I believe all the solutions on this page are ground-referenced. That is, if there are long wires, less than ideal ground impedance, high ground currents or something they won't be precise.

But this can be eliminated by so-called <strong>Howland current pump</strong>. The key benefit here is that it does not require the load to be connected to any power rail. Essentially, it's a differential amplifier. Please have a look on the following schematic.

R5 is the current sensing resistor, R6 is the load. The rest (R1-R4) are part of the differential amplifier. The voltage across R5 is equal to V3. Hence, the current through it will be \$I_{R5} ~= V3 / R5\$. If R6 << R4, then \$I_{R5} \approx I_{R6}\$. In this schematic 
there is always a small output error because part of the output current flows through R4 as well. But this can be eliminated using, e.g., voltage buffer between R4 and R5.

[![Howland charge pump (minimal version)[1]][1]

For more information see the following materials:

["What is a Howland Current Pump?"](https://www.youtube.com/watch?v=_x4aqCCz_Iw) by Kelvin Le

["Difference Amplifier Forms Heart of Precision Current Source"](http://www.analog.com/library/analogDialogue/cd/vol43n3.pdf#page=22)

["AN-1515 A Comprehensive Study of the Howland Current Pump"](http://www.ti.com/lit/an/snoa474a/snoa474a.pdf)

["Tame those versatile current source circuits"](http://e2e.ti.com/cfs-file/__key/telligent-evolution-components-attachments/00-18-01-00-01-18-14-27/Improved-Howland-Article.pdf)

["EEVblog #579 - Precision Low Current Source"](http://www.youtube.com/watch?v=OYOYI_IPKGY) (not Howland pump, but interesting)  

  [1]: https://i.sstatic.net/XldAs.png