Consider a list of lists in this form (with a shape $ m \times n \times 3 $):

 {{a1, R1, c11}, {a2, R1, c12}, {a3, R1, c13}, ..., {an, R1, c1n}},
 {{a1, R2, c21}, {a2, R2, c22}, {a3, R2, c23}, ..., {an, R2, c2n}},
 {{a1, Rm, cm1}, {a2, Rm, cm2}, {a3, Rm, cm3}, ..., {an, Rm, cmn}}

where in each outer list, the 2nd element $ R_i $ is fixed ($ i = 1, 2, ..., m $), and the 1st element changes from $ a_1 $ to $ a_n $, the 3rd element $ c_{ij} $ is normally a complex and its imaginary part can change from positive to negative or from negative to positive for several times. Here is a sample data for test.

I want to pick out the neighbor lists whenever the imaginary part of $ c_{ij} $ changes its sign, say, for $ R_2 $, the selected lists are something like $ \{a_j, R_2, c_{2j}\} $ and $ \{a_{j+1}, R_2, c_{2,j+1}\} $, where $ \text{Im} c_{2,j} < 0 $ and $ \text{Im} c_{2,j+1} > 0 $. More generally, for $ R_p $ I pick out $ \{a_j, R_p, c_{pj}\} $ and $ \{a_{j+1}, R_p, c_{p,j+1}\} $, and then to plot a curve with ListLinePlot[{{R1, a01}, {R2, a02}, ..., {Rp, a0p}, ..., {Rm, a0m}}], in which $ a_{0j} = (a_j + a_{j+1}) / 2 $. In other words, I what to plot a parameter curve w.r.t the 1st and 2nd elements, across which the imaginary part of the 3rd element changes sign.

I tried Cases, Select and ParametricPlot, but I am still having trouble to find all the pairs of the neighboring lists when the imaginary part of $ c_{ij} $ changes its sign.

  • 1
    $\begingroup$ Could you please provide a specific small dummy data (without which tests cannot be conducted), as well as the desired result? $\endgroup$ Commented Feb 20, 2020 at 5:11
  • $\begingroup$ @ΑλέξανδροςΖεγγ Thank you for the reminding. I have attached a sample data for testing. $\endgroup$
    – user55777
    Commented Feb 20, 2020 at 6:13

2 Answers 2


I imported your test data into a file called test.csv. On my Mac, this resulted in a set of strings, so I had to do a quick fixup to get data in the format you mentioned.

Would something like this be what you want?

ClearAll[testData, testDataFixed, theValues, calculatedValues];
testData = Import["~/Downloads/test.csv"];
testDataFixed = Map[ToExpression, testData, {2}];
theValues = 
  Cases[Partition[#, 2, 1], {{a_, b_, c_}, {d_, e_, f_}} /; 
      Sign[Im[c]] != Sign[Im[f]]] & /@ testDataFixed;
calculatedValues = 
  Flatten[Map[{#[[1, 2]], (#[[1, 1]] + #[[2, 1]])/2} &, 
    theValues, {2}], 1];

This gives the following plot: enter image description here I will note that due to the test for Sign, if the first element has no imaginary part and the second does, then this will also get flagged. I can modify this if you want to make sure that there really is an imaginary component that is changing sign.

A minor change to eliminate the cases when there is no imaginary part:

theValues = 
  Cases[Partition[#, 2, 
      1], {{a_, b_, c_}, {d_, e_, 
        f_}} /; (Sign[Im[c]] != Sign[Im[f]]) && (Sign[Im[c]] != 
         0) && (Sign[Im[f]] != 0)] & /@ testDataFixed;

This eliminates the values that were showing up at the bottom, giving this graph: enter image description here

OK, one more edit to split into clusters:


Giving: enter image description here

  • $\begingroup$ thank you. please modify to make sure that there indeed is an imaginary part that is reversing sign. Btw, any idea to connect the points to highlight the trend. Using ListLinePlot gives zigzag mess. In the figure, it seems two curves with a junction. $\endgroup$
    – user55777
    Commented Feb 20, 2020 at 7:59
  • $\begingroup$ Change for imaginary added. Will look into connecting the points. $\endgroup$
    – Mark R
    Commented Feb 21, 2020 at 3:23
  • $\begingroup$ Plot with connected points added based on FindClusters. $\endgroup$
    – Mark R
    Commented Feb 21, 2020 at 3:30

You can use SequenceCases to construct pairs based on sign changes in imaginary part of the third columns:

f = SequenceCases[#, {{a_, b_, c_}, {d_, e_, f_}} /; Sign[Im@c] != Sign[Im@f] :>        
     {b, (a + d)/2}, Overlaps -> True] & 

A data set with the structure described in OP:

m = 10; n = 5;
rr = Range[m];
aa = Range[n];
cc = Round[#, .01] & @ RandomComplex[{-5 - I, 5 + I}, {m, n}];
arc = Join[Transpose@Outer[List, aa, rr], Map[List, cc, {-1}], 3];

Grid[arc, Dividers -> All]

enter image description here

Use f on arc to get the desired output:

pairs = f /@ arc;


enter image description here

Grid[N @ pairs, Dividers -> All]

enter image description here

Use arc = sampledata to get

enter image description here


enter image description here

  • $\begingroup$ thank you. The data appear to form two curves with a junction. Any method to interpolate the data or connect the points into curve(s). Using ListLinePlot or Joined in ListPlotgives zigzag mess. $\endgroup$
    – user55777
    Commented Feb 20, 2020 at 15:08

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