Trying to imitate Excel's SUMPRODUCT function:

```
SUMPRODUCT(v1, v2, ..., vN) =
v1[0]*v2[0]*...*vN[0] + v1[1]*v2[1]*...*vN[1] + ... + v1[n]*v2[n]*...*vN[n]
```

where n is the number of elements in each vector.

This is similar to dot product, but for multiple vectors. I read the very detailed discussion of the regular dot product, but I don't know how to cleanly extend it to multiple vectors. For reference, I'm copying the optimized code proposed there, which I ported (trivially) to Python 3. BTW, for dot product, the last approach still wins in P3K.

```
def d0(v1,v2):
"""
d0 is Nominal approach:
multiply/add in a loop
"""
out = 0
for k in range(len(v1)):
out += v1[k] * v2[k]
return out
def d1(v1,v2):
"""
d1 uses a map
"""
return sum(map(mul,v1,v2))
def d3(v1,v2):
"""
d3 uses a starmap (itertools) to apply the mul operator on an zipped (v1,v2)
"""
return sum(starmap(mul,zip(v1,v2)))
```

```
import operator
def sumproduct(*lists):
return sum(reduce(operator.mul, data) for data in zip(*lists))
```

for python 3

```
import operator
import functools
def sumproduct(*lists):
return sum(functools.reduce(operator.mul, data) for data in zip(*lists))
```

What about good old list comprehensions? (As mentioned by @Turksarama this only works for two lists)

```
sum([x * y for x, y in zip(*lists)])
```

Testing in Python 3.6:

```
In [532]: import random
In [534]: x = [random.randint(0,100) for _ in range(100)]
In [535]: y = [random.randint(0,100) for _ in range(100)]
In [536]: lists = x, y
```

Using list comprehensions

```
In [543]: %timeit(sum([x * y for x, y in zip(*lists)]))
8.73 µs ± 24.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
```

Note that "tuple" comprehensions are slower

```
In [537]: %timeit(sum(x * y for x, y in zip(*lists)))
10.5 µs ± 170 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
```

Using `map`

```
In [539]: %timeit(sum(map(lambda xi, yi: xi * yi, x, y)))
12.3 µs ± 144 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
```

Using `functools.reduce`

```
In [542]: %timeit(sum(functools.reduce(operator.mul, data) for data in zip(*lists)))
38.6 µs ± 330 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
```

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