Using Names() Function Deletes CSV-CodePudding

I'm am doing a programming assignment and have the following segments of code which should provide context to the data I'm using:

d <- read.csv(file='data6a.csv') # read and display data from CSV
d

This is what is displayed :

d = str(d)
d$colname=as.factor(d$colname)

The issue arises here. The assignment has given me this line of code to rename column x11 to y:

names(d)[names(d) == 'old.var.name'] <- 'new.var.name'

I have modified the code do to so:

names(d)[names(d) == 'x11'] <- 'y'

However, when I check to see that the column names are updated using str(d) as the assignment directs me, I am given the following output:

List of 1

$ colname: Factor w/ 0 levels:

This seems super straightforward yet I'm unsure why this is occurring. The code has basically been given to me other than that section I modified. If someone could point me in the right direction, that would be great.

CodePudding user response：

Things are getting messy from the point you're assigning d <-str(d) because str will return the summary of of the output but should not be assigned to an object, from that point d is not a dataframe anymore.

Just run the code as following

d <- read.csv(file='data6a.csv')
names(d)[names(d) == 'x11'] <- 'y'

If you want to view the data just run str(d) but do not assign it to anything

CodePudding user response：

Thanks to the help of @akrun, the issue was identified.

When performing d = str(d), d is set to null as str() does not return anything.

Changing the code to the following still has issues:

str(d)
d$colname=as.factor(d$colname)

as it returns:

Error in `$<-.data.frame`(`tmp`, colname, value = integer(0)) : >replacement has 0 rows, data has 394

Because this assignment requires me to redefine each categorical variable as a factor, we cannot skip over that segment and just change the column names using names(d)[names(d) == 'x11'] <- 'y.`

The solution that worked for me was the following:

temp = str(d)
temp$colname=as.factor(temp$colname)

The code above displayed the categorical columns as factors in the variable temp so we can apply the line names(d)[names(d) == 'x11'] <- 'y' to d as it originally was.