Calling out specific Levels in a factor to use in Model

Hello I have a data set here that brings in the Day of the week as character

I change the Vector to factor with two levels

levels(RData2019$Day)
[1] "Friday" "Saturday" "Sunday" "Monday"
[5] "Tuesday" "Wednesday" "Thursday"

I am trying to create a LM() with just Friday Saturday and Sunday as independent variables any clue on how to do this I have been searching for three days with know luck.

you have 7 levels not 2.
It seems like you want to remove rows of the dataframe RData2019 which have a day outiside of Saturday and Sunday; its popular on this forum to use dplyr::filter() for those sorts of operations; but this could be addressed with base R function subset

perhaps your best order of operations is to filter/subset reducing the frame to only the two types of rows of interest and then transform that dat column from character to factor, and it will naturally have 2 levels.

hello I am having some trouble getting this to work I truly do appreciate your help

here is the code I came up with, out of research

RData2019$Day <- as.factor(RData2019$Day)

filter(RData2019, Day == "monday", "Tuesday", "Wednesday", "Thursday")

this is the error I received
Error in filter():
! Problem while computing ..2 = "Tuesday".
:heavy_multiplication_x: Input ..2 must be a logical vector, not a character.
Run rlang::last_error() to see where the error occurred.

rlang:: last_error

  1. dplyr::filter(...)
  2. dplyr:::filter.data.frame(RData2019, Day == "monday", "Tuesday", "Wednesday",
    "Thursday")

in R we can contain together multiple items of the same type in a c(ombination) using c() or of mixed types in a list list().
Also filter is used to specify what is kept (the inverse of what is removed)

filter(RData2019, Day %in% c("Saturday","Sunday"))

Got the code to run but i do not think we achieved the result we were looking for

Code

filter(RData2019, Day %in% c("Friday","Saturday","Sunday"))

Result
Game Hour` Date COMP DOLLAR GROUP

1 2019-03-30 19:00:00 2019-03-30 00:00:00 NA 8 12544
2 2019-03-31 15:00:00 2019-03-31 00:00:00 8 11 4797
3 2019-04-12 19:00:00 2019-04-12 00:00:00 1 NA 1631
4 2019-04-14 14:00:00 2019-04-14 00:00:00 66 291 3657
5 2019-04-19 19:00:00 2019-04-19 00:00:00 12 8 4245
6 2019-04-20 19:00:00 2019-04-20 00:00:00 NA 14 4803
7 2019-04-21 14:00:00 2019-04-21 00:00:00 51 NA 611
8 2019-05-03 19:00:00 2019-05-03 00:00:00 NA NA 4932
9 2019-05-04 19:00:00 2019-05-04 00:00:00 14 15 4946
10 2019-05-05 14:00:00 2019-05-05 00:00:00 158 318 3493

… with 31 more rows, and 17 more variables: NGHTLY ,

SEASON , SINGLE , Day , OpeningDay ,

OpeningWeekend , PreASB , BOSNYY ,

Holiday , DayGame , WeekdayDayGame ,

Bobblehead , Wearable , OtherGiveaway ,

Kids , Concert , SpecEvent

:information_source: Use print(n = ...) to see more rows, and colnames() to see all variable names

Then I ran a check on the Variable to see if I only had Friday Saturday and Sunday Values left

Code:

RData2019$Day

Result
[1] Thursday Saturday Sunday Monday Tuesday
[6] Wednesday Friday Sunday Monday Tuesday
[11] Wednesday Friday Saturday Sunday Tuesday
[16] Wednesday Friday Saturday Sunday Friday
[21] Saturday Sunday Monday Tuesday Wednesday
[26] Thursday Friday Saturday Sunday Tuesday
[31] Wednesday Thursday Friday Saturday Saturday
[36] Sunday Monday Tuesday Wednesday Thursday
[41] Friday Saturday Sunday Tuesday Wednesday
[46] Thursday Thursday Friday Saturday Sunday
[51] Tuesday Wednesday Tuesday Wednesday Friday
[56] Saturday Sunday Thursday Friday Saturday
[61] Sunday Monday Tuesday Tuesday Wednesday
[66] Thursday Friday Saturday Sunday Tuesday
[71] Wednesday Thursday Friday Saturday Sunday
[76] Tuesday Wednesday Thursday Friday Saturday
[81] Sunday
7 Levels: Friday Monday Saturday Sunday ... Wednesday

Thank you for this. just as a reminder I am trying to isolate Friday, Saturday, and Sunday Values to see what
effect these independent days or variables would have on the model.