
Re: Delete elements from list..
Posted:
Feb 24, 2011 6:29 AM


Hallo,
Thanks everybody who replied to my questions.
The real problem I have is just a bit more complex than my simplified example. My list is in fact a numerical 2D list like mylist1 == {{x_0, y_0}, {x_1,y_1},... {x_i, y_i}, ...{x_N, y_N}}. The xi are strictly increasing and the yi should be as well. Due to some measurement errors it can happen that this is not the case. I simply want to delete the {xi, yi} pairs where y_i <== y_i1. That way I end up with a list, mylist2, where also the y_i are strictly increasing. (that way I can make an Interpolation[Reverse/@mylist2] in order to have a function x_i(y_i)).
I have not had the time to study your answers in this view, but from a first look and the variety of the answers it seems that there is definitely something which should help.
Thanks for your help.
Maarten
Original Message From: Achilleas Lazarides [mailto:achilleas.lazarides@gmx.com] Sent: Wednesday, 23 February, 2011 11:22 Subject: Re: Delete elements from list..
Do you really want the resulting list to only have increasing elements? then something like
Module[{tmpl == mylist[[1]]}, Scan[(If[# > tmpl, Sow[#]; tmpl == #]) &, mylist]; // Reap]
should work (except that it eats the first element).
If you actually meant that you want to delete any element that is not strictly larger than the previous, then
Module[{tmpl == mylist[[1]]}, Scan[(If[# > tmpl, Sow[#]]; tmpl == #) &, mylist]; // Reap]
(which again eats the first element).
On Feb22, 2011, at 1:30 AM, Maarten van der Burgt wrote:
> Hallo, > > I have a list like: > > mylist == {1, 2, 3, 4, 5, 6, 4, 5, 7, 8, 9} > > I want to delete any element which is not strictly larger than the > previous element, until my new list has only increasing values. > > This means in mylist above I want to delete the 4 and the 5 at position > 7 and 8. > > Any elegant and fast way for doing this? > > In practice I want to do this for a large amount (1000) of very large > lists (1000). So speed is important. > > Thanks for your help. > > Maarten

