Okay, think that gets my heading the right direction. Again, Thank You!

Interesting solutions, but there is a built-in solution, that is much easier (and faster!):

Ratios[ts] - 1

Hi there,

not that the first value of the time series is 0.818908 and the second one is 0.355672. So that means that today is T=0.355672 and yesterday is Y=0.818908; assuming that yesterday is before today. That is why I said:

(0.355672 - .818908)/0.355672= -1.30242

If you want to get it the other way around you can simply use:

Differences[#]/Drop[#, -1] &@ts
(*{-0.565675, 0.958068, -0.437622, -0.837792, 10.4098, -0.233148, \
-0.66615, 2.1765, -0.187475, -0.207362, -0.800643, 3.16858, 1.83259, \
-0.187731, -0.814965, 0.57331, 2.31025, 0.0579287, 0.188419}*)

The first value of this gives the -56% change that you calculate.

Yes, Differences is a built-in function and it is usually a good idea to use them because they are more efficient. You can, however, program it yourself. There are many ways of achieving this:

Drop[#, 1] - Drop[#, -1] &@ts

or

Table[ts[[i + 1]] - ts[[i]], {i, 1, Length[ts] - 1}]

or

(ts - RotateRight[ts])[[2 ;;]]

or

Drop[ts - RotateRight[ts], 1]

or

Reap[Do[Sow[ts[[i + 1]] - ts[[i]]], {i, 1, Length[ts] - 1}]][[2, 1]]

or whatever you like.

Cheers,

M.

Thank you so much Marco, this points me in the right direction. However if; Today=0.818908 and Yesterday=0.355672 then it would be a +130.242% daily increase. And if; Today=0.355672 and Yesterday=0.818908 then it would be a -56.567% daily decrease.

The broader question was, like in your example you used the function "Differences" is there any way to see the math/code behind that function? Or is that proprietary code from Wolfram?

Again, thanks. K.