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] >>
