Laurens,

I looked at your code but I had difficulty in getting the replace to work properly because your list of data may have a new year within it. In this case, the replacement rule would be different in one part of the list vs the other. Under those circumstances, you would need to break the list or do a Map[] operation that handles each date individually. The code became difficult to make it reliably work (which I suppose you found on your own). I think you are best with the approach using one of the built-in date functions posted above.

Regards,

Neil

You are perfectly right, Neil. Your code does the job quite well. Thanks a lot! Since I have spent so much time on Replace, I am still anxious to hear from you hoewever why my approach with Replace did not work. Is a conditional Replace possible at all? Beste regards, Laurens

Dear Neil, I am afraid that things become too complicated if one keeps trying to just use Differences, since then one keeps the problem at the end of months where the last work day of a week is either the 29th, the 30th, or the 31st. Therefore I decided to start working with yearly serial numbers of days instead of monthly serial numbers. For this I invented the rules method. For normal years one has to use rules1 and for leap years rules 2. To find out if a year is a leap year I do the test Mod[year,4]==0. The problem that I am left with is now how to do a Replace with one of the rules sets (i.e. rules1 for normal years and rules2 for leap years) depending on the result of this test. I could easily get what I want with a bit of procedural programming in Mathematica, but I see it as a challenge to get it done with functional programming. In fact my problem with the leap years boils down to the question if a Replace with either one of two sets of rules can be done based on the result of a test.

Dear Neal and Vitaliy, I did find a method to convert the montly serial numbers of days to yearly serial numbers, but one problem remains. The following code gives the yearly serial numbers along with the prices of each day:

Pick[Most[
  Partition[
   Riffle[Replace[data[[All, 1, 2]], rules, All] + data[[All, 1, 3]], 
    data[[All, 2]]], 2]], 
 Thread[Differences[
    Partition[
      Riffle[Replace[data[[All, 1, 2]], rules, All] + 
        data[[All, 1, 3]], data[[All, 2]]], 2][[All, 1]]] >= 3]]

where e.g.

data = FinancialData["GE", "Close", {2018, 6, 1}]

and

rules1 = {1 -> 0, 2 -> 31, 3 -> 59, 4 -> 90, 4 -> 120, 6 -> 151, 
   7 -> 181, 8 -> 212, 9 -> 243, 10 -> 273, 11 -> 304, 12 -> 334};

In the same way the following code gives consecutively the year, the number of the month, the serial number of the day in the month, the price, and the serial number of the day in the year.

Pick[Most[
  Partition[
   Flatten[Riffle[data, 
     Replace[data[[All, 1, 2]], rules, All] + data[[All, 1, 3]]]], 
   5]], Thread[
  Differences[
    Partition[
      Flatten[Riffle[data, 
        Replace[data[[All, 1, 2]], rules, All] + data[[All, 1, 3]]]], 
      5][[All, 5]]] >= 3]]

As I said, one problem remains. That is that in leap years the rules should be:

rules2 = {1 -> 0, 2 -> 31, 3 -> 60, 4 -> 91, 4 -> 121, 6 -> 152, 
   7 -> 182, 8 -> 213, 9 -> 244, 10 -> 274, 11 -> 305, 12 -> 335};

So the replacement rules depend on the fact if there is a leap year. I have tried:

If[Mod[data[[All, 1, 1]], 4] == 0, 
 Replace[data[[All, 1, 2]], rules2, All], 
 Replace[data[[All, 1, 2]], rules1, All]]

but that does not work. So I suspected that this was caused by the fact that the elements that had to be changed did not contain the information needed for the decision which rules to use. So I tried to bring all information in one and the same object in the following numerical way. I start from:

data = FinancialData["GE", "Close", {{2015, 12, 20}, {2016, 3, 10}}]

and

list = data[[All, 1, 1]]*100000 + data[[All, 1, 2]]*1000 + 
  Replace[data[[All, 1, 2]], rules1, All]

The replacement action does work:

Replace[1000*(list/1000 - Floor[list/1000]), rules3, All]

where

rules3 = {59 -> 60, 120 -> 121, 151 -> 152, 181 -> 182, 212 -> 213, 
      243 -> 244, 273 -> 274, 304 -> 305, 334 -> 335};

but the conditional replacement does not:

 If[Mod[Floor[list/100000], 4] == 0, 
     Replace[1000*(list/1000 - Floor[list/1000]), rules3, All]]

Also using /; instead of If did not work. What did I do wrong????

Laurens,

I tried what I posted and it works for what you want.

Extract[data, Position[Differences[data[[All, 1, 3]]], _?(# > 1 &)]]

to get

{{{2018,6,1},14.1},{{2018,6,8},13.93},{{2018,6,15},13.3},{{2018,6,22},13.05},{{2018,7,3},13.37},{{2018,7,6},13.85},{{2018,7,13},13.89},{{2018,7,20},13.12},{{2018,7,27},13.06},{{2018,8,3},13.14},{{2018,8,10},12.77},{{2018,8,17},12.3},{{2018,8,24},12.5}}

Note that there are some issues. When you change month (for example jump from 6/29 to July you can't just look at the day of the month for the differences. Also, note that July 4 is a holiday so that came up as an "end of week".

Regards

Neil