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

Hi Sander, yes Thank you, that works well.

Actually looked at Ratios and Differences in the documentation prior to posting. Being new at this and not speaking the language was too hung up on the semantics, not really seeking a ratio (how many times one number contains another) or difference (subtracting one number from another) and did not think or know to use the modifiers to the statement.

Really trying feverishly to learn the language, but If I may be so bold as to ask another question; how do I get past the date? Import from excel looks like this:

1981-01-05 1596.6 1981-01-12 1600.1 1981-01-19 1606.8 1981-01-26 1609.4 1981-02-02 1608.6

Thank you for the help, kind regards, Kelly.

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.