Renaming integers within a data.frame - r

Question

Using R...
I have a data.frame with five variables.
One of the variables colr has values ranging from 1 to 5.
Defined as an integer with values 1, 2, 3, 4, and 5.
Problem: I would like to build a regression model where the values within colr, the integers 1,2,3,4, and 5 are reported as independent variables with the following names.
1 = Silver,
2 = Blue,
3 = Pink,
4 = Other than Silver, Blue or Pink,
5 = Color Not Reported.
Question: Is there a way to extract or rename these values in a way that is different from the following (as this process does not rename, eg. 1 to Silver in the summary regression output):
lm(dependent variable ~ + I(colr.f == 1) +
I(colr.f == 2) +
I(colr.f == 3) +
I(colr.f == 4) +
I(colr.f == 5),
data = df)
I am open to any method that would allow me to create and name these different values independently but would prefer to see if there is a way to do so using the tidyverse or dplyr as this is something I have to do frequently when building multivariate models.
Thank you for any help.

Answer 1

If you have this:
df <- data.frame(int = sample(5, 20, TRUE), value = rnorm(20))
df
#> int value
#> 1 3 -0.62042198
#> 2 4 0.85009260
#> 3 5 -1.04971518
#> 4 1 -2.58255471
#> 5 1 0.62357772
#> 6 4 0.00286785
#> 7 4 -0.05981318
#> 8 4 0.72961261
#> 9 4 -0.03156315
#> 10 1 -2.05486209
#> 11 5 1.77099554
#> 12 1 1.02790956
#> 13 1 -0.70354012
#> 14 1 0.27353731
#> 15 2 -0.04817215
#> 16 2 0.17151374
#> 17 5 -0.54824346
#> 18 2 0.41123284
#> 19 5 0.05466070
#> 20 1 -0.41029986
You can do this:
library(tidyverse)
df <- df %>% mutate(color = factor(c("red", "green", "orange", "blue", "pink"))[int])
df
#> int value color
#> 1 3 -0.62042198 orange
#> 2 4 0.85009260 blue
#> 3 5 -1.04971518 pink
#> 4 1 -2.58255471 red
#> 5 1 0.62357772 red
#> 6 4 0.00286785 blue
#> 7 4 -0.05981318 blue
#> 8 4 0.72961261 blue
#> 9 4 -0.03156315 blue
#> 10 1 -2.05486209 red
#> 11 5 1.77099554 pink
#> 12 1 1.02790956 red
#> 13 1 -0.70354012 red
#> 14 1 0.27353731 red
#> 15 2 -0.04817215 green
#> 16 2 0.17151374 green
#> 17 5 -0.54824346 pink
#> 18 2 0.41123284 green
#> 19 5 0.05466070 pink
#> 20 1 -0.41029986 red
Which allows a regression like this:
lm(value ~ color, data = df) %>% summary()
#>
#> Call:
#> lm(formula = value ~ color, data = df)
#>
#> Residuals:
#> Min 1Q Median 3Q Max
#> -2.03595 -0.33687 -0.00447 0.46149 1.71407
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) 0.2982 0.4681 0.637 0.534
#> colorgreen -0.1200 0.7644 -0.157 0.877
#> colororange -0.9187 1.1466 -0.801 0.436
#> colorpink -0.2413 0.7021 -0.344 0.736
#> colorred -0.8448 0.6129 -1.378 0.188
#>
#> Residual standard error: 1.047 on 15 degrees of freedom
#> Multiple R-squared: 0.1451, Adjusted R-squared: -0.0829
#> F-statistic: 0.6364 on 4 and 15 DF, p-value: 0.6444
Created on 2020-02-16 by the reprex package (v0.3.0)

Answer 2

I'm not sure I'm understanding your question the right way, but can't you just use
library(dplyr)
df <- df %>%
mutate(color=factor(colr.f, levels=c(1:5), labels=c("silver", "blue", "pink", "not s, b, p", "not reported"))
and then just run the regression on color only.
/edit for clarification. Making up some data:
df <- data.frame(
x=rnorm(100),
color=factor(rep(c(1,2,3,4,5), each=20),
labels=c("Silver", "Blue", "Pink", "Not S, B, P", "Not reported")),
y=rnorm(100, 4))
m1 <- lm(y~x+color, data=df)
m2 <- lm(y~x+color-1, data=df)
summary(m1)
Call:
lm(formula = y ~ x + color, data = df)
Residuals:
Min 1Q Median 3Q Max
-1.96394 -0.59647 0.00237 0.56916 2.13392
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.93238 0.19312 20.362 <2e-16 ***
x 0.13588 0.09856 1.379 0.171
colorBlue -0.07862 0.27705 -0.284 0.777
colorPink -0.02167 0.27393 -0.079 0.937
colorNot S, B, P 0.15238 0.27221 0.560 0.577
colorNot reported 0.14139 0.27230 0.519 0.605
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.8606 on 94 degrees of freedom
Multiple R-squared: 0.0268, Adjusted R-squared: -0.02496
F-statistic: 0.5177 on 5 and 94 DF, p-value: 0.7623
summary(m2)
Call:
lm(formula = y ~ x + color - 1, data = df)
Residuals:
Min 1Q Median 3Q Max
-1.96394 -0.59647 0.00237 0.56916 2.13392
Coefficients:
Estimate Std. Error t value Pr(>|t|)
x 0.13588 0.09856 1.379 0.171
colorSilver 3.93238 0.19312 20.362 <2e-16 ***
colorBlue 3.85376 0.19570 19.692 <2e-16 ***
colorPink 3.91071 0.19301 20.262 <2e-16 ***
colorNot S, B, P 4.08477 0.19375 21.083 <2e-16 ***
colorNot reported 4.07377 0.19256 21.156 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.8606 on 94 degrees of freedom
Multiple R-squared: 0.9578, Adjusted R-squared: 0.9551
F-statistic: 355.5 on 6 and 94 DF, p-value: < 2.2e-16
The first model is a model with intercept, therefore one of the factor levels must be dropped to avoid perfect multicollinearity. In this case, the "effect" of silver is the value of the intercept, while the "effect" of the other colors is the intercept coefficient value + their respective coefficient value.
The second model is estimated without intercept (without constant), so you can see the individual effects. However, you should probably know what you are doing before estimating the model without intercept.

Answer 3

With base R.
labels <- c("Silver", "Blue", "Pink", "Other Color", "Color Not Reported")
df$colr.f2 <- factor(colr.f, labels = labels, levels = seq_along(labels))