R：错误：出现意外的“ =”

Question

I have a data frame (df), a vector of column names (foo), and a function (spaces) that calculates a value for all rows in a specified column of a df. I am trying to accomplish the following:

1. Private foo as input to spaces
2. Spaces operates on each element of foo matching a column name in df
3. For each column spaces operates on, store the output of spaces in a new column of df with a column name produced by concatenating the name of the original column and ".counts".

I keep receiving Error:

> Error: unexpected '=' in:
>"        new[i] <- paste0(foo[i],".count")  # New variable name 
>    data <- transform(data, new[i] ="
>    }
> Error: unexpected '}' in "    }"

Below is my code. Note: spaces does what I want when provided an input of a single variable of the form df$x but using transform() should allow me to forego including the prefix df$ for each variable.

# Create data for example
a <- c <- seq(1:5)
b <- c("1","1 2", "1 2 3","1 2 3 4","1 2 3 4 5")
d <- 10
df <- data.frame(a,b,c,d)  # data fram df
foo <- c("a","b")  # these are the names of the columns I want to provide to spaces

# Define function: spaces
spaces <- function(s) { sapply(gregexpr(" ", s), function(p) { sum(p>=0) } ) }

# Initialize vector with new variable names
new <- vector(length = length(foo))

# Create loop with following steps:
  # (1) New variable name 
  # (2) Each element (e.g. "x") of foo is fed to spaces
  #     a new variable (e.g. "x.count") is created in df,
  #     this new df overwrites the old df

for (i in 1:length(foo)) {
    new[i] <- paste0(foo[i],".count")  # New variable name 
    df <- transform(df, new[i] = spaces(foo[i]))  # Function and new df
}

Answer 1

transform(df, new[i] = spaces(foo[i])) is not valid syntax. You cannot call argument names by an index. Create a temporary character string and use that.

for (i in 1:length(foo)) {
  new[i] <- paste0(foo[i],".count")  # New variable name 
  tmp <- paste0(new[i], ".counts")
  df <- transform(df, tmp = spaces(foo[i]))  # Function and new df
}