# Statistics for Hackers

## Motivation¶

There's no shortage of absolutely magnificent material out there on the topics of data science and machine learning for an autodidact, such as myself, to learn from. In fact, so many great resources exist that an individual can be forgiven for not knowing where to begin their studies, or for getting distracted once they're off the starting block. I honestly can't count the number of times that I've started working through many of these online courses and tutorials only to have my attention stolen by one of the multitudes of amazing articles on data analysis with Python, or some great new MOOC on Deep Learning. But this year is different! This year, for one of my new year's resolutions, I've decided to create a personalized data science curriculum and stick to it. This year, I promise not to just casually sign up for another course, or start reading yet another textbook to be distracted part way through. This year, I'm sticking to the plan.

As part of my personalized program of study, I've chosen to start with Harvard's Data Science course. I'm currently on week 3 and one of the suggested readings for this week is Jake VanderPlas' talk from PyCon 2016 titled "Statistics for Hackers". As I was watching the video and following along with the slides, I wanted to try out some of the examples and create a set of notes that I could refer to later, so I figured why not create a Jupyter notebook. Once I'd finished, I realized I'd created a decently-sized resource that could be of use to others working their way through the talk. The result is the article you're reading right now, the remainder of which contains my notes and code examples for Jake's excellent talk.

So, enjoy the article, I hope you find this resource useful, and if you have any problems or suggestions of any kind, the full notebook can be found on github, so please send me a pull request, or submit an issue, or just message me directly on Twitter.

## Preliminaries¶

```
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
import seaborn as sns
# Suppress all warnings just to keep the notebook nice and clean.
# This must happen after all imports since numpy actually adds its
# RankWarning class back in.
import warnings
warnings.filterwarnings("ignore")
# Setup the look and feel of the notebook
sns.set_context("notebook",
font_scale=1.5,
rc={"lines.linewidth": 2.5})
sns.set_style('whitegrid')
sns.set_palette('deep')
# Create a couple of colors to use throughout the notebook
red = sns.xkcd_rgb['vermillion']
blue = sns.xkcd_rgb['dark sky blue']
from IPython.display import display
%matplotlib inline
%config InlineBackend.figure_format = 'retina'
```

## Warm-up¶

The talk starts off with a motivating example that asks the question "If you toss a coin 30 times and see 22 heads, is it a fair coin?"

We all know that a fair coin should come up heads roughly 15 out of 30 tosses, give or take, so it does seem unlikely to see so many heads. However, the skeptic might argue that even a fair coin could show 22 heads in 30 tosses from time-to-time. This could just be a chance event. So, the question would then be "how can you determine if you're tossing a fair coin?"

### The Classic Method¶

The classic method would assume that the skeptic is correct and would then test the hypothesis (i.e., the *Null Hypothesis*) that the observation of 22 heads in 30 tosses could happen simply by chance. Let's start by first considering the probability of a single coin flip coming up heads and work our way up to 22 out of 30.

$$ P(H) = \frac{1}{2} $$

As our equation shows, the probability of a single coin toss turning up heads is exactly 50% since there is an equal chance of either heads or tails turning up. Taking this one step further, to determine the probability of getting 2 heads in a row with 2 coin tosses, we would need to multiply the probability of getting heads by the probability of getting heads again since the two events are independent of one another.

$$ P(HH) = P(H) \cdot P(H) = P(H)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} $$

From the equation above, we can see that the probability of getting 2 heads in a row from a total of 2 coin tosses is 25%. Let's now take a look at a slightly different scenario and calculate the probability of getting 2 heads and 1 tails with 3 coin tosses.

$$ P(HHT) = P(H)^2 \cdot P(T) = \left(\frac{1}{2}\right)^2 \cdot \frac{1}{2} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} $$

The equation above tells us that the probability of getting 2 heads and 1 tails in 3 tosses is 12.5%. This is actually the exact same probability as getting heads in all three tosses, which doesn't sound quite right. The problem is that we've only calculated the probability for a single permutation of 2 heads and 1 tails; specifically for the scenario where we only see tails on the third toss. To get the actual probability of tossing 2 heads and 1 tails we will have to add the probabilities for all of the possible permutations, of which there are exactly three: HHT, HTH, and THH.

$$ P(2H,1T) = P(HHT) + P(HTH) + P(THH) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8} $$

Another way we could do this is to calculate the total number of permutations and simply multiply that by the probability of each event happening. To get the total number of permutations we can use the binomial coefficient. Then, we can simply calculate the probability above using the following equation.

$$ P(2H,1T) = \binom{3}{2} \left(\frac{1}{2}\right)^{3} = 3 \left(\frac{1}{8}\right) = \frac{3}{8} $$

While the equation above works in our particular case, where each event has an equal probability of happening, it will run into trouble with events that have an unequal chance of taking place. To deal with those situations, you'll want to extend the last equation to take into account the differing probabilities. The result would be the following equation, where $N$ is number of coin flips, $N_H$ is the number of expected heads, $N_T$ is the number of expected tails, and $P_H$ is the probability of getting heads on each flip.

$$ P(N_H,N_T) = \binom{N}{N_H} \left(P_H\right)^{N_H} \left(1 - P_H\right)^{N_T} $$

Now that we understand the classic method, let's use it to test our null hypothesis that we *are* actually tossing a fair coin, and that this is just a chance occurrence. The following code implements the equations we've just discussed above.

```
def factorial(n):
"""Calculates the factorial of `n`
"""
vals = list(range(1, n + 1))
if len(vals) <= 0:
return 1
prod = 1
for val in vals:
prod *= val
return prod
def n_choose_k(n, k):
"""Calculates the binomial coefficient
"""
return factorial(n) / (factorial(k) * factorial(n - k))
def binom_prob(n, k, p):
"""Returns the probability of see `k` heads in `n` coin tosses
Arguments:
n - number of trials
k - number of trials in which an event took place
p - probability of an event happening
"""
return n_choose_k(n, k) * p**k * (1 - p)**(n - k)
```

Now that we have a method that will calculate the probability for a specific event happening (e.g., 22 heads in 30 coin tosses), we can calculate the probability for every possible outcome of flipping a coin 30 times, and if we plot these values we'll get a visual representation of our coin's probability distribution.

```
# Calculate the probability for every possible outcome of tossing
# a fair coin 30 times.
probabilities = [binom_prob(30, k, 0.5) for k in range(1, 31)]
# Plot the probability distribution using the probabilities list
# we created above.
plt.step(range(1, 31), probabilities, where='mid', color=blue)
plt.xlabel('number of heads')
plt.ylabel('probability')
plt.plot((22, 22), (0, 0.1599), color=red);
plt.annotate('0.8%',
xytext=(25, 0.08),
xy=(22, 0.08),
multialignment='right',
va='center',
color=red,
size='large',
arrowprops={'arrowstyle': '<|-',
'lw': 2,
'color': red,
'shrinkA': 10});
```

The visualization above shows the probability distribution for flipping a fair coin 30 times. Using this visualization we can now determine the probability of getting, say for example, 12 heads in 30 flips, which looks to be about 8%. Notice that we've labeled our example of 22 heads as 0.8%. If we look at the probability of flipping exactly 22 heads, it looks likes to be a little less than 0.8%, in fact if we calculate it using the `binom_prob`

function from above, we get 0.5%

```
print("Probability of flipping 22 heads: %0.1f%%" % (binom_prob(30, 22, 0.5) * 100))
```

So, then why do we have 0.8% labeled in our probability distribution above? Well, that's because we are showing the probability of getting **at least** 22 heads, which is also known as the p-value.

#### What's a p-value?¶

In statistical hypothesis testing we have an idea that we want to test, but considering that it's very hard to prove something to be true beyond doubt, rather than test our hypothesis directly, we formulate a competing hypothesis, called a null hypothesis, and then try to disprove it instead. The null hypothesis essentially assumes that the effect we're seeing in the data could just be due to chance.

In our example, the null hypothesis assumes we have a fair coin, and the way we determine if this hypothesis is true or not is by calculating how often flipping this fair coin 30 times would result in 22 or more heads. If we then take the number of times that we got 22 or more heads and divide that number by the total of all possible permutations of 30 coin tosses, we get the probability of tossing 22 or more heads with a fair coin. This probability is what we call the p-value.

The p-value is used to check the validity of the null hypothesis. The way this is done is by agreeing upon some predetermined upper limit for our p-value, below which we will assume that our null hypothesis is false. In other words, if our null hypothesis were true, and 22 heads in 30 flips could happen often enough by chance, we would expect to see it happen more often than the given threshold percentage of times. So, for example, if we chose 10% as our threshold, then we would expect to see 22 or more heads show up at least 10% of the time to determine that this is a chance occurrence and not due to some bias in the coin. Historically, the generally accepted threshold has been 5%, and so if our p-value is less than 5%, we can then make the assumption that our coin may not be fair.

The `binom_prob`

function from above calculates the probability of a single event happening, so now all we need for calculating our p-value is a function that adds up the probabilities of a given event, or a more extreme event happening. So, as an example, we would need a function to add up the probabilities of getting 22 heads, 23 heads, 24 heads, and so on. The next bit of code creates that function and uses it to calculate our p-value.

```
def p_value(n, k, p):
"""Returns the p-value for the given the given set
"""
return sum(binom_prob(n, i, p) for i in range(k, n+1))
print("P-value: %0.1f%%" % (p_value(30, 22, 0.5) * 100))
```

Running the code above gives us a p-value of roughly 0.8%, which matches the value in our probability distribution above and is also less than the 5% threshold needed to reject our null hypothesis, so it does look like we may have a biased coin.

### The Easier Method¶

That's an example of using the classic method for testing if our coin is fair or not. However, if you don't happen to have at least some background in statistics, it can be a little hard to follow at times, but luckily for us, there's an easier method...

Simulation!

The code below seeks to answer the same question of whether or not our coin is fair by running a large number of simulated coin flips and calculating the proportion of these experiments that resulted in at least 22 heads or more.

```
M = 0
n = 50000
for i in range(n):
trials = np.random.randint(2, size=30)
if (trials.sum() >= 22):
M += 1
p = M / n
print("Simulated P-value: %0.1f%%" % (p * 100))
```

The result of our simulations is 0.8%, the exact same result we got earlier when we calculated the p-value using the classical method above. So, it definitely looks like it's possible that we have a biased coin since the chances of seeing 22 or more heads in 30 tosses of a fair coin is less than 1%.

## Four Recipes for Hacking Statistics¶

We've just seen one example of how our hacking skills can make it easy for us to answer questions that typically only a statistician would be able to answer using the classical methods of statistical analysis. This is just one possible method for answering statistical questions using our coding skills, but Jake's talk describes four recipes in total for "hacking statistics", each of which is listed below. The rest of this article will go into each of the remaining techniques in some detail.

In the Warm-up section above, we saw an example direct simulation, the first recipe in our tour of statistical hacks. The next example uses the Shuffling method to figure out if there's a statistically significant difference between two different sample populations.

### Shuffling¶

In this example, we look at the Dr. Seuss story about the Star-belly Sneetches. In this Seussian world, a group of creatures called the Sneetches are divided into two groups: those with stars on their bellies, and those with no "stars upon thars". Over time, the star-bellied sneetches have come to think of themselves as better than the plain-bellied sneetches. As researchers of sneetches, it's our job to uncover whether or not star-bellied sneetches really are better than their plain-bellied cousins.

The first step in answering this question will be to create our experimental data. In the following code snippet we create a dataframe object that contains a set of test scores for both star-bellied and plain-bellied sneetches.

```
import pandas as pd
df = pd.DataFrame({'star': [1, 1, 1, 1, 1, 1, 1, 1] +
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
'score': [84, 72, 57, 46, 63, 76, 99, 91] +
[81, 69, 74, 61, 56, 87, 69, 65, 66, 44, 62, 69]})
df
```

If we then take a look at the average scores for each group of sneetches, we will see that there's a difference in scores of **6.6** between the two groups. So, on average, the star-bellied sneetches performed better on their tests than the plain-bellied sneetches. But, the real question is, is this a significant difference?

```
star_bellied_mean = df[df.star == 1].score.mean()
plain_bellied_mean = df[df.star == 0].score.mean()
print("Star-bellied Sneetches Mean: %2.1f" % star_bellied_mean)
print("Plain-bellied Sneetches Mean: %2.1f" % plain_bellied_mean)
print("Difference: %2.1f" % (star_bellied_mean - plain_bellied_mean))
```

To determine if this is a signficant difference, we could perform a t-test on our data to compute a p-value, and then just make sure that the p-value is less than the target 0.05. Alternatively, we could use simulation instead.

Unlike our first example, however, we don't have a generative function that we can use to create our probability distribution. So, how can we then use simulation to solve our problem?

Well, we can run a bunch of simulations where we randomly shuffle the labels (i.e., star-bellied or plain-bellied) of each sneetch, recompute the difference between the means, and then determine if the proportion of simulations in which the difference was at least as extreme as 6.6 was less than the target 5%. If so, we can conclude that the difference we see is, in fact, one that doesn't occur strictly by chance very often and so the difference is a significant one. In other words, if the proportion of simulations that have a difference of 6.6 or greater is less than 5%, we can conclude that the labels really do matter, and so we can conclude that star-bellied sneetches are "better" than their plain-bellied counterparts.

```
df['label'] = df['star']
num_simulations = 10000
differences = []
for i in range(num_simulations):
np.random.shuffle(df['label'])
star_bellied_mean = df[df.label == 1].score.mean()
plain_bellied_mean = df[df.label == 0].score.mean()
differences.append(star_bellied_mean - plain_bellied_mean)
```

Now that we've ran our simulations, we can calculate our p-value, which is simply the proportion of simulations that resulted in a difference greater than or equal to 6.6.

$$ p = \frac{N_{>6.6}}{N_{total}} = \frac{1512}{10000} = 0.15 $$

```
p_value = sum(diff >= 6.6 for diff in differences) / num_simulations
print("p-value: %2.2f" % p_value)
```

The following code plots the distribution of the differences we found by running the simulations above. We've also added an annotation that marks where the difference of 6.6 falls in the distribution along with its corresponding p-value.

```
plt.hist(differences, bins=50, color=blue)
plt.xlabel('score difference')
plt.ylabel('number')
plt.plot((6.6, 6.6), (0, 700), color=red);
plt.annotate('%2.f%%' % (p_value * 100),
xytext=(15, 350),
xy=(6.6, 350),
multialignment='right',
va='center',
color=red,
size='large',
arrowprops={'arrowstyle': '<|-',
'lw': 2,
'color': red,
'shrinkA': 10});
```

We can see from the histogram above---and from our simulated p-value, which was greater than 5%---that the difference that we are seeing between the populations can be explained by random chance, so we can effectively dismiss the difference as not statistically significant. In short, star-bellied sneetches are no better than the plain-bellied ones, at least not from a statistical point of view.

For further discussion on this method of simulation, check out John Rauser's keynote talk "Statistics Without the Agonizing Pain" from Strata + Hadoop 2014. Jake mentions that he drew inspiration from it in his talk, and it is a really excellent talk as well; I wholeheartedly recommend it.

### Bootstrapping¶

In this example, we'll be using the story of Yertle the Turtle to explore the bootstrapping recipe. As the story goes, in the land of Sala-ma-Sond, Yertle the Turtle was the king of the pond and he wanted to be the most powerful, highest turtle in the land. To achieve this goal, he would stack turtles as high as he could in order to stand upon their backs. As observers of this curious behavior, we've recorded the heights of 20 turtle towers and we've placed them in a dataframe in the following bit of code.

```
df = pd.DataFrame({'heights': [48, 24, 51, 12, 21,
41, 25, 23, 32, 61,
19, 24, 29, 21, 23,
13, 32, 18, 42, 18]})
```

The questions we want to answer in this example are: what is the mean height of Yertle's turtle stacks, and what is the uncertainty of this estimate?

#### The Classic Method¶

The classic method is simply to calculate the sample mean...

$$ \bar{x} = \frac{1}{N} \sum_{i=1}^{N} x_i = 28.9 $$

...and the standard error of the mean.

$$ \sigma_{\bar{x}} = \frac{1}{ \sqrt{N}}\sqrt{\frac{1}{N - 1} \sum_{i=1}^{N} (x_i - \bar{x})^2 } = 3.0 $$

But, being hackers, we'll be using simulation instead.

Just like in our last example, we are once again faced with the problem of not having a generative model, but unlike the last example, we're not comparing two groups, so we can't just shuffle around labels here, instead we'll use something called bootstrap resampling.

Bootstrap resampling is a method that simulates several random sample distributions by drawing samples from the current distribution with replacement, i.e., we can draw the same data point more than once. Luckily, pandas makes this super easy with its `sample`

function. We simply need to make sure that we pass in `True`

for the `replace`

argument to sample from our dataset with replacement.

```
sample = df.sample(20, replace=True)
display(sample)
print("Mean: %2.2f" % sample.heights.mean())
print("Standard Error: %2.2f" % (sample.heights.std() / np.sqrt(len(sample))))
```

More than likely the mean and standard error from our freshly drawn sample above didn't exactly match the one that we calculated using the classic method beforehand. But, if we continue to resample several thousand times and take a look at the average (mean) of all those sample means and their standard deviation, we should have something that very closely approximates the mean and standard error derived from using the classic method above.

```
xbar = []
for i in range(10000):
sample = df.sample(20, replace=True)
xbar.append(sample.heights.mean())
print("Mean: %2.1f" % np.mean(xbar))
print("Standard Error: %2.1f" % np.std(xbar))
```

### Cross Validation¶

For the final example, we dive into the world of the Lorax. In the story of the Lorax, a faceless creature sales an item that (presumably) all creatures need called a Thneed. Our job as consultants to Onceler Industries is to project Thneed sales. But, before we can get started forecasting the sales of Thneeds, we'll first need some data.

Lucky for you, I've already done the hard work of assembling that data in the code below by "eyeballing" the data in the scatter plot from the slides of the talk. So, it may not be exactly the same, but it should be close enough for our example analysis.

```
df = pd.DataFrame({
'temp': [22, 36, 36, 38, 44, 45, 47,
43, 44, 45, 47, 49,
52, 53, 53, 53, 54, 55, 55, 55, 56, 57, 58, 59,
60, 61, 61.5, 61.7, 61.7, 61.7, 61.8, 62, 62, 63.4, 64.6,
65, 65.6, 65.6, 66.4, 66.9, 67, 67, 67.4, 67.5, 68, 69,
70, 71, 71, 71.5, 72, 72, 72, 72.7, 73, 73, 73, 73.3, 74, 75, 75,
77, 77, 77, 77.4, 77.9, 78, 78, 79,
80, 82, 83, 84, 85, 85, 86, 87, 88,
90, 90, 91, 93, 95, 97,
102, 104],
'sales': [660, 433, 475, 492, 302, 345, 337,
479, 456, 440, 423, 269,
331, 197, 283, 351, 470, 252, 278, 350, 253, 253, 343, 280,
200, 194, 188, 171, 204, 266, 275, 171, 282, 218, 226,
187, 184, 192, 167, 136, 149, 168, 218, 298, 199, 268,
235, 157, 196, 203, 148, 157, 213, 173, 145, 184, 226, 204, 250, 102, 176,
97, 138, 226, 35, 190, 221, 95, 211,
110, 150, 152, 37, 76, 56, 51, 27, 82,
100, 123, 145, 51, 156, 99,
147, 54]
})
```

Now that we have our sales data in a pandas dataframe, we can take a look to see if any trends show up. Plotting the data in a scatterplot, like the one below, reveals that a relationship does seem to exist between temperature and Thneed sales.

```
# Grab a reference to fig and axes object so we can reuse them
fig, ax = plt.subplots()
# Plot the Thneed sales data
ax.scatter(df.temp, df.sales)
ax.set_xlim(xmin=20, xmax=110)
ax.set_ylim(ymin=0, ymax=700)
ax.set_xlabel('temperature (F)')
ax.set_ylabel('thneed sales (daily)');
```

We can see what looks like a relationship between the two variables temperature and sales, but how can we best model that relationship so we can accurately predict sales based on temperature?

Well, one measure of a model's accuracy is the Root-Mean-Square Error (RMSE). This metric represents the sample standard deviation between a set of predicted values (from our model) and the actual observed values.

```
def rmse(predictions, targets):
return np.sqrt(((predictions - targets)**2).mean())
```

We can now use our `rmse`

function to measure how well our models' accurately represent the Thneed sales dataset. And, in the next cell, we'll give it a try by creating two different models and seeing which one does a better job of fitting our sales data.

```
# 1D Polynomial Fit
d1_model = np.poly1d(np.polyfit(df.temp, df.sales, 1))
d1_predictions = d1_model(range(111))
ax.plot(range(111), d1_predictions,
color=blue, alpha=0.7)
# 2D Polynomial Fit
d2_model = np.poly1d(np.polyfit(df.temp, df.sales, 2))
d2_predictions = d2_model(range(111))
ax.plot(range(111), d2_predictions,
color=red, alpha=0.5)
ax.annotate('RMS error = %2.1f' % rmse(d1_model(df.temp), df.sales),
xy=(75, 650),
fontsize=20,
color=blue,
backgroundcolor='w')
ax.annotate('RMS error = %2.1f' % rmse(d2_model(df.temp), df.sales),
xy=(75, 580),
fontsize=20,
color=red,
backgroundcolor='w')
display(fig);
```

In the figure above, we plotted our sales data along with the two models we created in the previous step. The first model (in blue) is a simple linear model, i.e., a first-degree polynomial. The second model (in red) is a second-degree polynomial, so rather than a straight line, we end up with a slight curve.

We can see from the RMSE values in the figure above that the second-degree polynomial performed better than the simple linear model. Of course, the question you should now be asking is, is this the best possible model that we can find?

To find out, let's take a look at the RMSE of a few more models to see if we can do any better.

```
rmses = []
for deg in range(15):
model = np.poly1d(np.polyfit(df.temp, df.sales, deg))
predictions = model(df.temp)
rmses.append(rmse(predictions, df.sales))
plt.plot(range(15), rmses)
plt.ylim(45, 70)
plt.xlabel('number of terms in fit')
plt.ylabel('rms error')
plt.annotate('$y = a + bx$',
xytext=(14.2, 70),
xy=(1, rmses[1]),
multialignment='right',
va='center',
arrowprops={'arrowstyle': '-|>',
'lw': 1,
'shrinkA': 10,
'shrinkB': 3})
plt.annotate('$y = a + bx + cx^2$',
xytext=(14.2, 64),
xy=(2, rmses[2]),
multialignment='right',
va='top',
arrowprops={'arrowstyle': '-|>',
'lw': 1,
'shrinkA': 35,
'shrinkB': 3})
plt.annotate('$y = a + bx + cx^2 + dx^3$',
xytext=(14.2, 58),
xy=(3, rmses[3]),
multialignment='right',
va='top',
arrowprops={'arrowstyle': '-|>',
'lw': 1,
'shrinkA': 12,
'shrinkB': 3});
```

We can see, from the plot above, that as we increase the number of terms (i.e., the degrees of freedom) in our model we decrease the RMSE, and this behavior can continue indefinitely, or until we have as many terms as we do data points, at which point we would be fitting the data perfectly.

The problem with this approach though, is that as we increase the number of terms in our equation, we simply match the given dataset closer and closer, but what if our model were to see a data point that's not in our training dataset?

As you can see in the plot below, the model that we've created, though it has a very low RMSE, it has so many terms that it matches our current dataset too closely.

```
# Remove everything but the datapoints
ax.lines.clear()
ax.texts.clear()
# Changing the y-axis limits to match the figure in the slides
ax.set_ylim(0, 1000)
# 14 Dimensional Model
model = np.poly1d(np.polyfit(df.temp, df.sales, 14))
ax.plot(range(20, 110), model(range(20, 110)),
color=sns.xkcd_rgb['sky blue'])
display(fig)
```

The problem with fitting the data too closely, is that our model is so finely tuned to our specific dataset, that if we were to use it to predict future sales, it would most likely fail to get very close to the actual value. This phenomenon of too closely modeling the training dataset is well known amongst machine learning practitioners as overfitting and one way that we can avoid it is to use cross-validation.

Cross-validation avoids overfitting by splitting the training dataset into several subsets and using each one to train and test multiple models. Then, the RMSE's of each of those models are averaged to give a more likely estimate of how a model of that type would perform on unseen data.

So, let's give it a try by splitting our data into two groups and randomly assigning data points into each one.

```
df_a = df.sample(n=len(df)//2)
df_b = df.drop(df_a.index)
```

We can get a look at the data points assigned to each subset by plotting each one as a different color.

```
plt.scatter(df_a.temp, df_a.sales, color='red')
plt.scatter(df_b.temp, df_b.sales, color='blue')
plt.xlim(0, 110)
plt.ylim(0, 700)
plt.xlabel('temprature (F)')
plt.ylabel('thneed sales (daily)');
```

Then, we'll find the best model for each subset of data. In this particular example, we'll fit a second-degree polynomial to each subset and plot both below.

```
# Create a 2-degree model for each subset of data
m1 = np.poly1d(np.polyfit(df_a.temp, df_a.sales, 2))
m2 = np.poly1d(np.polyfit(df_b.temp, df_b.sales, 2))
fig, (ax1, ax2) = plt.subplots(nrows=1, ncols=2,
sharex=False, sharey=True,
figsize=(12, 5))
x_min, x_max = 20, 110
y_min, y_max = 0, 700
x = range(x_min, x_max + 1)
# Plot the df_a group
ax1.scatter(df_a.temp, df_a.sales, color='red')
ax1.set_xlim(xmin=x_min, xmax=x_max)
ax1.set_ylim(ymin=y_min, ymax=y_max)
ax1.set_xlabel('temprature (F)')
ax1.set_ylabel('thneed sales (daily)')
ax1.plot(x, m1(x),
color=sns.xkcd_rgb['sky blue'],
alpha=0.7)
# Plot the df_b group
ax2.scatter(df_b.temp, df_b.sales, color='blue')
ax2.set_xlim(xmin=x_min, xmax=x_max)
ax2.set_ylim(ymin=y_min, ymax=y_max)
ax2.set_xlabel('temprature (F)')
ax2.plot(x, m2(x),
color=sns.xkcd_rgb['rose'],
alpha=0.5);
```

Finally, we'll compare models across subsets by calculating the RMSE for each model using the training set for the other model. This will give us two RMSE scores which we'll then average to get a more accurate estimate of how well a second-degree polynomial will perform on any unseen data.

```
print("RMS = %2.1f" % rmse(m1(df_b.temp), df_b.sales))
print("RMS = %2.1f" % rmse(m2(df_a.temp), df_a.sales))
print("RMS estimate = %2.1f" % np.mean([rmse(m1(df_b.temp), df_b.sales),
rmse(m2(df_a.temp), df_a.sales)]))
```

Then, we simply repeat this process for as long as we so desire.

The following code repeats the process described above for polynomials up to 14 degrees and plots the average RMSE for each one against the non-cross-validated RMSE's that we calculated earlier.

```
rmses = []
cross_validated_rmses = []
for deg in range(15):
# df_a the model on the whole dataset and calculate its
# RMSE on the same set of data
model = np.poly1d(np.polyfit(df.temp, df.sales, deg))
predictions = model(df.temp)
rmses.append(rmse(predictions, df.sales))
# Use cross-validation to create the model and df_a it
m1 = np.poly1d(np.polyfit(df_a.temp, df_a.sales, deg))
m2 = np.poly1d(np.polyfit(df_b.temp, df_b.sales, deg))
p1 = m1(df_b.temp)
p2 = m2(df_a.temp)
cross_validated_rmses.append(np.mean([rmse(p1, df_b.sales),
rmse(p2, df_a.sales)]))
plt.plot(range(15), rmses, color=blue,
label='RMS')
plt.plot(range(15), cross_validated_rmses, color=red,
label='cross validated RMS')
plt.ylim(45, 70)
plt.xlabel('number of terms in fit')
plt.ylabel('rms error')
plt.legend(frameon=True)
plt.annotate('Best model minimizes the\ncross-validated error.',
xytext=(7, 60),
xy=(2, cross_validated_rmses[2]),
multialignment='center',
va='top',
color='blue',
size=25,
backgroundcolor='w',
arrowprops={'arrowstyle': '-|>',
'lw': 3,
'shrinkA': 12,
'shrinkB': 3,
'color': 'blue'});
```

According to the graph above, going from a 1-degree to a 2-degree polynomial gives us quite a large improvement overall. But, unlike the RMSE that we calculated against the training set, when using cross-validation we can see that adding more degrees of freedom to our equation quickly reduces the effectiveness of the model against unseen data. This is overfitting in action! In fact, from the looks of the graph above, it would seem that a second-degree polynomial is actually our best bet for this particular dataset.

#### 2-Fold Cross-Validation¶

Several different methods for performing cross-validation exist, the one we've just seen is called 2-fold cross-validation since the data is split into two subsets. Another close relative is a method called $k$-fold cross-validation. It differs slightly in that the original dataset is divided into $k$ subsets (instead of just 2), one of which is reserved strictly for testing and the other $k - 1$ subsets are used for training models. This is just one example of an alternate cross-validation method, but more do exist and each one has advantages and drawbacks that you'll need to consider when deciding which method to use.

## Conclusion¶

Ok, so that covers nearly everything that Jake covered in his talk. The end of the talk contains a short overview of some important areas that he didn't have time to cover, and there's a nice set of Q&A at the end, but I'll simply direct you to the video for those parts of the talk. Hopefully, this article/notebook has been helpful to anyone working their way through Jake's talk, and if for some reason, you've read through this entire article and haven't watched the video of the talk yet, I encourage you to take 40 minutes out of your day and go watch it now---it really is a fantastic talk!