# Scikit-Learn: First Steps With log_loss

# Scikit-Learn: First Steps With log_loss

### This article covers Python SciKit-Learn machine learning with log_loss algorithm for solving hard kaggle machine learning problems.

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Over the last week, I’ve spent a little bit of time playing around with the data in the Kaggle TalkingData Mobile User Demographics competition and came across a notebook written by dune_dweller showing how to run a logistic regression algorithm on the dataset.

The metric used to evaluate the output in this competition is multi-class logarithmic loss, which is implemented by the log_loss function in the scikit-learn library.

I’ve not used it before so I created a small example to get to grips with it.

Let’s say we have 3 rows to predict and we happen to know that they should be labelled ‘bam’, ‘spam’, and ‘ham’ respectively:

`>>> actual_labels = ["bam", "ham", "spam"]`

To work out the log loss score, we need to make a prediction for what we think each label actually is. We do this by passing an array containing a probability between 0-1 for each label

E.g. if we think the first label is definitely ‘bam’ then we’d pass [1, 0, 0], whereas if we thought it had a 50-50 chance of being ‘bam’ or ‘spam’ then we might pass [0.5, 0, 0.5]. As far as I can tell the values get sorted into (alphabetical) order so we need to provide our predictions in the same order.

Let’s give it a try. First, we’ll import the function:

`>>> from sklearn.metrics import log_loss`

Now let’s see what score we get if we make a perfect prediction:

```
>>> log_loss(actual_labels, [[1, 0, 0], [0, 1, 0], [0, 0, 1]])
2.1094237467877998e-15
```

What about if we make a completely wrong prediction?

```
>>> log_loss(actual_labels, [[0, 0, 1], [1, 0, 0], [0, 1, 0]])
34.538776394910684
```

We can reverse engineer this score to work out the probability that we’ve predicted the correct class.

If we look at the case where the average log loss exceeds 1, it is when log(pij)

This is the formula of logloss:

In which y_{ij} is 1 for the correct class and 0 for other classes and p_{ij} is the probability assigned for that class.

The interesting thing about this formula is that we only care about the correct class. The y_{ij} value of 0 cancels out the wrong classes.

In our two examples so far we actually already know the probability estimate for the correct class – 100% in the first case and 0% in the second case, but we can plug in the numbers to check we end up with the same result.

First, we need to work out what value would have been passed to the log function which is easy in this case. The value of y_{ij} is:

```
# every prediction exactly right
>>> math.log(1)
0.0
>>> math.exp(0)
1.0
```

```
# every prediction completely wrong
>>> math.log(0.000000001)
-20.72326583694641
>>> math.exp(-20.72326583694641)
1.0000000000000007e-09
```

Let’s try another example where we have less certainty:

```
>>> print log_loss(actual_labels, [[0.8, 0.1, 0.1], [0.3, 0.6, 0.1], [0.15, 0.15, 0.7]])
0.363548039673
```

```
# 0.363548039673 = -1/3 * (log(0.8) + log(0.6) + log(0.7)
>>> print log_loss(actual_labels, [[0.8, 0.1, 0.1], [0.3, 0.6, 0.1], [0.15, 0.15, 0.7]])
0.363548039673
```

In this case, on average our probability estimate would be:

```
# we put in the negative value since we multiplied by -1/N
>>> math.exp(-0.363548039673)
0.6952053289772744
```

We had 60%, 70%, and 80% accuracy for our 3 labels so an overall probability of 69.5% seems about right.

One more example. This time, we’ll make one more very certain (90%) prediction for ‘spam’:

```
>>> print log_loss(["bam", "ham", "spam", "spam"], [[0.8, 0.1, 0.1], [0.3, 0.6, 0.1], [0.15, 0.15, 0.7], [0.05, 0.05, 0.9]])
0.299001158669
>>> math.exp(-0.299001158669)
0.741558550213609
```

74% accuracy overall, sounds about right!

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