"The circuit is not one I'd recommend. It won't be very efficient, as evidenced by the large (in relation to lamp wattage) resistor. That component is to turn "spare" electricity into heat."

Don't be misled by the resistor wattage.
That is a "sense" resistor, and it's less than half an ohm.

That component is used to create a voltage drop.
V. = I. x R.
0.47 Ω at 2A. = 0.94V.

It is a tad oversized to limit it's heating.
Heat increases resistance.

That voltage drop is applied to the sense lead of a voltage regulator.
But, because the sense voltage tracks the current, we have tricked a cheap voltage regulator into performing current regulation.

In short, very little power is wasted in that resistor.
Less than 2 Watts at 2 Amps
P = I. x V.
2.0A. x 0.94V. = 1.88W.
This regulator circuit works very well with battery power.

Just a couple 'lectronic pennies.

Aloha,
Weezard