Michaels function works, but it does not return the modified data.frame. You need to add return(x).

Here is a one-line solution:

addWeekDay

to my mind, an "eval(parse(..))" construct is in most cases not necessary and should be avoided, as is discussed here http://stackoverflow.com/questions/13649979/what-specifically-are-the-dangers-of-evalparse

a simple solution would be

colName

Great. Thank you all for the answers, and the tips on df and eval. They are helpful.

@Eik Vettorazzi, Thank you - The within method is working fine directly - but not inside a function.

For example, "orders" is my data frame and it has 'Order.Date' column. So, if we do it directly as below, your method works:

> str(within(orders, assign('weekday',as.factor(weekdays(Order.Date)))))

However, if we wrap it inside a function as I would like, then the function would be:

addWeekDay str(addWeekDay(orders, 'Order.Date'))

> str(addWeekDay(orders, Order.Date))

In one case it says it cannot apply weekdays to character and in the next case it says it cannot find the 'Order.Date' object.

Alright. In continuation to my above post, the within method works if we explicitly state the field name in the command, as below:

> str(addWeekDay(orders, orders$Order.Date))

This is not really to my liking because the usage is not flexible.,Like my original function or Michael Moffat ' s method - one should be able to either supply the column name or column number.

I like Michael Moffat 's method - but would prefer a single line direct return mechanism like Eik Vettorazzi 's, so mixed and matched and finally this is my solution:

addWeekDay str(addWeekDay(orders, 'Order.Date'))

> str(addWeekDay(orders, 2))

both works fine (where 2 is column number).

Thank you all for the help. I have Upvoted all answers.

All the above can be made in a more readable way.

Use transform, a base R function:

transform(dFrame, weekdays = as.factor(weekdays(Order.Date)))

Or mutate from the plyr package (see ?mutate).

mutate(dFrame, weekdays = as.factor(weekdays(Order.Date)))

You can add dateField as a function parameter :

addWeekDay

Good solution!

It is also possible with within():

within(df, {weekdays = as.factor(weekdays(dateField))})

I would do it simply

CN