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2

You can interactively resize the histogram Graphics with the drag handles: Or programmatically, using the part spec as in bbgodfrey's second answer: MapAt[Show[#, ImageSize -> 350] &, pci, {2, 1, 1, 1}]


2

In response to the Comment by @JEP, an alternative is to extract the two graphics and the caption from the original quantity, pci, in the Question p1 = pci[[1]] p2 = pci[[2, 1, 1, 1]] caption = pci[[2, 1, 1, 2]] and then combine them as desired. For instance, Grid[{{Show[p1, ImageSize -> 360], Show[p2, ImageSize -> 250]}, {SpanFromAbove, ...


6

Rather than making the Histogram larger, I suggest you measure vals in terms of $1000. Replace your last line by pci = GeoRegionValuePlot[Thread[counties -> Quantity[QuantityMagnitude[vals/1000], IndependentUnit["$K per year"]]], PlotLegends -> Histogram] Dividing vals by 1000 expresses quantities in units of $1000. IndependentUnit["$K per ...


2

An alternate approach is to truncate the labels. However, this does not center the labels. nmcounties = AdministrativeDivisionData[ Entity["AdministrativeDivision", {"NewMexico", "UnitedStates"}], "Subdivisions"]; nmpopdensity = AdministrativeDivisionData[#, "PopulationDensity"] & /@ nmcounties; Note that you can only need Thread with ...


4

This required a bit more work than I initially anticipated. To your second question first, per the documentation, GeoRegionValuePlot is quite versatile in what it accepts, and when you are working with a common, queryable propery, you should use the form GeoRegionValuePlot[enityList -> "property"] or GeoRegionValuePlot[EntityClass -> "property"] ...


7

One can use the periodicity over $\theta$ and add one periodic copy of the data. In this case FindCurvePath works much better. I also add an interpolation of the result arrayData = Flatten[Thread@{#, Join[{##2}, {##2} + 2 π]} & @@@ originalData, 1]; curvesPosition = FindCurvePath@arrayData; {t, θ} = Interpolation@Transpose@{Range[0., 1, 1/(Length@# - ...


7

You can post-process the graphics output to shift the arrows: vp1 = VectorPlot[{-1 - x^2 + y, 1 + x - y^2}, {x, -2, 2}, {y, -2, 2}, VectorPoints -> points, VectorScale -> {.5, .4}, ImageSize -> 400, Prolog -> {Yellow, Opacity[.5], Rectangle[{-1, -1}, {1, 1}], Opacity[1], Red, PointSize[Large], Point[points]}]; vp1b = vp1 ...


15

Method 1: FindCurvePath (as mentioned by Yves Klett). This method is simple, but unfortunately, there are small issues (as shown in plots), that the curves are not identified perfectly. arrayData = Flatten[Function[{lst}, {First @ lst, #} & /@ Rest[lst]] /@ originalData, 1]; curvesPosition = FindCurvePath[arrayData]; ListPlot[curves = ...



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