Tuesday, January 29, 2019

Python string format quick guide

This is a quick summary guide for when you just want to remind yourself how to format a value with Python's 'string'.format().


Curly brackets are used to refer to parameter values in the 'format' method. On their own, the first pair of curly brackets will refer to the first parameter, second to the second, etc.

'{} {} {}'.format('a', 1, True)
a 1 True

Alternatively you can specify which parameter you want to use where using indexes.

'{0} {0} {2}'.format('a', 1, True)
a a True

You can even use names instead of indexes.

'{a} {b} {c}'.format(a='a', b=1, c=True)
a 1 True

And even refer to indexes in lists and to instance variables in objects.

'{a[0]} {b.numerator}'.format(a=[3,1,2], b=fractions.Fraction(1,2))
3 1


Adding a colon inside the curly brackets allows you to format the values before they are added to the string. You can combine these with any of the above methods of referencing. The following formatting meta characters are to be used in the following order:
{:[alignment] [grouping] [precision] [data type]}

Aligning/padding/leading zeros
An alignment meta character is either >, <, or ^. The number after it is the field width in which to align.

'|{:>3}|{:<3}|{:^3}|'.format('a', 'b', 'c')
|  a|b  | c |

Any character before the alignment meta character is used instead of spaces.
'{:_>5} {:0>5}'.format('a', 12)
____a 00012

You can group digits in numbers into triples with commas automatically by just putting a comma in the curly brackets.


When formatting floats you can specify how many digits to keep after the point by adding a dot and a number. By default the format of the number is in scientific notation which means that your floating point number will look like 12e-4 with the precision referring to the number of digits in the mantissa (the number before 'e'). See the next section to format your number in fixed point notation.


Data types
This is the most important part. The data type always comes at the end and is a single letter.

You can use it to change the numeric base of a whole number.
'dec-{:d} hex-{:X} oct-{:o} bin-{:b}'.format(16, 16, 16, 16)
dec-16 hex-10 oct-20 bin-10000

You can use it to convert unicode values to characters.

You can use it to convert numbers into scientific notations,

or fixed point notations,

or percentages.

Mixed examples





Thursday, December 27, 2018

Language models and the unknown token: Why perplexity gets worse as the corpus size increases

Unless you're making use of a character-based language model somehow, you're going to have a finite vocabulary of words that you can handle and so you need a way to handle out-of-vocabulary words. One common way to handle such words is by replacing them all with the unknown token, a pseudo word that replaces all out-of-vocabulary words. The unknown token has several advantages: Infrequent words in the training corpus will probably either not be learned by the neural network or not be used when generating sentences. Ignoring them will save a lot in model size as the embedding layer and softmax take a lot of parameters. Plus all the infrequent words put together will occur very frequently so the language model will definitely learn to use the unknown token. It also makes it possible to handle new words at test time (that are not found even once in the training corpus) as they will be replaced by the unknown token.

The problem with this approach is what happens when measuring the probability or perplexity of a sentence based on the probabilities of individual words. If you're comparing language models to see which ones make the best predictions, you usually use them all on a corpus to see how well they predict the words in the corpus. The higher the probabilities assigned to those sentences, the better the language model. This is usually measured using language model perplexity. But see what happens when you vary the vocabulary size. You will find that smaller vocabulary sizes lead to better language models, even though this makes no sense.

It turns out that if you just multiply all the probabilities of individual words as-is, including that of the unknown token, then your probability will be sensitive to the vocabulary size. Let's say that you have a vocabulary size of 0, that is, you're considering all the words to be out-of-vocabulary and hence all of them will be replaced by the unknown token. This means that during training, the language model will learn that after every unknown token there is (probably) another unknown token, unless its the end of the sentence. This will make the language model give very high probabilities for the unknown token. High word probabilities mean high sentence probabilities which mean good perplexities.

Now If we add another word to the vocabulary then we'll be introducing some uncertainty into the language model as now it has to decide between using the unknown token or the known word. Even in a perfect language model, the same prefix of words can be followed by either of the two words so there is no way to correctly assign 100% of the probability to one or the other. This means that the probabilities will be split between the two words, leading to an overall decrease in probabilities, leading to a worse perplexity. Adding more words to the vocabulary makes this even worse, which means that language models with smaller vocabularies have a huge unfair advantage over language models that actually do their job and correctly predict the right word.

We can't do away with the unknown token but we can strip away the unknown token's power. Assuming that all the language models are being evaluated on the same corpus, then different vocabularies will have different words being turned into the unknown token. Let's say that your language model considers 1000 different words in its vocabulary but the corpus you're evaluating it on has 500 different words that are out-of-vocabulary. So in reality your language model is predicting one of 1500 different words; it's just that 500 of those words are assumed to be a single word with a single probability. But really there should be 500 separate probabilities for those out-of-vocabulary words and not just one. If we avoid merging all those probabilities into one, then all the language models will have a fair comparison all they will all have the same vocabulary and they will all have the same amount of uncertainty about which word should come next. The question is how to distribute that single unknown token probability between the 500 out-of-vocabulary words. The simplest solution is to assume a uniform distribution and just give each word the same slice of probability from the whole. So if the unknown token has a probability of $p$, then each out-of-vocabulary word gets a probability of $\frac{p}{500}$.

Now every time you encounter the unknown token in the evaluation corpus you know that the token is being used in place of one of those 500 words. But you don't know which one it is. Not a problem, just divide the probability by 500 and it's as if all words in the corpus are in the vocabulary. Do this to every unknown token probability and now you have a fair measure of perplexity. Let's see an example.

Let's say that we want to find the probability of the following sentence:
the loud dog barked at the loud man

and let's say that the language model we're using to do that has the following vocabulary:
the at dog man

this means that the sentence is now changed to:
the UNK dog UNK at the UNK man

Now the naive way to get the probability of the sentence is as follows:

P(\text{the UNK dog UNK at the UNK man}) = \\
p(\text{the}|\text{}) \times p(\text{UNK}|\text{the}) \times p(\text{dog}|\text{the UNK}) \times p(\text{UNK}|\text{the UNK dog}) \\
\times p(\text{at}|\text{the UNK dog UNK}) \times p(\text{the}|\text{the UNK dog UNK at}) \times p(\text{UNK}|\text{the UNK dog UNK at the}) \\
\times p(\text{man}|\text{the UNK dog UNK at the UNK}) \times p(\text{}|\text{the UNK dog UNK at the UNK man})

But now with the new way we'll divide the unknown token's probabilities by 2, the number of different out of vocabulary words ('loud' and 'barked'):

P(\text{the UNK dog UNK at the UNK man}) = \\
p(\text{the}|\text{}) \times p(\text{UNK}|\text{the})/2 \times p(\text{dog}|\text{the UNK}) \times p(\text{UNK}|\text{the UNK dog})/2 \\
\times p(\text{at}|\text{the UNK dog UNK}) \times p(\text{the}|\text{the UNK dog UNK at}) \times p(\text{UNK}|\text{the UNK dog UNK at the})/2 \\
\times p(\text{man}|\text{the UNK dog UNK at the UNK}) \times p(\text{}|\text{the UNK dog UNK at the UNK man})

Of course we can leave the re-weighting till the end of the equation by dividing the first equation by the number of different out-of-vocabulary words for as many times as there are unknown tokens, like this:

P(\text{the UNK dog UNK at the UNK man}) = \\
p(\text{the}|\text{}) \times p(\text{UNK}|\text{the}) \times p(\text{dog}|\text{the UNK}) \times p(\text{UNK}|\text{the UNK dog}) \\
\times p(\text{at}|\text{the UNK dog UNK}) \times p(\text{the}|\text{the UNK dog UNK at}) \times p(\text{UNK}|\text{the UNK dog UNK at the}) \\
\times p(\text{man}|\text{the UNK dog UNK at the UNK}) \times p(\text{}|\text{the UNK dog UNK at the UNK man})/ 2^3

Now the sentence probability goes up as the vocabulary size increases!

Thursday, November 29, 2018

Explainable AI (XAI): Sensitivity Analysis

Imagine you have a neural network that can classify images well. At test time, would it be enough to just know what the class of the image is? Or would you also want to know why the image was classified the way it was? This is one of the goals of explainable AI and in this post we'll see what sensitivity analysis is.

Sensitivity analysis is a way to measure the importance of different parts of an input to one part of an output. In other words, you want to know which pixels were most important to give a high probability for a particular class. The way sensitivity analysis measures this is by measuring how sensitive the output is to each pixel, that is, which pixels, when changed, will change the output the most. Most pixels should leave the output unchanged when they themselves are changed, but some pixels would be critical to the particular output that the neural network has given. To measure this, we simply find the following:


that is, the sensitivity of output $i$ to input $j$ is the absolute value of the gradient of the output with respect to the input. In Tensorflow we can compute this as follows:

sensitivity = tf.abs(tf.gradients([outputs[0, output_index]], [images])[0][0])

where 'output_index' is a placeholder that is a scalar of type tf.int64, 'outputs' is a softmax with probabilities for each class and where the first index is the batch index and the second index is the class probability, and 'images' is the pixels of images where the first index is batch index and the second index is the image in vector form. This code also assumes that only the first image in the batch is to be analysed. This is because Tensorflow can only find the gradient of a single scalar so we can only find the sensitivity of a single output of a single image.

Here are some examples I got when I tried it on a simple fully connected two layer neural network trained to classify MNIST handwritten digits.

These heat maps show which pixels have the highest sensitivity score for the digit they were classified as. We can see how the empty space in the middle is important for classifying the zero. The four and the nine can be easy confused for each other were it not for the little bit in the top left corner. The seven can be confused for a nine if we complete the top left loop and the five can be confused with a six or eight if we draw a little diagonal line.

Wednesday, October 24, 2018

Mentally estimating square roots

What's $\sqrt{10}$? I'd reckon it's about $3 + \frac{10-9}{2 \times 3} = 3.166$. The actual answer is 3.162. What about $\sqrt{18}$? Should be about $4 + \frac{18-16}{2 \times 4} = 4.250$. The actual answer is 4.242. I found out about this calculation from this video but there was no explanation for why it works. Here's an explanation.

So the method here is as follows.
  1. Let the number you want to find the square root of be $a^2$.
  2. Let the largest square number which is less than $a^2$ be $b^2$ such that $b^2 <= a^2 < (b+1)^2$. For example if $a^2$ is 10 then $b^2$ is 9, if $a^2$ is 18 then $b^2$ is 16.
  3. The square root of $a^2$ is approximately $b + \frac{a^2 - b^2}{2 b}$.

This method is easy to carry out mentally but why does it work? The trick here is that the graph of the square root function grows so slowly that we can approximate the curve between two adjacent square numbers as a line.

We can use the line to approximate the square root of any number between two square numbers. The first thing we need to know is the gradient of the line. The vertical distance between two adjacent square numbers on the square root curve is 1, since the two square numbers are the squares of two consecutive numbers. The horizontal distance changes and becomes larger as the adjacent square numbers become larger but we can calculate it as follows:

$$(b+1)^2 - b^2 = b^2 + 2b + 1 - b^2 = 2b + 1$$

So the horizontal distance is twice the square root of the smaller square number plus one. Therefore the gradient of the line is $\frac{1}{2b+1}$. Once we know by how much the line grows vertically for every horizontal unit, we can then determine how much higher than $b$ the point on the line will be at $a$ by multiplying the gradient by $a^2-b^2$, as shown below:

Since the difference in height is less than 1, it is going to be the part of the square root that comes after the decimal point, with the whole number part being $b$.

It might be hard to mentally divide by an odd number in $\frac{a^2-b^2}{2b+1}$ so we further approximate it as $\frac{a^2-b^2}{2b}$ instead. And that's why this method works.

Saturday, September 29, 2018

Comparing numpy scalars directly is time consuming, use .tolist() before a comparison

This is something that I found out about recently when going through the elements of a numpy array in order to do some checks on each numbers. Turns out you shouldn't just do this

for x in nparr:
    if x == 0:
        something something

as that uses a lot more time than doing this

for x in nparr.tolist():
    if x == 0:
        something something

This is because a for loop iterating over a numpy array does not result in a sequence of Python constants but in a sequence of numpy scalars which would result in comparing a numpy array to a constant. Converting the array into a list first before the for loop will then result in a sequence of constants.

Here is some profiling I've done using cProfile to check different ways to do an 'if' on a numpy array element:

import cProfile
import numpy as np

runs = 1000000

print('Comparing numpy to numpy')
x = np.array(1.0, np.float32)
y = np.array(1.0, np.float32)
for _ in range(runs):
    if x == y:

print('Comparing numpy to constant')
x = np.array(1.0, np.float32)
for _ in range(runs):
    if x == 1.0:

print('Comparing constant to constant')
x = 1.0
for _ in range(runs):
    if x == 1.0:

print('Comparing numpy.tolist() to constant')
x = np.array(1.0, np.float32)
for _ in range(runs):
    if x.tolist() == 1.0:

print('Comparing numpy to numpy.array(constant)')
x = np.array(1.0, np.float32)
for _ in range(runs):
    if x == np.array(1.0, np.float32):

print('Comparing numpy.tolist() to numpy.tolist()')
x = np.array(1.0, np.float32)
y = np.array(1.0, np.float32)
for _ in range(runs):
    if x.tolist() == y.tolist():

Here are the results in order of speed:

Comparing constant to constant:0.088 seconds
Comparing numpy.tolist() to constant:0.288 seconds
Comparing numpy.tolist() to numpy.tolist():0.508 seconds
Comparing numpy to numpy:0.684 seconds
Comparing numpy to constant:1.192 seconds
Comparing numpy to numpy.array(constant):1.203 seconds

It turns out that it is always faster to first convert your numpy scalars into constants via .tolist() than to do anything with them as numpy scalars.

Thursday, August 30, 2018

The McLauren series and Taylor series (approximating complicated functions with simple polynomials)

The McLauren series

Imagine you had a function $f(x)$ that you knew was a polynomial, but whose details were unknown and you could only apply operations to the function without being able to read it. How could you find the coefficients of this polynomial? We know that for coefficients $a_i$:

$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + ...$

If we find $f(0)$ then we can find $a_0$.

$f(0) = a_0 + a_1 0 + a_2 0^2 + a_3 0^3 + a_4 0^4 + ... = a_0 + 0 + 0 + ... = a_0$

That was easy, but how can we find $a_1$? We need an operation that gets rid of $a_0$ and also the $x$ in the term $a_1 x$. That operation turns out to be differentiation with respect to $x$:

$f'(x) = a_1 + 2 a_2 x + 3 a_3 x^2 + 4 a_4 x^3 + ...$

Great! Now we can find $a_1$ by replacing $x$ with 0:

$f'(0) = a_1 + 2 a_2 0 + 3 a_3 0^2 + 4 a_4 0^3 + ... = a_1$

We can find $a_2$ by repeating these two steps:

$f''(x) = 2 a_2 + 2 \cdot 3 a_3 x + 3 \cdot 4 a_4 x^2 + ...$
$f''(0) = 2 a_2 + 2 \cdot 3 a_3 0 + 3 \cdot 4 a_4 0^2 + ... = 2 a_2$

What we found is twice of $a_2$ which means that we need to divide by 2 to get $a_2$. The differentiation operation is multiplying constants by each coefficient and the constants get bigger and bigger the more times we apply differentiation. You might notice that what's happening is that $a_0$ was multiplied by 1, $a_1$ was also multiplied by 1, $a_2$ has been multiplied by 2, $a_3$ will be multiplied by 6, $a_4$ by 24, and so on. These are factorials, sequences of whole numbers multiplied together ($3! = 1 \times 2 \times 3 = 6$). Which means that we need to divide by the next factorial after each round of differentiation and substitution by zero.

In general we can find the $i$th coefficient in an unknown polynomial function by doing the following:

$a_i = \frac{f^i(0)}{i!}$

That's great. Now to test it. Let's see if a complex function is actually a polynomial in disguise. Take something simple such as $f(x) = e^x$. This doesn't look like a polynomial, but it may be represented by a polynomial with an infinite number of terms. Let's find what are the coefficients of the hidden polynomial in $e^x$.

$f(x) = e^x = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + ...$
$\frac{f(0)}{0!} = \frac{e^0}{1} = a_0$
$\frac{f(0)}{0!} = 1$

OK, so $a_0$ is 1. Let's find the rest of the coefficients.

$a_1 = \frac{f'(0)}{1!} = \frac{e^0}{1} = 1$
$a_2 = \frac{f''(0)}{2!} = \frac{e^0}{2} = \frac{1}{2}$
$a_3 = \frac{f'''(0)}{3!} = \frac{e^0}{6} = \frac{1}{6}$

So the first few terms of the polynomial hidden within $e^x$ are:
$f(x) = 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3 + ...$

Does this partial polynomial look anything like $e^x$ when plotted as a graph?

Pretty good within a boundary! Note how the curve has a perfect fit at $x = 0$ and that it gets worse as we move away from there. Adding more terms to the polynomial will enlarge the area around $x = 0$ that is close to the curve but it will always be radiating out from there.

Let's try for $f(x) = cos(x)$ now.

$a_0 = \frac{f(0)}{0!} = \frac{cos(0)}{1} = 1$
$a_1 = \frac{f'(0)}{1!} = \frac{-sin(0)}{1} = 0$
$a_2 = \frac{f''(0)}{2!} = \frac{-cos(0)}{2} = -\frac{1}{2}$
$a_3 = \frac{f'''(0)}{3!} = \frac{sin(0)}{6} = 0$

So the first few terms of the polynomial hidden within $cos(x)$ is
$f(x) = 1 - \frac{1}{2} x^2 + ...$

The Taylor series

As you can see, this "polynomialisation" of functions is a neat way to approximate a function we might not know how to implement exactly but know how to differentiate and how to find its value when $x$ is 0. But what if we don't know what $f(0)$ is such as in $ln(x)$ or $\frac{1}{x}$? A slight modification to our method allows us to use any value of $x$ and not just 0. Let's call this value $b$. By slightly modifying the polynomial we expect to be hiding inside the function, we can make the polynomial act the same way when $f(b)$ is used instead of $f(0)$:

$f(x) = a_0 + a_1 (x - b) + a_2 (x - b)^2 + a_3 (x - b)^3 + a_4 (x - b)^4 + ...$
$f(b) = a_0 + a_1 (b - b) + a_2 (b - b)^2 + a_3 (b - b)^3 + a_4 (b - b)^4 + ...$
$f(b) = a_0$

$f'(x) = a_1 + 2 a_2 (x - b) + 3 a_3 (x - b)^2 + 4 a_4 (x - b)^3 + ...$
$f'(b) = a_1$

$a_i = \frac{f^i(b)}{i!}$

The catch here is that we are now finding coefficients to the polynomial $a_0 + a_1 (x - b) + a_2 (x - b)^2 + ...$ and not of $a_0 + a_1 x + a_2 x^2 + ...$, but that's OK. Let's try this on $ln(x)$ with $b = 1$.

$a_0 = \frac{f(1)}{0!} = \frac{ln(1)}{1} = 0$
$a_1 = \frac{f'(1)}{1!} = \frac{\frac{1}{1}}{1} = 1$
$a_2 = \frac{f''(1)}{2!} = \frac{-\frac{1}{1^2}}{1} = -1$
$a_3 = \frac{f'''(1)}{3!} = \frac{\frac{2}{1^3}}{1} = 2$

So the first few terms of the polynomial hidden within $ln(x)$ is
$f(x) = (x - 1) - (x - 1)^2 + 2 (x - 1)^3 + ...$

Adding more terms will approximate the original function better and better but what if we didn't have to? Remember how I said in the previous section that the polynomial approximates the original function best close to $x = 0$. Well now we can approximate it best around any point $b$ and not just around 0. This means that if our function has multiple known values, such as $cos(x)$ which is known to be 1 at $x = 0$, 0 at $x = \frac{\pi}{2}$, -1 at $x = \pi$, etc., then we can use several short polynomials centered at different points in the function instead of one large polynomial that approximates it well over a large interval. Let's try approximating $cos(x)$ around $x = \pi$, which means that we'll set $b$ to $\pi$.

$a_0 = \frac{f(\pi)}{0!} = \frac{cos(\pi)}{1} = -1$
$a_1 = \frac{f'(\pi)}{1!} = \frac{-sin(\pi)}{1} = 0$
$a_2 = \frac{f''(\pi)}{2!} = \frac{-cos(\pi)}{2} = \frac{1}{2}$
$a_3 = \frac{f'''(\pi)}{3!} = \frac{sin(\pi)}{6} = 0$

So the first few terms of the polynomial hidden within $cos(x)$ which best approximates the area around $x = \pi$ is
$f(x) = -1 + \frac{1}{2} (x - \pi)^2 + ...$

This is useful when implementing mathematical functions on a computer. You keep several simple polynomials defined at different points in the domain of the function and then pick the closest one to the $x$ you need to evaluate. You can then compute an approximation that isn't too bad without requiring a lot of computational time.

Sunday, July 29, 2018

Hyperparameter tuning using Scikit-Optimize

One of my favourite academic discoveries this year was Scikit-Optimize, a nifty little Python automatic hyperparameter tuning library that comes with a lot of features I found missing in other similar libraries.

So as explained in an earlier blog post, automatic hyperparameter tuning is about finding the right hyperparameters for a machine learning algorithm automatically. Usually this is done manually using human experience but even simple MonteCarlo search random guesses can result in better performance than human tweaking (see here). So automatic methods were developed that try to explore the space of hyperparameters and their resulting performance after training the machine learning model and then try to home in on the best performing hyperparameters. Of course each time you want to evaluate a new hyperparameter combination you'd need to retrain and evaluate your machine learning model, which might take a very long time to finish, so it's important that a good hyperparameter combination is found with as little evaluations as possible. To do this we'll use Bayesian Optimization, a process where a separate simpler model is trained to predict the resulting performance of the whole hyperparameter space. We check this trained model to predict which hyperparameters will give the best resulting performance and actually evaluate them by training our machine learning model with them. The actual resulting performance is then used to update the hyperparameter space model so that it makes better predictions and then we'll get a new promising hyperparameter combination from it. This is repeated for a set number of times. Now the most common hyperparameter space model to use is a Gaussian Process which maps continuous numerical hyperparameters to a single number which is the predicted performance. This is not very good when your hyperparameters contain categorical data such as a choice of activation function. There is a paper that suggests that random forests are much better at mapping general hyperparameter combinations to predicted performance.

Now that we got the theory out of the way, let's see how to use the library. We'll apply it on a gradient descent algorithm that needs to minimize the squared function. For this we'll need 3 hyperparameters: the range of the initial value to be selected randomly, the learning rate, and the number of epochs to run. So we have two continuous values and one discrete integer value.

import random

def cost(x):
    return x**2

def cost_grad(x):
    return 2*x

def graddesc(learning_rate, max_init_x, num_epochs):
    x = random.uniform(-max_init_x, max_init_x)
    for _ in range(num_epochs):
        x = x - learning_rate*cost_grad(x)
    return x

Now we need to define the skopt optimizer:

import skopt

opt = skopt.Optimizer(
                skopt.space.Real(0.0, 10.0, name='max_init_x'),
                skopt.space.Real(1.0, 1e-10, 'log-uniform', name='learning_rate'),
                skopt.space.Integer(1, 20, name='num_epochs'),

The above code is specifying 3 hyperparameters:
  • the maximum initial value that is a real number (continuous) and that can be between 10 and 0
  • the learning rate that is also a real number but that is also on a logarithmic scale (so that you are equally likely to try very large values and very small values) and can be between 1 and 1e-10
  • the number of epochs that is an integer (whole number) and that can be between 1 and 20
It is also saying that the hyperparameter space model should be initialized based on 5 random hyperparameter combinations (you train the hyperparameter space model on 5 randomly chosen hyperparameters in order to be able to get the first suggested hyperparameter), that this model should be a random forest (RF), that the acquisition function (the function to decide which hyperparameter combination is the most promising to try next according to the hyperparameter space model) is the expected improvement (EI) of the hyperparameter combination, that the acquisition optimizer (the optimizer to find the next promising hyperparameter combination) is automatically chosen, and that the random state is set to a fixed number (zero) so that it always gives the same random values each time you run it.

Next we will use the optimizer to find good hyperparameter combinations.

best_cost = 1e100
best_hyperparams = None
for trial in range(5 + 20):
    [max_init_x, learning_rate, num_epochs] = opt.ask()
    [max_init_x, learning_rate, num_epochs] = [max_init_x.tolist(), learning_rate.tolist(), num_epochs.tolist()]
    next_hyperparams = [max_init_x, learning_rate, num_epochs]
    next_cost = cost(graddesc(max_init_x, learning_rate, num_epochs))
    if next_cost < best_cost:
        best_cost = next_cost
        best_hyperparams = next_hyperparams
    print(trial+1, next_cost, next_hyperparams)
    opt.tell(next_hyperparams, next_cost)
The nice thing about this library is that you can use an 'ask/tell' system where you ask the library to give you the next hyperparameters to try and then you do something with them in order to get the actual performance value and finally you tell the library what this performance value is. This lets you do some nifty things such as ask for another value if the hyperparameters resulted in an invalid state in the machine learning model or even to save your progress and continue later.

In the above code we're running a for loop to run the number of times we want to evaluate different hyperparameters. We need to run it for the 5 random values we specified before to initialize the hyperparameter space model plus another 20 evaluations to actually optimize the hyperameters. Now skopt does something funny which is that it returns not plain Python values for hyperparameters but rather each number is represented as a numpy scalar. Because of this we convert each numpy scalar back into a plain Python float or int using ".tolist()". We ask for the next hyperparamters to try, convert them to plain Python values, get their resulting cost after running gradient descent, store the best hyperparameters found up to now, and tell the library what the given hyperparameters' resulting performance was. At the end we print the best hyperparamter combination found.

Some extra stuff:
  • You can ask for categorical hyperparameters using "skopt.space.Categorical(['option1', 'option2'], name='options')" which will return one of the values in the list when calling "ask".
  • You can ask for a different hyperparameter combination in case of an invalid one by changing "ask" to give you several hyperparameter suggestions rather than just one and then trying each one of them until one works by using "opt.ask(num_hyperpars)" (you can also incrementally ask for more values and always take the last one).
  • You can save and continue by saving all the hyperparameters evaluated and their corresponding performance value in a text file. You then later resupply the saved hyperparameters and their performance using "tell" for each one. This is much faster than actually evaluating them on the machine learning model so straight supplying known values will be ready quickly. Just be careful that you also call "ask" before each "tell" in order to get the same exact behaviour from the optimizer or else the next "tell" will give different values from what it would have given had you not loaded the previous ones manually.