Skip to main content

list manipulation - How do I disable that Mathematica orders terms in lexicographic order?


I have matrices with entries where the order of the entries is important. Unfortunately Mathematica resorts terms in products in a lexicographic order. For example,


A = {{a1, a2}, {a3, a4}};
B = {{b1, b2}, {b3, b4}};
Z = {{z1, z2}, {z3, z4}};

Z.B.A

(* {{a1 (b1 z1 + b3 z2) + a3 (b2 z1 + b4 z2),

a2 (b1 z1 + b3 z2) + a4 (b2 z1 + b4 z2)}, {a1 (b1 z3 + b3 z4) +
a3 (b2 z3 + b4 z4), a2 (b1 z3 + b3 z4) + a4 (b2 z3 + b4 z4)}} *)

I would like that Mathematica keeps the order of the terms as they appear in the product, i.e.


(* z1 b1 a1 + ... *)

How can the lexicographic ordering be disabled?



Answer



This is related to the Orderless attribute of Times and Plus. These attributes could be removed permanently with some hacks, but that would break Mathematica.


If you only want to display the result in a certain way, but not do calculations with it, it may be safe to remove those attributes temporarily using Block.



Block[{Plus, Times}, With[{result = Z.B.A}, HoldForm[result]]]

Mathematica graphics




Update


Jacob is asking how to preserve some of the evaluation rules of Times, such as 0*x → 0 and 1*x → x, while still removing the Orderless attribute. This is possible with Internal`InheritedBlock. It works like Block in that any modifications to the blocked symbols will be temporary, but it does not remove blocked symbol definitions at the start. Thus we can selectively remove the Orderless attribute.


This is an undocumented function, so the usual caveats apply: it may change behaviour in future versions, it may behave unexpectedly, and it may even crash your kernel (this one has never crashed my kernel, but other internal functions have).


Internal`InheritedBlock[{Plus, Times},
Unprotect[Plus, Times];
ClearAttributes[{Plus, Times}, Orderless];

With[{result = Z.B.A}, HoldForm[result]]
]

Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...