character(no_outliers$weighted_volume_avg))
```

Then try your `lm`

again.