# How to replace Do loops by Inner

I need to replace Do loops by something else because it takes forever with big programs. I have been hinted towards Inner, but even though I tried, I did not manage to get what I need. Here's a toy example of my actual problem:

Input:

u = {{u1a, u1b}, {u2a, u2b}, {u3a, u3b}, {u4a, u4b}};
v = {{v1a, v1b}, {v2a, v2b}, {v3a, v3b}, {v4a, v4b}};
f = {f1, f2, f3, f4};


The output I am looking for:

d = {{u1a*v1a*f1 + u2a*v2a*f2 + u3a*v3a*f3 +u4a*v4a*f4},
{u1b*v1b*f1 + u2b*v2b*f2 + u3b*v3b*f3 + u4b*v4b*f4}}


It is important not to have any loop constructs; only Inner, Table, etc. which are generally faster than loops

• Like f.(u v)? (No Inner[] here, sorry.) – J. M.'s technical difficulties May 14 '13 at 18:19
• What do you mean by (u v)? This is the straight-forward wau to do it: – Mencia May 14 '13 at 18:23
• Do[d[[i]] = Sum[V[[l, i]]*U[[l, i]]*f[[l]], {l, 1, 4, 1}], {i, 1, 2, 1}]; – Mencia May 14 '13 at 18:24
• Why not execute that piece of code I gave to see what I mean? (On that note, please look up Dot[] in the docs.) – J. M.'s technical difficulties May 14 '13 at 18:24
• @J.M. thankyouuuuuuuuuuu!!! :)))) It definitely works! I appreciate it! – Mencia May 14 '13 at 18:29

To settle this question:

You certainly can use Inner[] for the task, like so:

Inner[Times, f, u v, Plus]


but you should know that Mathematica, conveniently enough, knows how to do a dot product. The function for doing this is named, appropriately enough, Dot[]:

Dot[f, u v]


but I find that a bit too long to write, so I use a syntax that is a bit more understandable, at least to me:

f.(u v)


where the parentheses are needed to give (Hadamard) multiplication higher priority than the dot product.

• Why community wiki, if I may ask? – Ajasja May 14 '13 at 18:39
• @Ajasja, I don't want to get reputation for this answer. (I'm choosing to interpret "CW" as "Credit Waived".) – J. M.'s technical difficulties May 14 '13 at 18:40
• Yes, I gathered that. But I was wondering why you don't want to get the reputation (or why you care about this). Is this answer to simple? (Personally I think that every useful answer deserves to be upvoted) – Ajasja May 14 '13 at 19:04
• The upvote would be for your expertise in diagnosing the problem, not the quantity of work visible in the answer, like in the old story of the man with a hammer... – cormullion May 14 '13 at 19:27
• @Ajasja, if I consider the (not few!) answers I wanted to get reputation for, but don't; versus the (also not few!) answers I really didn't want to get rep for, but did, this seems like a way I'd prefer to proceed. I have nicer answers languishing for lack of even a glance from other people; should I have to beg attention for those? I think not. – J. M.'s technical difficulties May 15 '13 at 1:53

One option is to use MapThread, which gets you there:

List /@ MapThread[Plus, u v f]


But the infix notation for the Dot product is more elegant (as J.M. proposed in the comment above):

List /@ (f.(u v))


Note that if you're doing a lot of these computations, that f.(u v) has a very slight computational edge, which you can test with the Timing command:

    List /@ MapThread[Plus, u v f] // Timing
List /@ (f.(u v)) // Timing

• Thanks for your reply and those edits @j-m. I'll try to (remember to) adhere to that styling in the future. – TransferOrbit May 15 '13 at 9:10