Python/Numpy array manipulation in Mathematica syntax

As a python programmer, numpy tools usually come to my mind if I want to manipulate arrays (list, matrices, ...). Is there any reference (e.g., dictionary) of how to translate numpy syntax to Mathematica syntax? For instance, assuming a is an array, I found the following translations (python code as quotes, Mathematica code as code):

linspace / logspace

a = np.linspace(0,10,11)

a = Array[# &, 11, {0, 10}]
(* {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} *)

element selection based on condition

a[a>0]

Select[a, # > 0 &]

(* {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} *)

element selection based on other boolean array

s = a>0

a[s]

s = # > 0 & /@ a

Pick[a, s]
(* {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} *)

a[3:-2:2]

a[[4;;-3;;2]]
(* {3, 5, 7} *)

Such a correspondence list is what I am looking for. Does this already exist? If not, I believe it would be helpful to extend this Q&A post to continuously build such a list.

One thing I am trying to use just now is np.roll. Using the example from above, I am searching for a command that yields

EquivalentOfNpRoll[a,2]
(* {9, 10, 0, 1, 2, 3, 4, 5, 6, 7, 8}

How would I do this? Permute seems to be an overkill. At least it was far from obvious to me how to implement this rolling permutation in a general but compact way.

Update:

From the comments:

np.roll(a,2)

RotateRight[a, 2]
(* {9, 10, 0, 1, 2, 3, 4, 5, 6, 7, 8} *)

a=np.arange(0,11,1)

a = Range[0, 10, 1]

Comments

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...

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