Mathematica equivalent to gnuplot's fsteps option?

Question

In a ListLinePlot I can set InterpolationOrder -> 0 to get the equivalent of gnuplot's steps option, where the stepped line for each two points steps at the second point. Is there a Mma equivalent to gnuplot's fsteps option, where the stepped line for each two points steps at the first point? (In tikz terminology, I want |- instead of -| to connect successive points.)

Example application: Verhulst diagrams.

Edit 1:

Answers to the question referenced by @MarcoB show that ListLinePlot with InterpolationOrder->0 constructs a standard Manhatten path (alternating horizontal and vertical segments between points). This is equivalent to gnuplot's steps option. I'm after the equivalent of gnuplot's fsteps option: alternating vertical and horizontal segments.

To be clear: I am not asking how to produce this plot by creating new data through list manipulations. (See posts on cobweb plots with Mma for that.) I'm looking for built-in Mma functionality. (I accept that this may not exist.)

Edit 2:

logisticPath = NestList[2.75*#*(1 - #) &, 0.4, 10]
ListStepPlot[Transpose[{logisticPath, logisticPath}], "Left", 
 PlotRange -> All,
 Prolog -> Line[{{0.4, 0.4}, {0.8, 0.8}}]]

Produces

This is almost right. I just need to get rid of that initial segment, which is outside the data. What I want looks like this (using the same data):

cobwebPoints = Partition[Riffle[logisticPath, logisticPath], 2, 1]
ListLinePlot[cobwebPoints, PlotRange -> All,
 Prolog -> {Line[{{0, 0}, {1, 1}}]},
 PlotRange -> All]

Answer 1

Fresh v10.2 introduces ListStepPlot which provides just the right facility for this task:

pts = {{1.81515, 1.46912}, {4.99276, 1.87137}, {9.12065, 
    6.83238}, {9.9288, 5.30858}, {5.6387, 4.30558}, {4.45562, 
    6.44036}, {0.0721842, 9.82819}, {2.88155, 4.80602}, {7.01912, 
    9.91699}, {0.277599, 5.36962}};

ListStepPlot[SortBy[First][pts], Left, Mesh -> Full, MeshStyle -> Red]

Answer 2

Alan, I think you can use Michael's stepFunctions from the answer I linked to in the comments to solve your problem. Here is a link to MichaelE2's original answer: https://mathematica.stackexchange.com/a/30103/27951. Below is an adaptation to your problem using the step* functions defined in Michael's original answer. Note that this approach requires no modification to your existing points.

---

First of all, let's have some random points to play with:

pts = {{1.81515, 1.46912}, {4.99276, 1.87137}, {9.12065, 6.83238}, {9.9288, 5.30858}, 
       {5.6387, 4.30558}, {4.45562, 6.44036}, {0.0721842, 9.82819}, {2.88155, 4.80602}, 
       {7.01912, 9.91699}, {0.277599, 5.36962}};

ListLinePlot[
 SortBy[First][pts],
 InterpolationOrder -> 0, PlotRange -> All,
 Epilog -> {PointSize[0.02], Red, Point@pts}
]

Above is the native behavior of InterpolationOrder -> 0 that is undesirable to you in this case.

---

It seems to me that the function you are looking for is stepFloor in the post linked above. You use it as follows:

Plot[
 stepFloor[pts][x], {x, 0, 11},
 Epilog -> {PointSize[0.02], Red, Point@pts}
]

---

You can also compare all the functions Michael had proposed. Note that stepCeiling repoduces the behavior of the native InterpolationOrder -> 0 in ListLinePlot.

original = ListLinePlot[
   Sort@pts,
   InterpolationOrder -> 0,
   Epilog -> {PointSize[0.02], Red, Point@pts},
   PlotLegends -> LineLegend[
     {Blue, Directive[Black, Dashed], Directive[Darker@Green, Dashed], Directive[Darker@Red, Dashed]},
     {"original", "stepFloor", "stepCeiling", "stepNearest"}
     ]
  ];

floor = Plot[stepFloor[pts][x], {x, -1, 10}, PlotStyle -> {Thick, Dashed, Black}];
ceiling = Plot[stepCeiling[pts][x], {x, -1, 10}, PlotStyle -> {Thick, Dashed, Darker@Green}];
nearest = Plot[stepNearest[pts][x], {x, -1, 10}, PlotStyle -> {Thick, Dashed, Darker@Red}];

Show[original, floor, ceiling, nearest]

Answer 3

One way is to simply shift the data:

data = {{0, 1}, {1, -1}, {2, 2}, {3, 1}, {4, 3/2}};
ListLinePlot[ data, InterpolationOrder -> 0]
ListLinePlot[ 
   Transpose[{data[[All, 1]], 
     Append[data[[All, 2]][[2 ;;]], data[[-1, 2]]]}], 
        InterpolationOrder -> 0]