You have the right idea for a basic unregulated supply. A transformer, four diodes, and as large a cap as you can manage will serve well enough for a lot of purposes, but isn't appropriate for all.
There are two main problems with such a unregulated supply. First, the voltage is not known well. Even with ideal components, so that the AC coming out of the transformer is a fixed fraction of the AC going in, you still have variations in that AC input. Wall power can vary by around 10%, and that's without considering unusual situations like brownouts. Then you have the impedance of the transformer. As you draw current, the output voltage of the transformer will drop.
Second, there will be ripple, possibly quite significant ripple. That cap is charged twice per line cycle, or every 8.3 ms. In between the line peaks, the cap is supplying the output current. This decreases the voltage on the cap. The only way to decrease this ripple in this type of design is to use a bigger cap or draw less current.
And don't even think about power factor. The power factor a full wave bridge presents to the AC line is "not nice". The transformer will smooth that out a little, but you will still have a crappy power factor regardless of what the load does. Fortunately, power factor is of little concern for something like a bench supply. Your refrigerator probably treats the power line worse than your bench supply ever will. Don't worry about it.
Some things you can't do with this supply is run a anything that has a tight voltage tolerance. For example, many digital devices will want 5.0 V or 3.3 V ± 10%. You're supply won't be able to do that. What you should probably do is aim for 7.5 V lowest possible output under load, with the lowest valid line voltage in, and at the bottom of the ripples. If you can guarantee that, you can use a 7805 regulator to make a nice and clean 5 V suitable for digital circuits.
Note that after you account for all the reasons the supply voltage might drop, that the nominal output voltage may well be several volts higher. If so, keep the dissipation of the regulator in mind. For example, if the nominal supply output is 9 V, then the regulator will drop 4 V. That 4 V times the current is the power that will heat the regulator. For example, if this is powering a digital circuit that draws 200 mA, then the dissipation in the regulator will be 4V x 200mA = 800mW. That's will get a 7805 in free air quite hot, but it will probably still be OK. Fortunately, 7805 regulators contain a thermal shutdown circuit, so they will just shut off the output for a while instead of allowing themselves to get cooked.
The 50Hz noise is mains hum. Any time you pick up noise at the same frequency as your mains hum, assume it's from the mains (once, maybe twice in a lifetime it'll be a wrong assumption, but every other time it's right)
Pretty much the only way to get rid of picked-up noise (mains hum, emi/rfi, whatever) is to ground it out.
From your updated question post, it looks like the noise is appearing on a (possibly long) wire carrying a non-differential signal & coming as data feedback from a (mains powered?) "slave" device.
Either the hum is originating from within the "slave" device due to some power supply noise to it (likely due to poor wiring & other noisy devices on same circuit), or it is being picked up by the wire as EMI/RFI.
If it is being picked up as EMI/RFI (some part of it almost surely is), then switching to differential signalling could "fix it."
If it is coming from the "slave" device somehow, then there may be a need to add some 50Hz filtering &/or grounding in that device, to eliminate the noise before it gets onto the data line.
Best Answer
http://archive.ics.uci.edu/ml/
This is an archive for machine learning datasets..There is not the one which you want but there are datasets of some other Power Substations one in India.