Python – RuntimeWarning: invalid value encountered in greater

numpypythonsoftmax

I tried to implement soft-max with the following code (out_vec is a numpy vector of floats):

numerator = np.exp(out_vec)
denominator = np.sum(np.exp(out_vec))
out_vec = numerator/denominator

However, I got an overflow error because of np.exp(out_vec). Therefore, I checked (manually) what the upper limit of np.exp() is, and found that np.exp(709) is a number, but np.exp(710) is considered to be np.inf. Thus, to try to avoid the overflow error, I modified my code as follows:

out_vec[out_vec > 709] = 709 #prevent np.exp overflow
numerator = np.exp(out_vec)
denominator = np.sum(np.exp(out_vec))
out_vec = numerator/denominator

Now, I get a different error:

RuntimeWarning: invalid value encountered in greater out_vec[out_vec > 709] = 709

What's wrong with the line I added? I looked up this specific error and all I found is people's advice on how to ignore the error. Simply ignoring the error won't help me, because every time my code encounters this error it does not give the usual results.

Best Answer

Your problem is caused by the NaN or Inf elements in your out_vec array. You could use the following code to avoid this problem:

if np.isnan(np.sum(out_vec)):
    out_vec = out_vec[~numpy.isnan(out_vec)] # just remove nan elements from vector
out_vec[out_vec > 709] = 709
...

or you could use the following code to leave the NaN values in your array:

out_vec[ np.array([e > 709 if ~np.isnan(e) else False for e in out_vec], dtype=bool) ] = 709

Python 3.7+ or CPython 3.6

Dicts preserve insertion order in Python 3.7+. Same in CPython 3.6, but it's an implementation detail.

>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}

>>> dict(sorted(x.items(), key=lambda item: item[1]))
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}

Older Python

It is not possible to sort a dictionary, only to get a representation of a dictionary that is sorted. Dictionaries are inherently orderless, but other types, such as lists and tuples, are not. So you need an ordered data type to represent sorted values, which will be a list—probably a list of tuples.

For instance,

import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(1))

sorted_x will be a list of tuples sorted by the second element in each tuple. dict(sorted_x) == x.

And for those wishing to sort on keys instead of values:

import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(0))

In Python3 since unpacking is not allowed we can use

x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=lambda kv: kv[1])

If you want the output as a dict, you can use collections.OrderedDict:

import collections

sorted_dict = collections.OrderedDict(sorted_x)

Best Answer

Related Solutions

Python – How to sort a list of dictionaries by a value of the dictionary

Python – How to sort a dictionary by value

Python 3.7+ or CPython 3.6

Older Python

Related Topic