Battery circuit power drain
It is not very efficient using a resistor to drop the current of a high power LED.

In your example 3.7/6 of the power is used by the LED and 2.3/6 of the power will be consumed by the resistor which it will have to dissipate as heat.

Something like one of these components would do the job: 

http://www.jaycar.com.au/productView.asp?ID=AA0593&amp;form=CAT2&amp;SUBCATID=976#1

It takes a variable battery voltage range, and has an input for a potentiometer so you could adjust the brightness.