plotting - Filling option failing when ListLinePlot given my data

Context

I want to do Filled plot such as:

dat1 = {Table[{i^2, i + 1}, {i, 5}], Table[{i^2, i}, {i, 5}]}
ListLinePlot[dat1, PlotStyle -> Red, Filling -> {1 -> {2}}]

Mathematica graphics

But when I apply this to my own data:

dat = 
{
  {{3.0779456558039593`, 3.842486584798708`}, 
   {1.2951669675555262`, 4.05402324780319`}, 
   {0.9196161755899819`, 3.939944632202657`}, 
   {0.5722748507275665`, 3.8875332289711`}, 
   {0.29373287932080444`, 3.7317102841786385`}},
  {{3.0779456558039593`, 7.125115694053534`}, 
   {1.2951669675555262`, 5.8724983311100605`}, 

   {0.9196161755899819`, 5.707473500231869`}, 
   {0.5722748507275665`, 5.114513590881306`}, 
   {0.29373287932080444`, 4.240800591615147`}}
}

It fails to fill the region:

ListLinePlot[dat, PlotStyle -> Red, Filling -> {1 -> {2}}]

Mathematica graphics

Question

Am I missing something obvious?

Answer

The cause is your data is out of order. So, the workaround is

ListLinePlot[SortBy[First] /@ dat, PlotStyle -> Red, 
  Filling -> {1 -> {2}}]

I believe it is not a bug. Or, at least, it is an issue that is tangentially discussed in the ListLinePlot documentation. So, while this is certainly disconcerting to see, unusual behavior with unsorted data is expected.

Comments

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I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.

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