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df <- data.frame(
    name   = c("Alice", "Bob", "Carol", "Dave", "Eve"),
    age    = c(28, 34, 22, 41, 30),
    salary = c(55000, 72000, 48000, 95000, 61000),
    dept   = c("HR", "IT", "IT", "Finance", "HR")
)

# Dollar-sign: extract a column as a vector
df$name       # character vector of names
df$salary     # numeric vector of salaries

# Double-bracket: programmatic column access
col <- "age"
df[[col]]     # same as df$age

# Single-bracket: row/column matrix-style
df[1, ]           # first row (a 1-row data frame)
df[ , 2]          # second column (drops to vector by default)
df[ , 2, drop = FALSE]  # keep as data frame
df[1:3, c("name", "age")]   # rows 1-3, named columns

# Logical row filtering
df[df$age > 30, ]                      # rows where age > 30
df[df$dept == "IT", ]                  # IT employees
df[df$salary >= 60000 & df$active != FALSE, ]   # compound condition

# subset() — readable syntax (avoid inside functions)
subset(df, age < 35, select = c(name, salary))

# Negative indexing (drop columns by position)
df[ , -1]          # all columns except the first
df[ , -c(1, 4)]    # drop columns 1 and 4

# which() to get row indices
high_earners <- which(df$salary > 60000)
df[high_earners, ]

# Updating values in place via subsetting
df$salary[df$dept == "IT"] <- df$salary[df$dept == "IT"] * 1.05