Function over each value in Python Array (without using def)

Question

The input array is x with dimensions (1 x 3) and the output array is 3 x 3 (column of input x column of input). The output array's diagonals are the values^2. If row != column, then the formula is x(row)+x(col) for each value. Currently for 1 x 3 but should assume a variety of dimensions as input. Cannot use 'def'. The current code does not work, what would you recommend?

x = np.array([[0, 5, 10]])
output array formulas = 
[[i^2,   x(row)+x(col),  x(row)+x(col)]
 [x(row)+x(col), i^2,    x(row)+x(col)]
 [x(row)+x(col), x(row)+x(col),   i^2]]

# where row and column refer to the output matrix row, column. For example, the value in (1,2) is x(1)+x(2)= 5

ideal output = 
[[0 5 10]
 [5 25  15]
 [10 15 100]]

Code Attempted:

x = np.array([[0, 5, 10]])
r, c = np.shape(x)
results = np.zeros((c, c))
g[range(c), range(c)] = x**2
for i in x:
    for j in i:
        results[i,j] = x[i]+x[j]

why not just something like: [n[i][j]**2 if i == j else n[i][j] for j in i for i in n]. (obviously you would have to get the actual indexes of i and j so this code is just an algorithm example. But you could use enumerate()) — Eli Harold
– Eli Harold, Commented Jan 19, 2022 at 16:48
@EliHarold, I tried that but getting error: ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all(). Can you show an example with the given x array? — maximus
– maximus, Commented Jan 19, 2022 at 16:50

juanpa.arrivillaga · Accepted Answer · 2022-01-19 16:55:15Z

1

Learn to use numpy methods and broadcasting:

>>> x
array([[ 0,  5, 10]])
>>> x.T
array([[ 0],
       [ 5],
       [10]])
>>> x.T + x
array([[ 0,  5, 10],
       [ 5, 10, 15],
       [10, 15, 20]])
>>> result = x.T + x
>>> result
array([[ 0,  5, 10],
       [ 5, 10, 15],
       [10, 15, 20]])

Then this handy built-in:

>>> np.fill_diagonal(result, x**2)
>>> result
array([[  0,   5,  10],
       [  5,  25,  15],
       [ 10,  15, 100]])

Can replace the results[range(c), range(c)] = x**2

answered Jan 19, 2022 at 16:55

juanpa.arrivillaga

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Salvatore Daniele Bianco · Accepted Answer · 2022-01-19 17:00:03Z

1

Try this:

x.repeat(x.shape[1], axis=0)
x = x+x.T
x[np.arange(len(x)),np.arange(len(x))] = (np.diag(x)/2)**2

answered Jan 19, 2022 at 17:00

Salvatore Daniele Bianco

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