you can do the python example as: a[[1, 4] + range(7, 17, 2)] (ignoring the issues that this is not the same range as julia since python uses 0-based indices ...)
you don't need to index with an ndarray, and that way you can get the nice use of the + operator for concatenate. Lacking the : for range does make some code more verbose, but you get used to it. For the julia example isn't matlab like concatenation using ';' being removed from julia? As to the overall topic I don't think it is fair to have to poo-poo python to also feel that julia does it well. I really like both. Arguing about 0 vs 1 based indexing is like emacs vs vi ... it leads to nothing but madness ... R + matlab + julia + fortran + Mathematica use 1 based and the world hasn't ended sometimes 0 based is nice, but you get used to not having it, just like you get used to having it. As for the truncating integer div ... I do hate that that in python and do "from __future__ import division" always ;) On Monday, April 4, 2016 at 11:55:40 PM UTC-7, 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] >> >>
