Introducing Retentioneer, a retention analysis script written in R

In an effort to continue learning R and gain a deeper understanding of how various metrics are calculated, I’ve been working on a few scripts to analyze user behavior. The first was the script that let you visualize your active users by signup cohort and today I’m happy to open source a new one, a retention analysis script that I’m calling Retentioneer (cause everything needs a cool name, right?).

Retentioneer is a script that lets you measure how long your app’s users remain active after signing up broken down by which month or year they signed up in. You can check it out on Github for complete instructions on how to use it and the configuration options available.

If you’ve never seen a retention curve before, here’s an example that you might make it clearer:

My friend Adam Weeks cofounded Brewski Me, a popular iPhone app which helps users keep track of the craft beer they drink. By running Brewski Me’s activity data through Retentioneer we get the following yearly retention curves:

brewskime-retention

By combining a little bit of knowledge about the history of the app with this chart, we can learn a lot:

From 2011 through 2013 Brewski Me had about a 34% 90-day retention rate, meaning that more than 1 in 3 users used it for longer than 3 months (side note: you can configure the script to count only users who were active exactly n-days days after signing up or count them if they were active on or after n-days).

Why the drop in 2014 in 2015? Three possibilities:

  • In mid-2014, Brewski Me changed from a paid app to a free app. We suspect that users who paid up front were more likely to be engaged long term compared to later users who could try it out for free.
  • Because we’re counting a user as retained if they were active at any point after 30, 60, 90 days etc, users who signed up in 2011 – 2013 have simply had more time to come back to the app compared to users who signed up in 2014 and 2015.
  • Finally, the rise of its main competitor Untappd and the social network effects it created may have led some users to switch away from Brewski Me.

In future posts I’ll go into more detail on some of the things I learned working on this including the impact of using activities measured to the second vs measured by just the date and how segmenting your users before running their activity data through this script might make more sense depending on the type of app.

Visualizing Your SaaS App’s Monthly Active Users Broken Down by Signup Cohort

This week at Automattic I’ve been helping with a tool that will allow us to visualize the number of active WordPress.com users each month broken down by when those users signed up for an account. I think this type of chart and what you can learn from it are incredibly valuable so I wanted to show you all how to quickly create one for your own service.

Here’s an example of what this type of chart looks like courtesy of Buffer’s Joel Gascoigne:

What I really like about it is that for each month you can see how many active users there are and when those users signed up for an account. This not only gives you a sense how long ago your active users signed up, but also of your service’s ability to retain users over time.

If you’d like to create a similar chart to visualize your SaaS app’s active users, I put together a small R script on Github that will help you do just that.

The only thing that you need to provide the script is a CSV file that contains user IDs and dates that those users performed an action in your app.

For example, the test data set that comes with it contains user IDs and actions performed by users of one of my apps (Preceden, a web-based timeline maker) for the first year that the site existed (as determined by the automatically set created_at and updated_at values on the Ruby on Rails Active Record objects that each user is associated with):

2   2010-03-28
2   2010-04-09
2   2010-04-10
2   2010-05-16
3   2010-01-31
3   2014-05-07
3   2014-09-30
3   2015-04-11
4   2010-01-31
4   2010-10-06
...

In this example user IDs 2 and 3 each performed actions on four dates and user ID 4 performed actions on 2 dates. The script will analyze that data to figure out which cohort the user belongs to based on the earliest date the user performed an action and count that user toward the active users for each subsequent month that he or she performed an action:

monthly

As you can see there was a huge spike at the beginning of the year when Preceden launched on HackerNews and was subsequently covered on other tech sites, but by December only a fraction of those users were still active. On that note, I encourage you to strive to build a service like Buffer that delivers long term value so your chart doesn’t wind up looking like this one :).

If you have any questions or need help customizing the script in any way, please don’t hesitate to drop me a note.

Thanks Joel Martinez and Rob Felty for providing feedback on the code.

Lean Domain Search at 3½

Lean Domain Search, despite almost no work since its acquisition by Automattic two years ago, has continued to thrive, now handling more than 160,000 searches per month:

lean-domain-search-3.5-black

It’s monthly growth rate works out to be about 6.5%. Not huge, but not bad for maintenance mode. :)

I think its growth is still driven almost entirely by word of mouth so if you’ve ever shared it with anyone (I’m looking at you, Jay Neely), thanks!

TetriNET Bot Source Code Published on Github

A few years ago I wrote about a bot I built in high school to play an online multiplayer Tetris game called TetriNET. The tl;dr is that I got into TetriNET with some friends, built a bot to automate the play, and eventually entered my school’s science fair with it and wound up making it to internationals. As you can imagine, I was pretty cool in high school…

Anyway, when I wrote the post Github was just getting off the ground and it didn’t even occur to me at the time to open source the code there; instead I just zipped up the Visual Basic Project (go VB6!) and linked to it from the post.

Happily, I have gained a little bit more experience with Git and Github since then so I took some to clean up the code (converting CRLF line endings to LF, spaces to tabs, etc) and finally published it.

You can check it out here if you’re curious: https://github.com/mattm/tetrinet-bot.

On that note, this is pretty much exactly what I looked like while checking out out my old code so keep in mind it was a long time ago :)…

A Step by Step Backpropagation Example

Background

Backpropagation is a common method for training a neural network. There is no shortage of papers online that attempt to explain how backpropagation works, but few that include an example with actual numbers. This post is my attempt to explain how it works with a concrete example that folks can compare their own calculations to in order to ensure they understand backpropagation correctly.

If this kind of thing interests you, you should sign up for my newsletter where I post about AI-related projects that I’m working on.

Backpropagation in Python

You can play around with a Python script that I wrote that implements the backpropagation algorithm in this Github repo.

Overview

For this tutorial, we’re going to use a neural network with two inputs, two hidden neurons, two output neurons. Additionally, the hidden and output neurons will include a bias.

Here’s the basic structure:

neural_network (7)

In order to have some numbers to work with, here’s are the initial weights, the biases, and training inputs/outputs:

neural_network (9)

The goal of backpropagation is to optimize the weights so that the neural network can learn how to correctly map arbitrary inputs to outputs.

For the rest of this tutorial we’re going to work with a single training set: given inputs 0.05 and 0.10, we want the neural network to output 0.01 and 0.99.

The Forward Pass

To begin, lets see what the neural network currently predicts given the weights and biases above and inputs of 0.05 and 0.10. To do this we’ll feed those inputs forward though the network.

We figure out the total net input to each hidden layer neuron, squash the total net input using an activation function (here we use the logistic function), then repeat the process with the output layer neurons.

Total net input is also referred to as just net input by some sources.

Here’s how we calculate the total net input for h_1:

net_{h1} = w_1 * i_1 + w_2 * i_2 + b_1 * 1

net_{h1} = 0.15 * 0.05 + 0.2 * 0.1 + 0.35 * 1 = 0.3775

We then squash it using the logistic function to get the output of h_1:

out_{h1} = \frac{1}{1+e^{-net_{h1}}} = \frac{1}{1+e^{-0.3775}} = 0.593269992

Carrying out the same process for h_2 we get:

out_{h2} = 0.596884378

We repeat this process for the output layer neurons, using the output from the hidden layer neurons as inputs.

Here’s the output for o_1:

net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1

net_{o1} = 0.4 * 0.593269992 + 0.45 * 0.596884378 + 0.6 * 1 = 1.105905967

out_{o1} = \frac{1}{1+e^{-net_{h1}}} = \frac{1}{1+e^{-1.105905967}} = 0.75136507

And carrying out the same process for o_2 we get:

out_{o2} = 0.772928465

Calculating the Total Error

We can now calculate the error for each output neuron using the squared error function and sum them to get the total error:

E_{total} = \sum \frac{1}{2}(target - output)^{2}

Some sources refer to the target as the ideal and the output as the actual.
The \frac{1}{2} is included so that exponent is cancelled when we differentiate later on. The result is eventually multiplied by a learning rate anyway so it doesn’t matter that we introduce a constant here [1].

For example, the target output for o_1 is 0.01 but the neural network output 0.75136507, therefore its error is:

E_{o1} = \frac{1}{2}(target_{o1} - out_{o1})^{2} = \frac{1}{2}(0.01 - 0.75136507)^{2} = 0.274811083

Repeating this process for o_2 we get:

E_{o2} = 0.023560026

The total error for the neural network is the sum of these errors:

E_{total} = E_{o1} + E_{o2} = 0.274811083 + 0.023560026 = 0.298371109

The Backwards Pass

Our goal with backpropagation is to update each of the weights in the network so that they cause the actual output to be closer the target output, thereby minimizing the error for each output neuron and the network as a whole.

Output Layer

Consider w_5. We want to know how much a change in w_5 affects the total error, aka \frac{\partial E_{total}}{\partial w_{5}}.

\frac{\partial E_{total}}{\partial w_{5}} is read as “the partial derivative of E_{total} with respect to w_{5}“. You can also say “the gradient with respect to w_{5}“.

By applying the chain rule we know that:

\frac{\partial E_{total}}{\partial w_{5}} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial w_{5}}

Visually, here’s what we’re doing:

output_1_backprop (4)

We need to figure out each piece in this equation.

First, how much does the total error change with respect to the output?

E_{total} = \frac{1}{2}(target_{o1} - out_{o1})^{2} + \frac{1}{2}(target_{o2} - out_{o2})^{2}

\frac{\partial E_{total}}{\partial out_{o1}} = 2 * \frac{1}{2}(target_{o1} - out_{o1})^{2 - 1} * -1 + 0

\frac{\partial E_{total}}{\partial out_{o1}} = -(target_{o1} - out_{o1}) = -(0.01 - 0.75136507) = 0.74136507

-(target - out) is sometimes expressed as out - target
When we take the partial derivative of the total error with respect to out_{o1}, the quantity \frac{1}{2}(target_{o2} - out_{o2})^{2} becomes zero because out_{o1} does not affect it which means we’re taking the derivative of a constant which is zero.

Next, how much does the output of o_1 change with respect to its total net input?

The partial derivative of the logistic function is the output multiplied by 1 minus the output:

out_{o1} = \frac{1}{1+e^{-net_{o1}}}

\frac{\partial out_{o1}}{\partial net_{o1}} = out_{o1}(1 - out_{o1}) = 0.75136507(1 - 0.75136507) = 0.186815602

Finally, how much does the total net input of o1 change with respect to w_5?

net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1

\frac{\partial net_{o1}}{\partial w_{5}} = 1 * out_{h1} * w_5^{(1 - 1)} + 0 + 0 = out_{h1} = 0.593269992

Putting it all together:

\frac{\partial E_{total}}{\partial w_{5}} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial w_{5}}

\frac{\partial E_{total}}{\partial w_{5}} = 0.74136507 * 0.186815602 * 0.593269992 = 0.082167041

You’ll often see this calculation combined in the form of the delta rule:

\frac{\partial E_{total}}{\partial w_{5}} = -(target_{o1} - out_{o1}) * out_{o1}(1 - out_{o1}) * out_{h1}

Alternatively, we have \frac{\partial E_{total}}{\partial out_{o1}} and \frac{\partial out_{o1}}{\partial net_{o1}} which can be written as \frac{\partial E_{total}}{\partial net_{o1}}, aka \delta_{o1} (the Greek letter delta) aka the node delta. We can use this to rewrite the calculation above:

\delta_{o1} = \frac{\partial E_{total}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} = \frac{\partial E_{total}}{\partial net_{o1}}

\delta_{o1} = -(target_{o1} - out_{o1}) * out_{o1}(1 - out_{o1})

Therefore:

\frac{\partial E_{total}}{\partial w_{5}} = \delta_{o1} out_{h1}

Some sources extract the negative sign from \delta so it would be written as:

\frac{\partial E_{total}}{\partial w_{5}} = -\delta_{o1} out_{h1}

To decrease the error, we then subtract this value from the current weight (optionally multiplied by some learning rate, eta, which we’ll set to 0.5):

w_5^{+} = w_5 - \eta * \frac{\partial E_{total}}{\partial w_{5}} = 0.4 - 0.5 * 0.082167041 = 0.35891648

Some sources use \alpha (alpha) to represent the learning rate, others use \eta (eta), and others even use \epsilon (epsilon).

We can repeat this process to get the new weights w_6, w_7, and w_8:

w_6^{+} = 0.408666186

w_7^{+} = 0.511301270

w_8^{+} = 0.561370121

We perform the actual updates in the neural network after we have the new weights leading into the hidden layer neurons (ie, we use the original weights, not the updated weights, when we continue the backpropagation algorithm below).

Hidden Layer

Next, we’ll continue the backwards pass by calculating new values for w_1, w_2, w_3, and w_4.

Big picture, here’s what we need to figure out:

\frac{\partial E_{total}}{\partial w_{1}} = \frac{\partial E_{total}}{\partial out_{h1}} * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}

Visually:

nn-calculation

We’re going to use a similar process as we did for the output layer, but slightly different to account for the fact that the output of each hidden layer neuron contributes to the output (and therefore error) of multiple output neurons. We know that out_{h1} affects both out_{o1} and out_{o2} therefore the \frac{\partial E_{total}}{\partial out_{h1}} needs to take into consideration its effect on the both output neurons:

\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial out_{h1}} + \frac{\partial E_{o2}}{\partial out_{h1}}

Starting with \frac{\partial E_{o1}}{\partial out_{h1}}:

\frac{\partial E_{o1}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial out_{h1}}

We can calculate \frac{\partial E_{o1}}{\partial net_{o1}} using values we calculated earlier:

\frac{\partial E_{o1}}{\partial net_{o1}} = \frac{\partial E_{o1}}{\partial out_{o1}} * \frac{\partial out_{o1}}{\partial net_{o1}} = 0.74136507 * 0.186815602 = 0.138498562

And \frac{\partial net_{o1}}{\partial out_{h1}} is equal to w_5:

net_{o1} = w_5 * out_{h1} + w_6 * out_{h2} + b_2 * 1

\frac{\partial net_{o1}}{\partial out_{h1}} = w_5 = 0.40

Plugging them in:

\frac{\partial E_{o1}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial net_{o1}} * \frac{\partial net_{o1}}{\partial out_{h1}} = 0.138498562 * 0.40 = 0.055399425

Following the same process for \frac{\partial E_{o2}}{\partial out_{o1}}, we get:

\frac{\partial E_{o2}}{\partial out_{h1}} = -0.019049119

Therefore:

\frac{\partial E_{total}}{\partial out_{h1}} = \frac{\partial E_{o1}}{\partial out_{h1}} + \frac{\partial E_{o2}}{\partial out_{h1}} = 0.055399425 + -0.019049119 = 0.036350306

Now that we have \frac{\partial E_{total}}{\partial out_{h1}}, we need to figure out \frac{\partial out_{h1}}{\partial net_{h1}} and then \frac{\partial net_{h1}}{\partial w} for each weight:

out_{h1} = \frac{1}{1+e^{-net_{h1}}}

\frac{\partial out_{h1}}{\partial net_{h1}} = out_{h1}(1 - out_{h1}) = 0.59326999(1 - 0.59326999 ) = 0.241300709

We calculate the partial derivative of the total net input to h_1 with respect to w_1 the same as we did for the output neuron:

net_{h1} = w_1 * i_1 + w_2 * i_2 + b_1 * 1

\frac{\partial net_{h1}}{\partial w_1} = i_1 = 0.05

Putting it all together:

\frac{\partial E_{total}}{\partial w_{1}} = \frac{\partial E_{total}}{\partial out_{h1}} * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}

\frac{\partial E_{total}}{\partial w_{1}} = 0.036350306 * 0.241300709 * 0.05 = 0.000438568

You might also see this written as:

\frac{\partial E_{total}}{\partial w_{1}} = (\sum\limits_{o}{\frac{\partial E_{total}}{\partial out_{o}} * \frac{\partial out_{o}}{\partial net_{o}} * \frac{\partial net_{o}}{\partial out_{h1}}}) * \frac{\partial out_{h1}}{\partial net_{h1}} * \frac{\partial net_{h1}}{\partial w_{1}}

\frac{\partial E_{total}}{\partial w_{1}} = (\sum\limits_{o}{\delta_{o} * w_{ho}}) * out_{h1}(1 - out_{h1}) * i_{1}

\frac{\partial E_{total}}{\partial w_{1}} = \delta_{h1}i_{1}

We can now update w_1:

w_1^{+} = w_1 - \eta * \frac{\partial E_{total}}{\partial w_{1}} = 0.15 - 0.5 * 0.000438568 = 0.149780716

Repeating this for w_2, w_3, and w_4

w_2^{+} = 0.19956143

w_3^{+} = 0.24975114

w_4^{+} = 0.29950229

Finally, we’ve updated all of our weights! When we fed forward the 0.05 and 0.1 inputs originally, the error on the network was 0.298371109. After this first round of backpropagation, the total error is now down to 0.291027924. It might not seem like much, but after repeating this process 10,000 times, for example, the error plummets to 0.000035085. At this point, when we feed forward 0.05 and 0.1, the two outputs neurons generate 0.015912196 (vs 0.01 actual) and 0.984065734 (vs 0.99 actual).

If you’ve made it this far and found any errors in any of the above or can think of any ways to make it clearer for future readers, don’t hesitate to drop me a note. Thanks!

Introducing ABTestCalculator.com, an Open Source A/B Test Significance Calculator

I spent some time recently working on a small side project that I’m excited to share with you all today. It’s an A/B test significance calculator and you can check it out at ABTestCalculator.com.

abtestcalculator

What’s an A/B test significance calculator, you ask? At a high level, A/B testing is a technique that allows you to improve your website by showing visitors one of several versions of something and then measuring the impact each version has on some other event. For example, you might A/B test the wording of a button on your homepage and find that it increases the number of people who sign up by 10%. An A/B test significance calculator helps you analyze the results of an A/B test to determine whether there is a statistically significant change that is not just the result of random chance.

The math is somewhat complicated which is why a number of A/B test calculators exist, including isvalid.org by Evan Solomon, another by KISSmetrics, another by VWO, and many more.

Why build another? Three reasons: to learn the math, to get better at JavaScript, and to build a tool that makes the results of an A/B test easier to understand.

I think most of these other tools do users a disservice by not clearly explaining how to interpret the results. They tend to throw around the percentage improvement and significance figures without explaining what they mean which in the past has led me to make uninformed and often wrong decisions. Worse, most don’t tell you when you don’t have enough participants or conversions in your test and will happily apply statistical analysis to your results even when those methods don’t apply.

It is my hope with this tool that users leave with a clearer understand of how to interpret the results. A few features:

  • An executive summary that provides an overview in plain English about how to interpret the results
  • One graph showing where the true conversion rate for each variation falls (using something called a Wald approximation) and another showing the percentage change between those two distributions
  • It handles ties as well as tests where there aren’t enough participants or conversions to come to a conclusion
  • Results are significant when there is at least a 90% chance that one variation will lead to an improvement
  • The ability to copy a URL for the results to make them easier to share

If you have any suggestions on how to make it better please don’t hesitate to let me know.

On the coding site of things, most of the JavaScript I’ve done in the past (including Preceden and Lean Domain Search) has been with lots and lots of messy jQuery. A lot of new JavaScript technologies have come out in the last few years and I was put on a project at Automattic not too long ago that uses many of them. I fumbled around with it, getting stuff done but not really understanding the fundamentals.

I’m happy to say that this tool uses React for the view layer, NPM and Browserify for dependency management, Gulp for its build system, parts of ES6 for the code (courtesy of Babel), JSHint for code analysis, Mocha for testing, and Github Pages for hosting — all of which I had little to no experience with when I started this project. If you’re interested in checking it out, all of the code is open source (my first!) so you can view it on Github.

This project is the best JavaScript I know how to do so if you do check out the code, please let me know if you have any suggestions on how to improve it.

One last note in case you were wondering about the domain: the former owner had a simple A/B test calculator up on it, but wasn’t actively working on it so I found his email via WHOIS, offered him $200 for it, he countered with $700, I countered with $300 and we had a deal. Normally I wouldn’t pay someone for a domain (I heard there is this amazing service to help people find available domains…), but the price was right and I figured it was worth it for the credibility and SEO value it adds. When I showed him the new site recently all he responded with was “I’m pretty glad I sold the domain now!” which was nice :).

Thanks for checking it out!

Visualizing the Sampling Distribution of a Proportion with R

In yesterday’s post, we showed that a binomial distribution can be approximated by a normal distribution and some of the math behind it.

Today we’ll take it a step further, showing how those results can help us understand the distribution of a sample proportion.

Consider the following example:

Out of the last 250 visitors to your website, 40 signed up for an account.

The conversion rate for that group is 16%, but it’s possible (and likely!) that the true conversion rate differs from this. Statistics can help us determine a range (a confidence interval) for what the true conversion rate actually is.

Recall that in the last post we said that the mean of the binomial distribution can be approximated with a normal distribution with a mean and standard deviation calculated by:

\mu = np

\sigma = \sqrt{npq}

For a proportion, we want to figure out the mean and standard deviation on a per-trial basis so we divide each formula by n, the number of trials:

\mu = \frac{np}{n} = p

\sigma = \frac{\sqrt{npq}}{n} = \sqrt{\frac{npq}{n^2}} = \sqrt{\frac{pq}{n}}

With the mean and standard deviation of the sample proportion in hand, we can plot the distribution for this example:

sample-proportion-distribution

As you can see, the most likely conversion rate is 16% (which is no surprise), but the true conversion rate can fall anywhere under that curve with it being less and less likely as you move farther away.

Where it gets really interesting is when you want to compare multiple proportions.

Let’s say we’re running an A/B test and the original had 40 conversions out of 250 like the example above and the experimental version had 60 conversions out of 270 participants. We can plot the distribution of both sample proportions with this code:

Here’s the result:

multiple-sample-proportion-distribution

What can we determine from this? Is the experimental version better than the original? What if the true proportion for the original is 20% (towards the upper end of its distribution) and the true proportion for the experimental version is 16% (towards the lower end of its distribution)?

We’ll save the answer for a future post :)