Skip to main content

plotting - Listplot imaginary part of complex numbers


I have the following list


w={{0.01,99 +0.00001414 I},{0.15,6.6370108 +0.003144129 I},{0.25,3.9515722 +0.00854493297 I},{6,0.10041 +0.28132187 I}}

and I want to ListPlot the imaginary part but with the command Im[w] I get the list



{{0,0.00001414},{0,0.00314413},{0,0.00854493},{0,0.281322}}

This way I basically lose the x axis values 0.01, 0.15, 0.25 and 6. How can I get in a list the imaginary part and the x axis values? (With Re[w] I can get the real part right, the 0.01, 0.15 etc don't change)



Answer



Almost a dozen alternatives with timings:


ClearAll[f1, f2, f3, f4, f5, f6, f7, f8, f9, f10, f11]; 
f1[w_] := w /. {a_, Complex[_, b_]} :> {a, b};
f2[w_] := {#, Im@#2} & @@@ w; (* my favorite ... credit: ubpdqn *)
f3[w_] := w /. Complex[_, b_] :> b;
f4[w_] := MapAt[Im, w, {{All, -1}}];

f5[w_] := Module[{s = Im /@ w}, s[[All, 1]] = w[[All, 1]]; s];
f6[w_] := Module[{s = w}, s[[All, -1]] = Im /@ w[[All, -1]]; s];
f7[w_] := Transpose@{First /@ w, Im /@ Last /@ w};
f8[w_] := Transpose@{w[[All, 1]], Im /@ w[[All, -1]]};
f9[w_] := Through@{First, Composition[Im, Last]}@# & /@ w;
f10[w_] := Transpose@{First /@ w, Composition[Im, Last] /@ w};
f11[w_] := Replace[w, {x_, complex_} :> {x, Im@complex}, {1}];

f1@w
(* {{0.01`, 0.00001414`},{0.15`, 0.003144129`},{0.25`, 0.00854493297`},{6, 0.28132187`}}*)


functions = {f1, f2, f3, f4, f5, f6, f7, f8, f9, f10, f11};
results = (Through@functions@w);
Equal @@ results
(* True *)

Timings:


testdata1 = Transpose[{RandomReal[100, 10000], RandomComplex[100, 10000]}];
testdata2 = Transpose[{RandomReal[100, 100000], RandomComplex[100, 100000]}];
testdata3 = Transpose[{RandomReal[100, 1000000], RandomComplex[100, 1000000]}];


Using Mr.Wizard's timeAvg function:


SetAttributes[timeAvg, HoldFirst]  
timeAvg[func_]:= Do[If[#>0.3, Return[#/5^i]] &@@ AbsoluteTiming@Do[func,{5^i}], {i, 0, 15}]

TableForm[{#, timeAvg[#@testdata1;]} & /@ functions]

enter image description here


results = Through@functions@testdata1;
Equal @@ results

(* True *)

TableForm[{#, timeAvg[#@testdata2;]} & /@ functions]

enter image description here


results = Through@functions@testdata2;
Equal @@ results
(* True *)

TableForm[{#, timeAvg[#@testdata3;]} & /@ functions]


enter image description here


results = Through@functions@testdata3;
Equal @@ results
(* True *)

Comments

Post a Comment

Popular posts from this blog

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...
Post a Comment
Read more

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...
Post a Comment
Read more

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...
Post a Comment
Read more