Hello Greg,

Thanks very much!

This helps!

Cheers,

Rebecca

From: Greg Snow [mailto:538...@gmail.com]
Sent: Friday, January 18, 2013 5:17 PM
To: Yuan, Rebecca
Cc: R help
Subject: Re: [R] A smart way to use "$" in data frame

The important thing to understand is that $ is a shortcut for [[ and you are 
moving into the realm where a shortcut is the longest distance between 2 points 
(see fortune(312)).

So your code can be something like:

state <- 'oldstate'
balance <- 'oldbalance'
dataa[[balance]][ dataa[[state]]=='AR' ]

You may also benefit from learning to use tools like "with" and "subset" 
(though subset has its own complications when used inside of other functions) 
or "grep" and "match" to find the columns of interest.

On Fri, Jan 18, 2013 at 12:40 PM, Yuan, Rebecca 
<rebecca.y...@bankofamerica.com<mailto:rebecca.y...@bankofamerica.com>> wrote:
Hello all,

I have a data frame dataa:

newdate newstate newid newbalance newaccounts
1 31DEC2001        AR         1                 1170           61
2 31DEC2001        VA         2                  4565           54
3 31DEC2001        WA         3                 2726           35
4 31DEC2001        AR         3                 2700           35

The following gives me the balance of state AR:

dataa$newbalance[data$newstate == 'AR']
1170
2700

Now, I have another different data frame datab, it is very similar to data, 
except that the name of the columns are different, and the order of the columns 
are different:

oldstate olddate oldbalance oldid oldaccounts
1 AR       31DEC2012        1234         7              40
2 WA     31DEC2012        2222         3              30
3 VA       31DEC2012        2345         5              23
3 AR       31DEC2012        5673         5              23

datab$oldbalance[datab$oldstate== 'AR' ]
1234
5673

Could I have a way to quote

data$balance[data$state == 'AR']

in general, where balance=oldbalance, state=oldstate when data=dataa, and 
balance = newbalance, state = newstate when data=datab ?

Thanks very much!

Cheers,

Rebecca

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