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

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...
Post a Comment
Read more

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...
Post a Comment
Read more

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...
Post a Comment
Read more