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`norm_fn=np.abs` would compute L1 norm as: `np.sqrt(np.sum(np.abs(x)))`, which is incorrect; the sqrt is redundant. `norm_fn=np.abs` will now work correctly. L2-norm case always worked correctly. For L2-norm, set `norm_fn = (np.sqrt, np.square) = (outer_fn, inner_fn)`, which will compute `outer_fn(sum(inner_fn(x)))`. Note that `norm_fn=np.square` will **no longer compute L2-norm correctly**. Pardon the mishap.
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from .inspect_gen import * | ||
from .inspect_rnn import * | ||
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__version__ = '1.14.3' | ||
__version__ = '1.14.4' |
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