如何从字符向量的数据帧列的所有值中删除$？

Question

I have a data frame in R that has information about NBA players, including salary information. All the data in the salary column have a "$" before the value and I want to convert the character data to numeric for the purpose of analysis. So I need to remove the "$" in this column. However, I am unable to subset or parse any of the values in this column. It seems that each value is a vector of 1. I've included below the structure of the data and what I have tried in my attempt at removing the "$".

> str(combined)

'data.frame':   588 obs. of  9 variables:
$ Player: chr  "Aaron Brooks" "Aaron Gordon" "Aaron Gray" "Aaron Harrison" ...
$ Tm    : Factor w/ 30 levels "ATL","BOS","BRK",..: 4 22 9 5 9 18 1 5 25 30 ...
$ Pos   : Factor w/ 5 levels "C","PF","PG",..: 3 2 NA 5 NA 2 1 1 4 5 ...
$ Age   : num  31 20 NA 21 NA 24 29 31 25 33 ...
$ G     : num  69 78 NA 21 NA 52 82 47 82 13 ...
$ MP    : num  1108 1863 NA 93 NA ...
$ PER   : num  11.8 17 NA 4.3 NA 5.6 19.4 18.2 12.7 9.2 ...
$ WS    : num  0.9 5.4 NA 0 NA -0.5 9.4 2.8 4 0.3 ...
$ Salary: chr  "$2000000" "$4171680" "$452059" "$525093" ...

combined[, "Salary"] <- gsub("$", "", combined[, "Salary"])

The last line of code above is able to run successfully but it doesn't augment the "Salary" column.

I am able to successfully augment it by running the code listed below, but I need to find a way to automize the replacement process for the whole data set instead of doing it row by row.

combined[, "Salary"] <- gsub("$2000000", "2000000", combined[, "Salary"])

How can I subset the character vectors in this column to remove the "$"? Apologies for any formatting faux pas ahead of time, this is my first time asking a question. Cheers,

Answer 1

The $ is a metacharacter which means the end of the string. So, we need to either escape ( \\\\$ ) or place it in square brackets ( "[$]" ) or use fixed = TRUE in the sub . We don't need gsub as there seems to be only a single $ character in each string.

 combined[, "Salary"] <- as.numeric(sub("$", "", combined[, "Salary"], fixed=TRUE))

Or as @gung mentioned in the comments, using substr would be faster

as.numeric(substr(d$Salary, 2, nchar(d$Salary)))