Thanks for the suggestion, but that returns a list, not an array. Also I wanted a mix of arbitrary indices and a range (which could be arbitrarily long). Using your idea, I then tried:
np.array([a[i] for i in [1,2,4, range(7,17,2)]]) but that didn't work. Also it's a bit awkward compared to the elegance of Julia. On Tuesday, April 5, 2016 at 10:32:40 AM UTC+2, Sisyphuss wrote: > > You can use list compression in python > > [a[i] for i in [1,2,4,7,9] ] > > On Apr 5, 2016 8:55 AM, "DNF" <> wrote: > > > > Typo, I meant to type: > > > > Python 3.5 > > a[i*len(a)//n:(i+1)*len(a)//n] > > > > Julia: > > a[1+i*end÷n:(i+1)end÷n] > > > > I'm just learning Python, and must say I find indexing in Python to be > very awkward compared to Julia or Matlab. Do you have any suggestion for > how I should do this in Python? > > a[[1; 4; 7:2:15]] > > So far I've got: > > a[np.concatenate(([1,4], np.arange(7,17,2)))] > > > > > > > > > > On Tuesday, April 5, 2016 at 8:46:57 AM UTC+2, DNF wrote: > >> > >> > >> On Saturday, April 2, 2016 at 1:55:55 PM UTC+2, Spiritus Pap wrote: > >>> > >>> A simple example why it makes my life hard: Assume there is an array > of size 100, and i want to take the i_th portion of it out of n. This is a > common scenario for research-code, at least for me and my friends. > >>> In python: > >>> a[i*100/n:(i+1)*100/n] > >>> In julia: > >>> a[1+div(i*100,n):div((i+1)*100,n)] > >>> > >>> A lot more cumbersome in julia, and it is annoying and unattractive. > This is just a simple example. > >> > >> > >> Python 3.5 > >> a[i*len(a)//n:(i+1)*len(a)//n] > >> > >> Julia: > >> a[1+i*end÷5:(i+1)end÷5] > >> >
