A 220 Ω resistor may be a bit on the low side. When discharging a 47 µF capacitor from 400 V the average power dissipated is 75 W:
I'd vote for 1 kΩ, but you can give the 220 Ω 2W resistor a try. If it blows, use a higher resistance.
Be careful when discharging the capacitor, use well insulated tools!
a 220Ω 2W resistor would be rated for use in about a 20VDC circuit, but for short durations it can be used in higher power and energy dissipation (e.g. rule of thumb from Dale, 5x for 5 seconds). The 400V is a lethal voltage level so that is a separate important concern.
So a typical 2W power resistor can survive a 10W pulse for 5 seconds, which is 50 Joules of energy. You didn't say what size capacitor, but you can calculate it's energy and compare, e.g. 47uF at 400V is about 4 Joules.
You may see a little spark but i doubt there will be molten metal, but wear safety glasses.
Thanks for all the replies, the 220Ω was just a low enough figure I thought would allow a fairly fast discharge without leaving the probes on for too long. I will use something in the low kΩ range and use insulated probes of course.
A neon is not going to be much use as an indicator as they require somewhere in the vicinity of 70 -90v to turn on.
The resistor I quoted works fine, simply leave it in circuit for maybe 10 seconds.
You will feel it get warm on the larger caps as it does it's job.
If in doubt, check with a voltmeter afterwards.