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My question is a lot similar to the one about Custom Scaling of x-axis in ListPlot, however since I am using ListLogPlot, the option of ScalingFunctions does not work. I have a data set:

data={{0.05, Around[1.2, 0.01]}, {2.05, Around[1.56, 0.01]}, {4.05,Around[2.8974, 0.0237]}, {5.05, Around[3.51, 0.02]}, {6.05, Around[3.98, 0.02]}, {8.05, Around[4.3068, 0.0287]}, {10.05, Around[4.42, 0.02]}, {15.05, Around[4.81, 0.0303]}, {25.05, 
  Around[5.34, 0.01]}, {35.05, Around[6.486, 0.035]}, {45.05, Around[8.93, 0.04]}}

Since I have more data points uptil 10.05, I want to amplify that data more, so in principle I want that part of the plot to take majority of the space and the rest of 4-5 points less space. In a ratio of 3/4, to 1/4 or so. However I am failing to do so. I am currently combining two plots manually by playing around with aspect ratios and such but that is not an elegant approach. Any help will be appreciated.

Also this is just a sample set, I generally have 5-6 plots of such nature in a single plot and therefore benefit from having the y axis in a Log as the y values change drastically.

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    $\begingroup$ If you simply want custom x-axis scaling together with a log y-axis, is there a reason you're not using ListPlot with ScalingFunctions->{xScaling, "Log"}? You might also be interested in this question in regards to combining two plots together. $\endgroup$
    – Lukas Lang
    Jan 22, 2023 at 20:02
  • $\begingroup$ Presumably you want the x-axis in log rather than the y-axis. ListPlot[data, ScalingFunctions -> {"Log", None}] $\endgroup$
    – Bob Hanlon
    Jan 22, 2023 at 20:12
  • $\begingroup$ @LukasLang Thank you for the suggestion, I think that should do it! Sorry for some reason that did not occur to me. $\endgroup$
    – WaleeK
    Jan 23, 2023 at 10:07
  • $\begingroup$ @BobHanlon, no I need the data in y in log as generally the graph is used to compare data sets and the differences are more evident and easier to illustrate in a log plot than in a linear plot and the region of interest is 0-10 on x axis and the rest is when the data from both data sets is more or less converges to the same value. However as I said, what Lukas suggested should work $\endgroup$
    – WaleeK
    Jan 23, 2023 at 10:10
  • $\begingroup$ There are certain issues when I try to do it with Log as a scaling function or use ListLog Plot. Because as the question stated ListLogPlot does allow scaling as log. The issues now is that when I use scaling as Log in Y and scale and inverse scale in x, I can't seem to get the frameticks for entries greater than 10. And if I join the plots using your row technique than I have to manually adjust the values on the frame, which again consumes a lot fo time as I have a lot of similar graphs to go over $\endgroup$
    – WaleeK
    Jan 24, 2023 at 11:29

1 Answer 1

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It is actually pretty easy if you know exactly what properties of graphics you need to prioritize. Options are pretty flexible. A few approaches.

APPROACH 1: arbitrary smooth X-scaling

People were advising X-scaling with "Log". But that, while often fine, is pretty limited. You can define arbitrary X-scaling with functions. For example below is power-scaling:

scaleX[a_][x_]:=x^a

Manipulate[
    ListPlot[data,
    Frame->True,GridLines->{Range[1,45,5],All},FrameTicks->{Range[1,45,5],All},
    ScalingFunctions->{{scaleX[a],Quiet@InverseFunction[scaleX[a]]},None}],
{{a,1},.1,2}]

enter image description here

APPROACH 2: arbitrary Piecewise X-scaling

You can even do this with Piecewise function which basically solves your problem in a pretty neat all-in-single plot way:

simpleX[x_]:=Piecewise[{{3x/4,x<=10.5},{1x/4+10.5/2,x>10.5}}]

ListPlot[data,
    Frame->True,GridLines->{Range[1,45,5],All},FrameTicks->{Range[1,45,5],All},
    ScalingFunctions->{{simpleX,InverseFunction[simpleX]},None}]

enter image description here

APPROACH 3: horizontal plot stacking

enter image description here

Here is an approach that can be packed into a function that takes multiple plots if you understand the mechanics of it. Using your data

data={{0.05,Around[1.2,0.01]},{2.05,Around[1.56,0.01]},{4.05,Around[2.8974,0.0237]},{5.05,Around[3.51,0.02]},{6.05,Around[3.98,0.02]},{8.05,Around[4.3068,0.0287]},{10.05,Around[4.42,0.02]},{15.05,Around[4.81,0.0303]},{25.05,Around[5.34,0.01]},{35.05,Around[6.486,0.035]},{45.05,Around[8.93,0.04]}};

Define split in data:

split=10.5;

Understand which properties are the SAME:

same={GridLines->All,ImageSize->{Automatic,300},BaseStyle->12};

and which and how they are DIFFERENT:

diff1={PlotRange->{{0,split},{0,10}},ImagePadding->16{{1,0},{1,1}},Frame->{{True,False},{True,True}},AspectRatio->1/GoldenRatio,FrameTicks->{Range[10],Automatic}};
diff2={PlotRange->{{split,47},{0,10}},ImagePadding->16{{0,1},{1,1}},Frame->{{False,True},{True,True}},AspectRatio->3 1/GoldenRatio,FrameTicks->All};

Then build in a manner that can be easily generalized:

Row[{
    ListPlot[Select[data,#[[1]]<=split&],same~Join~diff1],
    ListPlot[Select[data,#[[1]]>split&],same~Join~diff2]
}]
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  • $\begingroup$ Thank you for the detailed answer! Combining it with the suggestion from Lukas about using Log as a scaling function for y I think it should do it $\endgroup$
    – WaleeK
    Jan 23, 2023 at 10:03
  • $\begingroup$ There are certain issues when I try to do it with Log as a scaling function or use ListLog Plot. Because as the question stated ListLogPlot does allow scaling as log. The issues now is that when I use sclaing as Log in Y and scale and inverse scale in x, I can't seem to get the frameticks for entries greater than 10. And if I join the plots using your row technique than I have to manually adjust the values on the frame, which again consumes a lot fo time as I have a lot of similar graphs to go over $\endgroup$
    – WaleeK
    Jan 24, 2023 at 9:45

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