Skip to main content

sorting - Sort data after specific ordering (ascending/descending) in multiple columns


I have some data:


data={{"a", 1, 1}, {"a", 1, 2}, {"a", 1, 3}, 
{"c", 2, 1}, {"b", 2, 2}, {"b", 2, 3},
{"c", 3, 1}, {"a", 3, 2}, {"a", 3, 3}}


When I use:


Sort[data]

I get the list ordered in incremental order first with column 1 then by 2 and last by the third. Suppose I want the first column in increasing order but the second in decreasing order and the third again in increasing order. How do I do that?


I tried with Sort and SortBy but I haven't managed to make it work.


The output should be:



a 3 2


a 3 3



a 1 1


a 1 2


a 1 3


b 2 2


b 2 3


c 3 1


c 2 1




Answer





Caveat lector: Incorrect results are generated by this solution, e.g.,


sortByColumn[{{"a", 1, 1}, {"b", 2, 3}, {"a", 3, 2}}, {1, 1, -1}]


returns


{{"a", 1, 1}, {"b", 2, 3}, {"a", 3, 2}}


when the correct result is obviously


{{"a", 1, 1}, {"a", 3, 2}, {"b", 2, 3}}


I've commented on the answer to bring it to the attention of the author, however seeing as they've not been here in some time, I'm also putting this here: I think a highly upvoted and accepted answer needs to be correct. - ciao



Here is my contribution, which has the following benefits over previous answers:




  • It sorts both numbers and non-numeric structures

  • You can sort any column (not just the first, followed by the second, etc)

  • You can sort in either direction (ascending / descending)

  • Original order is kept: if you sort on the second column, the first entry will follow the order of the original list. See the example with {0,-1}

  • Edit also allow specifying the priority of the columns. So given {-1,1} for the ordering, you can specify {1,2} to give the higher priority to the second column.


The code is as follows, including my usage code for my own comments.


Clear[sortByColumn]
sortByColumn::usage =
"Arguments: [Table, Direction, Priority]. Returns the list sorted \

by the directions for each column specified in `Direction`. For \
ascending order, use `1`, and for descending order, use `-1`. For \
sorting more than one column, input `Direction` as a list. For \
example, Direction={-1,1} will sort the first column in descending \
order followed by the second column in ascending order, ignoring any \
other column. To sort on the second column, use {0,1} for the syntax.

When sorting two or more columns, you can provide the `Priority` \
for which column should be sorted first. For example, \
`sortByColumn[data,{-1,1},{1,2}]` would sort first in ascending order \

on the second column (because it has a higher priority) and then in \
descending order on the first column.";

sortByColumn[list_?MatrixQ, dir : _Integer | {__Integer}, priority_: {}] :=
Module[{l = Length@list[[1, All]], w, p, d},
w = Reverse@Range@l;
p = If[Length@priority > 0, PadRight[Flatten@{priority}, l],
p = Range@l];
w = w[[Ordering@p]];
d = PadRight[Flatten@{dir}, l];

Sort[list, NonNegative@Total[(w d MapThread[Order, {##}])] &]]

For example, using the data set provided by Mr. Wizard:


data={{"a", 1, 1}, {"a", 1, 5}, {"a", 1, 3}, 
{"c", 2, 1}, {"b", 2, 2}, {"b", 2, 3},
{"c", 3, 1}, {"a", 3, 2}, {"a", 3, 3}};
data[[All, 2]] = data[[All, 2]] /. {1 -> "q", 2 -> "r", 3 -> "s"};

Here are the results of some trial runs. First the original:


{a,q,1}

{a,q,2}
{a,q,3}
{c,r,1}
{b,r,2}
{b,r,3}
{c,s,1}
{a,s,2}
{a,s,3}

The result of sortByColumn[data,-1].



{c,r,1}
{c,s,1}
{b,r,2}
{b,r,3}
{a,q,1}
{a,q,2}
{a,q,3}
{a,s,2}
{a,s,3}


Result of sortByColumn[data,{0,-1}]


{c,s,1}
{a,s,2}
{a,s,3}
{c,r,1}
{b,r,2}
{b,r,3}
{a,q,1}
{a,q,2}
{a,q,3}


And finally, the result the OP wanted, sortByColumn[data,{1,-1,1}]


{a,s,2}
{a,s,3}
{a,q,1}
{a,q,2}
{a,q,3}
{b,r,2}
{b,r,3}
{c,s,1}

{c,r,1}

An example showing the use of the priority argument: sortByColumn[data, {-1, 1}, {1, 2}]


{a,q,1}
{a,q,5}
{a,q,3}
{c,r,1}
{b,r,2}
{b,r,3}
{c,s,1}

{a,s,2}
{a,s,3}

Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

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.