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Given a matrix M=RandomReal[{0},{5,5,5}]; How can I identify elements with a list?

Expected

In[81]:= M[[5,5,5]]

Out[81]= 0.

M[[#]]&/@{{5,5,5}} Doesn't work, no matter how much Flatten[]

There has to be an easier way than

In[83]:= a={2,2,2}
Out[83]= {2,2,2}
In[87]:= M[[a[[1]],a[[2]],a[[3]]]]
Out[87]= 0.

Apologies, I know there is probably a extremely simple answer... Thanks for anyone who takes a look.

Edit* I would also like to adjust the identified element which has given a number of errors. M[[{5,5,5}]]=1 Using @J.M. s solution

In[7]:= M2[[##]] & @@ allcells[[-1]] = +1

During evaluation of In[7]:= Set::write: Tag Apply in (M2[[##1]]&)@@{5,5,5} is Protected.

Out[7]= 1
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  • 2
    $\begingroup$ Try M[[##]] & @@ {5,5,5}. $\endgroup$ – J. M. will be back soon Mar 23 '18 at 1:35
  • $\begingroup$ @J.M. Awh! Awesome! Thanks a lot just for complete I'm using M[[##]] & @@@ {{5,5,5},{4,4,4},{5,4,5}} but I doff my dunce cap' $\endgroup$ – Awkward Panda Mar 23 '18 at 1:38
  • $\begingroup$ If you figured out how it works, I invite you to write an answer to your own question. ;) $\endgroup$ – J. M. will be back soon Mar 23 '18 at 1:45
  • $\begingroup$ You might also want have a look at Extract. For example, Extract[M, {{5, 5, 5}}] returns {0.}. This might seem to be artificial, but Extract also works with lists of lists, e.g. idx = RandomInteger[{1, 5}, {10, 3}]; Extract[M, idx]. In this case, you may also use M[[##]] & @@@ idx, but this unpacks arrays and is thus several magnitudes slower than Extract. $\endgroup$ – Henrik Schumacher Mar 23 '18 at 9:30
  • $\begingroup$ @J.M. @HenrikSchumacher How can I edit the element? I seem to be getting an error In[7]:= M[[##]] & @@ a[[-1]] = +1 During evaluation of In[7]:= Set::write: Tag Apply in (M[[##1]]&)@@{5,5,5} is Protected. Out[7]= 1 $\endgroup$ – Awkward Panda Mar 23 '18 at 11:35
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The fundamental difference here, though subtle yet important, is between M[[a,b,c]] and M[[{a,b,c}]], the key part being (pun not intended) the curly brackets around a,b,c.

The first one gives what you want, and it's even in the form of your "expected" code, whereas the second one deals with a different functionality of Part, or rather the [[...]] code.

The M[[{a,b,c}]] code gives a list of elements of M at positions a,b,c, rather than the single element at the multidimensional position {a,b,c}.

Relating to your example, and as you already showed, the code M[[5,5,5]] gives the single element 0. at position {5,5,5}, whereas the code M[[{5,5,5}]] gives the list of elements each at position 5 in the first level of M.

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  • $\begingroup$ I actually laughed at 'the key Part being'... now I feel bad. $\endgroup$ – Awkward Panda Mar 24 '18 at 1:02

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