Iterating through two arrays using lambda

Question

I'm in the process of trying to teach myself lambda expressions in C# and I've seem to have stumped myself...

This is essentially what I am trying to accomplish, please assume that ar1 & ar2 will always have the same length.

    double sum = 0;

    for(int x=0; x<size; x++){
        sum += (ar1[x]*ar2[x]);
    }

Notice the Arrays are being multiplied.

Is there an Lambda function that can accomplish this in a single line?

Thank you

Suppose you had to invent a method that took a lambda and two sequences; could you write such a method? — Eric Lippert
– Eric Lippert, Commented Jan 7, 2017 at 2:58

Sergey Berezovskiy · Accepted Answer · 2017-01-07 02:55:35Z

6

double sum = ar1.Zip(ar2, (a1, a2) => a1 * a2).Sum();

Zip - applies lambda to corresponding elements of two sequences

Sum - computes sum of results

answered Jan 7, 2017 at 2:55

Sergey Berezovskiy

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6 Comments

Nefariis Over a year ago

I mean it works -- can walk me through "ar1.zip(ar2, (a1, a2) +> a1*a2)"? .... pretend Im a total noob and have no idea what's going on (pretend)...

Sergey Berezovskiy Over a year ago

@Nefariis well, what Zip does - it gets enumerators of both sequences and starts enumerating them simultaneously. I.e. each time it calls MoveNext on both sequences. On each step we have two Current values (a1 and a2). And on each step Zip applies lambda function which accepts these two current values and produces some result. In our case lambda produces multiplication of two Current values. Result of multiplication is yielded. So actually Zip produces sequence of multiplications of corresponding items.

Nefariis Over a year ago

wait, I think Im seeing it -- you are creating another array with a1 and a2 zipped. Then saying that the indices (a1,a2) will be multiplied => a1*a2 - and then you are just summing it up at the end... Clever - here I was trying to iterate through each one

Nefariis Over a year ago

Thank you very much, I have a feeling this will come in handy

Luc Morin Over a year ago

Linq is divine ;-)

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