Populating numpy array with most performance - numpy

Question

I have few arrays a,b,c and d as shown below and would like to populate a matrix by evaluating a function f(...) which consumes a,b,c and d.
with nested for loop this is obviously possible but I'm looking for more pythonic and fast way to do this.
So far I tried, np.fromfunction with no luck.
Thanks
PS: This function f has a conditional. I still can consider approaches which does not support conditionals but if the solution supports conditionals that would be fantastic.
example function in case helpful
def fun(a,b,c,c): return a+b+c+d if a==b else a*b*c*d
Also why fromfunction failed is shown below
>>> a = np.array([1,2,3,4,5])
>>> b = np.array([10,20,30])
>>> def fun(i,j): return a[i] * b[j]
>>> np.fromfunction(fun, (3,5))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "C:\Users\Anaconda3\lib\site-packages\numpy\core\numeric.py", line 1853, in fromfunction
return function(*args, **kwargs)
File "<stdin>", line 1, in fun
IndexError: arrays used as indices must be of integer (or boolean) type

Answer 1

The reason the function fails is that np.fromfunction passes floating-point values, which are not valid as indices. You can modify your function like this to make it work:
def fun(i,j):
return a[j.astype(int)] * b[i.astype(int)]
print(np.fromfunction(fun, (3,5)))
[[ 10 20 30 40 50]
[ 20 40 60 80 100]
[ 30 60 90 120 150]]

Answer 2

Jake has explained why your fromfunction approach fails. However, you don't need fromfunction for your example. You could simply add an axis to b and have numpy broadcast the shapes:
a = np.array([1,2,3,4,5])
b = np.array([10,20,30])
def fun(i,j): return a[j.astype(int)] * b[i.astype(int)]
f1 = np.fromfunction(fun, (3, 5))
f2 = b[:, None] * a
(f1 == f2).all() # True
Extending this to the function you showed that contains an if condition, you could just split the if into two operations in sequence: creating an array given by the if expression, and overwriting the relevant parts by the else expression.
a = np.array([1, 2, 3, 4, 5])
b = np.array([5, 4, 3, 2, 1])
c = np.array([100, 200, 300, 400, 500])
d = np.array([0, 1, 2, 3])
# Calculate the values at all indices as the product
result = d[:, None] * (a * b * c)
# array([[ 0, 0, 0, 0, 0],
# [ 500, 1600, 2700, 3200, 2500],
# [1000, 3200, 5400, 6400, 5000],
# [1500, 4800, 8100, 9600, 7500]])
# Calculate sum
sum_arr = d[:, None] + (a + b + c)
# array([[106, 206, 306, 406, 506],
# [107, 207, 307, 407, 507],
# [108, 208, 308, 408, 508],
# [109, 209, 309, 409, 509]])
# Set diagonal elements (i==j) to sum:
np.fill_diagonal(result, np.diag(sum_arr))
which gives the following result:
array([[ 106, 0, 0, 0, 0],
[ 500, 207, 2700, 3200, 2500],
[1000, 3200, 308, 6400, 5000],
[1500, 4800, 8100, 409, 7500]])