Skip to main content

programming - Modifying a List in a function in place



An example will be most specific:


func[list_, column_] := list[[All, column]] = Map[#*2 &, list[[All, column]]];

This throws errors.


I want to avoid doing something like this:


func2[list_] := Map[#*2 &; list];

list[[All, 2]] = func2[list[[All,2]]]

because nesting a couple of functions raises complexity unnecessarily, the output would have to be reassigned every time.


Thanks in advance.


As a followup, using HoldFirst works fine, but using the so defined function in a Map gives again errors.


The setup is as follows:


create a nested list


testList = Table[Table[{x y, x y 2}, {x, 1,3}], {y,1,3}]

define afunc with HoldFirst Attribute



afunc = Function[{list, col}, list[[All, col]] = Map[# * 2 &, list[[All, col]]], HoldFirst]

and another function using the first


bfunc[nestedList_, col_] := Map[afunc[#, col] &, nestedList]

now, a call to


bfunc[testList, 2]

should alter the 2'nd columns of the nested lists I'd expect, but it instead throws errors


i've tried to set Attribute HoldFirst on this function as well but it didn't work out as expected




Answer



You basically need a pass-by-reference semantics, which in Mathematica can be emulated with Hold-attributes. Add a HoldFirst attribute to your function:


SetAttributes[func,HoldFirst]

and you should be fine. Without it, list evaluates to its value before the assignment is attempted, and since expressions in Mathematica are immutable, you get an error.


To address your question in comments, the one-liner you asked for can be this:


func = Function[{list,column}, 
list[[All, column]] = Map[#*2 &, list[[All, column]]],
HoldFirst
]


Note however that, since this is a pure function, you can not do argument checks as elegantly as you can with patterns, and you can not overload your functions on different arguments as elegantly.


Note also that, while yet another way to do this is to keep your function as it is but rather wrap the first argument in Unevaluated in every function call, I would strongly advise against that. The reason is that it makes the function itself not self-contained, because it has to assume that the user will always remember to use Unevaluated (which it shouldn't), and there is no way to test whether or not the user actually did use it.


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 - Adding a thick curve to a regionplot

Suppose we have the following simple RegionPlot: f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}] Now I'm trying to change the curve defined by $y=g[x]$ into a thick black curve, while leaving all other boundaries in the plot unchanged. I've tried adding the region $y=g[x]$ and playing with the plotstyle, which didn't work, and I've tried BoundaryStyle, which changed all the boundaries in the plot. Now I'm kinda out of ideas... Any help would be appreciated! Answer With f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 You can use Epilog to add the thick line: RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, Epilog -> (Plot[g[x], {x, 0, 2}, PlotStyle -> {Black, Thick}][[1]]), PlotStyle -> {Directive[Yellow, Opacity[0.4]], Directive[Pink, Opacity[0.4]],