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Assuming this is a few small bulbs directly connected to power, here's a starting point based on Dale's diode method. This shows 3 type 1N4001 diodes where the more diodes the more dimming. I'd guess you'll need somewhere between 2 and 6. The so-called polarity bands of the diodes must be oriented as shown. Radio Shack sells a value-pack of 25 equivalent diodes for $2.99. Or if you're not in a rush, search eBay for "1N4001" and you can pick up 50 for $0.99 w/free shipping from Asia.

A single resistor could work and would be positioned in place of the diodes. Per above, it's hard to suggest a value without details. But if your lamps are incandescents, suitable values would be closer to 10 ohms rather than 1k ohms.

OK, LEDs. A single 1k will definitely drop the brightness and like you say you'll get more dimming with two. No harm in doing so. I'm guessing that 1k may actually be too large in which case take two 1k and tie them in parallel (side-by-side, rather than end-to-end) to effectively make a 500 ohm (=1k /2) resistor. And if that is still too dim, put 3 of the 1k side-by-side in parallel to make a 333 ohm (=1k /3) resistor.
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