Mask from max values in numpy array, specific axis
Create an identity matrix and select from its rows using argmax
on your array:
np.identity(a.shape[1], bool)[a.argmax(axis=1)]
# array([[False, True],
# [ True, False],
# [False, True]], dtype=bool)
Please note that this ignores ties, it just goes with the value returned by argmax
.
Method #1
Using broadcasting
, we can use comparison against the max values, while keeping dims to facilitate broadcasting

a.max(axis=1,keepdims=1) == a
Sample run 
In [83]: a
Out[83]:
array([[0, 1],
[2, 1],
[4, 8]])
In [84]: a.max(axis=1,keepdims=1) == a
Out[84]:
array([[False, True],
[ True, False],
[False, True]], dtype=bool)
Method #2
Alternatively with argmax
indices for one more case of broadcastedcomparison
against the range of indices along the columns 
In [92]: a.argmax(axis=1)[:,None] == range(a.shape[1])
Out[92]:
array([[False, True],
[ True, False],
[False, True]], dtype=bool)
Method #3
To finish off the set, and if we are looking for performance, use intialization and then advancedindexing

out = np.zeros(a.shape, dtype=bool)
out[np.arange(len(a)), a.argmax(axis=1)] = 1
You're already halfway in the answer. Once you compute the max along an axis, you can compare it with the input array and you'll have the required binary mask!
In [7]: maxx = np.amax(a, axis=1)
In [8]: maxx
Out[8]: array([1, 2, 8])
In [12]: a >= maxx[:, None]
Out[12]:
array([[False, True],
[ True, False],
[False, True]], dtype=bool)
Note: This uses NumPy broadcasting when doing the comparison between a
and maxx