min_value = np.iinfo(im.dtype).min
max_value = np.iinfo(im.dtype).max

docs:

• np.iinfo (machine limits for integer types)
• np.finfo (machine limits for floating point types)

Quoting from a numpy discussion list:

That information is available via numpy.finfo() and numpy.iinfo():

In [12]: finfo('d').max
Out[12]: 1.7976931348623157e+308

In [13]: iinfo('i').max
Out[13]: 2147483647

In [14]: iinfo('uint8').max
Out[14]: 255

Link here.

You can use a decorator:

import sys
def noInf(f):
  def wrapped(*args, **kwargs):
    res = f(*args, **kwargs)
    if res == float('inf'):
      return sys.float_info.max
    return res
  return wrapped

@noInf
def myMult(x, y):
  return x*y

print(myMult(sys.float_info.max, 2)) # prints 1.79769313486e+308

Upd: it was originally wrong, now this is just the essence of the the previous correct answer:

a = np.array([[0.0543275 , 0.51249827, 0.43317423],
              [0.07144389, 0.51152126, 0.41703486], 
              [0.0776112 , 0.48593384, 0.43645496]])

b = np.zeros_like(a)

b[np.arange(a.shape[0]), np.argmax(a, axis=1)] = 1

Since np.argmax() gives us indices of the max elements, we just use them for indexing directly. Now b contains desired output:

array([[0., 1., 0.],
       [0., 1., 0.],
       [0., 1., 0.]])

you can also do: b.astype(int) to turn to integers.

Can you rework the calculation so it works with the logarithms of the numbers instead?

That's pretty much how the built-in floats work in any case...

You would only convert the number back to linear for display, at which point you'd separate the integer and fractional parts; the fractional part gets exponentiated as normal to give the 8 digits of precision, and the integer part goes into the "×10ⁿ" or "×eⁿ" or "×2ⁿ" part of the output (depending on what base logarithm you use).