Another great turnout at the DataPhilly meetup last night. Was great to see all you random data nerds!
Code snippets to generate animated examples here.
Another great turnout at the DataPhilly meetup last night. Was great to see all you random data nerds!
Code snippets to generate animated examples here.
There was an amazing turnout at last night’s DataPhilly meetup (~200 people!). I was completely delighted by the turnout and people’s engagement level. Here are the slides of the talk I gave to set up the evening with a high-level introduction to machine learning.
— DataPhilly (@DataPhilly) February 19, 2016
6:00 PM to 9:00 PM
Abstract: Corey will present a brief introduction to machine learning. In his talk he will demystify what is often seen as a dark art. Corey will describe how we “teach” machines to learn patterns from examples by breaking the process into its easy-to-understand component parts. By using examples from fields as diverse as biology, health-care, astrophysics, and NBA basketball, Corey will show how data (both big and small) is used to teach machines to predict the future so we can make better decisions.
Bio: Corey Chivers is a Senior Data Scientist at Penn Medicine where he is building machine learning systems to improve patient outcomes by providing real-time predictive applications that empower clinicians to identify at risk individuals. When he’s not pouring over data, he’s likely to be found cycling around his adoptive city of Philadelphia or blogging about all things probability and data at bayesianbiologist.com.
Automating data science through tree-based pipeline optimization
Abstract: Over the past decade, data science and machine learning has grown from a mysterious art form to a staple tool across a variety of fields in business, academia, and government. In this talk, I’m going to introduce the concept of tree-based pipeline optimization for automating one of the most tedious parts of machine learning — pipeline design. All of the work presented in this talk is based on the open source Tree-based Pipeline Optimization Tool (TPOT), which is available on GitHub at https://github.com/rhiever/tpot.
Bio: Randy Olson is an artificial intelligence researcher at the University of Pennsylvania Institute for Biomedical Informatics, where he develops state-of-the-art machine learning algorithms to solve biomedical problems. He regularly writes about his latest adventures in data science at RandalOlson.com/blog, and tweets about the latest data science news at http://twitter.com/randal_olson.
Abstract: Bayesian optimization is a technique for finding the extrema of functions which are expensive, difficult, or time-consuming to evaluate. It has many applications to optimizing the hyperparameters of machine learning models, optimizing the inputs to real-world experiments and processes, etc. This talk will introduce the Gaussian process approach to Bayesian optimization, with sample code in Python.
Bio: Austin Rochford is a Data Scientist at Monetate. He is a former mathematician who is interested in Bayesian nonparametrics, multilevel models, probabilistic programming, and efficient Bayesian computation.
There’s a curious thing about unlikely independent events: no matter how rare, they’re most likely to happen right away.
Let’s get hypothetical
You’ve taken a bet that pays off if you guess the exact date of the next occurrence of a rare event (p = 0.0001 on any given day i.i.d). What day do you choose? In other words, what is the most likely day for this rare event to occur?
Setting aside for now why in the world you’ve taken such a silly sounding bet, it would seem as though a reasonable way to think about it would be to ask: what is the expected number of days until the event? That must be the best bet, right?
We can work out the expected number of days quite easily as 1/p = 10000. So using the logic of expectation, we would choose day 10000 as our bet.
Let’s simulate to see how often we would win with this strategy. We’ll simulate the outcomes by flipping a weighted coin until it comes out heads. We’ll do this 100,000 times and record how many flips it took each time.
The event occurred on day 10,000 exactly 35 times. However, if we look at a histogram of our simulation experiment, we can see that the time it took for the rare event to happen was more often short, than long. In fact, the event occurred 103 times on the very first flip (the most common Time to Event in our set)!
So from the experiment it would seem that the most likely amount of time to pass until the rare event occurs is 0. Maybe our hypothetical event was just not rare enough. Let’s try it again with p=0.0000001, or an event with a 1 in 1million chance of occurring each day.
While now our event is extremely unlikely to occur, it’s still most likely to occur right away.
What does this all have to do with seizing the day? Everything we do in a given day comes with some degree of risk. The Stanford professor Ronald A. Howard conceived of a way of measuring the riskiness of various day-to-day activities, which he termed the micromort. One micromort is a unit of risk equal to p = 0.000001 (1 in a million chance) of death. We are all subject to a baseline level of risk in micromorts, and additional activities may add or subtract from that level (skiing, for instance adds 0.7 micromorts per day).
While minimizing the risks we assume in our day-to-day lives can increase our expected life span, the most likely exact day of our demise is always our next one. So carpe diem!!
Don’t get too freaked out by all of this. It’s just a bit of fun that comes from viewing the problem in a very specific way. That is, as a question of which exact day is most likely. The much more natural way to view it is to ask, what is the relative probability of the unlikely event occurring tomorrow vs any other day but tomorrow. I leave it to the reader to confirm that for events with p < 0.5, the latter is always more likely.
With spring finally making it’s presence known, I thought I’d re-share this cycling data analysis and visualization I did with some great people a while back. Get out there and feel that wind in your hair!
The recent Hack/Reduce hackathon in Montreal was a tonne of fun. Our team tackled a data set of consisting of Bixi (Montreal’s bicycle share system) station states at one minute temporal resolution. We used Hadoop and mapreduce to pull out some features of user behaviours. One of the things we extracted was the flux at each station, which we defined as the number of bikes arriving and departing from a given station per unit time. When you plot the total system flux across all stations against time, you can see the pulse of the city. Here are the first few weeks of this year’s Bixi season.(click to enlarge)
A few things jump out: 1) There are clearly defined peaks at both the morning and evening rush hours, but it looks like the evening rush is typically a little stronger. I guess cycling home is a great way to relax after…
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Guest post by Matt Sundquist of plot.ly.
Plotly is a social graphing and analytics platform. Plotly’s R library lets you make and share publication-quality graphs online. Your work belongs to you, you control privacy and sharing, and public use is free (like GitHub). We are in beta, and would love your feedback, thoughts, and advice.
1. Installing Plotly
Let’s install Plotly. Our documentation has more details.
install.packages("devtools") library("devtools") devtools::install_github("R-api","plotly")
Then signup online or like this:
library(plotly) response = signup (username = 'yourusername', email= 'youremail')
Thanks for signing up to plotly! Your username is: MattSundquist Your temporary password is: pw. You use this to log into your plotly account at https://plot.ly/plot. Your API key is: “API_Key”. You use this to access your plotly account through the API.
2. Canadian Population Bubble Chart
library(plotly) p <- plotly(username="MattSundquist", key="4om2jxmhmn") library(maps) data(canada.cities) trace1 <- list(x=map(regions="canada")$x, y=map(regions="canada")$y) trace2 <- list(x= canada.cities$long, y=canada.cities$lat, text=canada.cities$name, type="scatter", mode="markers", marker=list( "size"=sqrt(canada.cities$pop/max(canada.cities$pop))*100, "opacity"=0.5) ) response <- p$plotly(trace1,trace2) url <- response$url filename <- response$filename browseURL(response$url)
In our graph, the bubble size represents the city population size. Shown below is the GUI, where you can annotate, select colors, analyze and add data, style traces, place your legend, change fonts, and more.
Editing from the GUI, we make a styled version. You can zoom in and hover on the points to find out about the cities. Want to make one for another country? We’d love to see it.
And, here is said meetup, in action:
You can also add in usa and us.cities:
3. Old Faithful and Multiple Axes
Ben Chartoff’s graph shows the correlation between a bimodal eruption time and a bimodal distribution of eruption length. The key series are: a histogram scale of probability, Eruption Time scale in minutes, and a scatterplot showing points within each bin on the x axis. The graph was made with this gist.
4. Plotting Two Histograms Together
Suppose you are studying correlations in two series (Popular Stack Overflow ?). You want to find overlap. You can plot two histograms together, one for each series. The overlapping sections are the darker orange, automatically rendered if you set barmode to ‘overlay’.
library(plotly) p <- plotly(username="Username", key="API_KEY") x0 <- rnorm(500) x1 <- rnorm(500)+1 data0 <- list(x=x0, name = "Series One", type='histogramx', opacity = 0.8) data1 <- list(x=x1, name = "Series Two", type='histogramx', opacity = 0.8) layout <- list( xaxis = list( ticks = "", gridcolor = "white",zerolinecolor = "white", linecolor = "white" ), yaxis = list( ticks = "", gridcolor = "white", zerolinecolor = "white", linecolor = "white" ), barmode='overlay', # style background color. You can set the alpha by adding an a. plot_bgcolor = 'rgba(249,249,251,.85)' ) response <- p$plotly(data0, data1, kwargs=list(layout=layout)) url <- response$url filename <- response$filename browseURL(response$url)
5. Plotting y1 and y2 in the Same Plot
library(plotly) p <- plotly(username="Username", key="API_KEY") # enter data x <- seq(-2, 2, 0.05) y1 <- pnorm(x) y2 <- pnorm(x,1,1) # format, listing y1 as your y. First <- list( x = x, y = y1, type = 'scatter', mode = 'lines', marker = list( color = 'rgb(0, 0, 255)', opacity = 0.5) ) # format again, listing y2 as your y. Second <- list( x = x, y = y2, type = 'scatter', mode = 'lines', opacity = 0.8, marker = list( color = 'rgb(255, 0, 0)') )
And a shot of the Plotly gallery, as seen at the Montreal meetup. Happy plotting!
A friend of mine just alerted me to a story on NPR describing a prize on offer from Warren Buffett and Quicken Loans. The prize is a billion dollars (1B USD) for correctly predicting all 63 games in the men’s Division I college basketball tournament this March. The facebook page announcing the contest puts the odds at 1:9,223,372,036,854,775,808, which they note “may vary depending upon the knowledge and skill of entrant”.
Being curious, I thought I’d see what the assumptions were that went into that number. It would make sense to start with the assumption that you don’t know a lick about college basketball and you just guess using a coin flip for every match-up. In this scenario you’re pretty bad, but you are no worse than random. If we take this assumption, we can calculate the odds as 1/(0.5)^63. To get precision down to a whole integer I pulled out trusty bc for the heavy lifting:
$ echo "scale=50; 1/(0.5^63)" | bc 9223372036854775808.000000
Well, that was easy. So if you were to just guess randomly, your odds of winning the big prize would be those published on the contest page. We can easily calculate the expected value of entering the contest as P(win)*prize, or 9,223,372,036ths of a dollar (that’s 9 nano dollars, if you’re paying attention). You’ve literally already spent that (and then some) in opportunity cost sunk into the time you are spending thinking about this contest and reading this post (but read on, ’cause it’s fun!).
But of course, you’re cleverer than that. You know everything about college basketball – or more likely if you are reading this blog – you have a kickass predictive model that is going to up your game and get your hands into the pocket of the Oracle from Omaha.
What level of predictiveness would you need to make this bet worth while? Let’s have a look at the expected value as a function of our individual game probability of being correct.
And if you think that you’re really good, we can look at the 0.75 to 0.85 range:
So it’s starting to look enticing, you might even be willing to take off work for a while if you thought you could get your model up to a consistent 85% correct game predictions, giving you an expected return of ~$35,000. A recent paper found that even after observing the first 40 scoring events, the outcome of NBA games is only predictable at 80%. In order to be eligible to win, you’ve obviously got to submit your picks before the playoff games begin, but even at this herculean level of accuracy, the expected value of an entry in the contest plummets down to $785.
Those are the odds for an individual entrant, but what are the chances that Buffet and co will have to pay out? That, of course, depends on the number of entrants. Lets assume that the skill of all entrants is the same, though they all have unique models which make different predictions. In this case we can get the probability of at least one of them hitting it big. It will be the complement of no one winning. We already know the odds for a single entrant with a given level of accuracy, so we can just take the probability that each one doesn’t win, then take 1 minus that value.
Just as we saw that the expected value is very sensitive to the predictive accuracy of the participant, so too is the probability that the prize will be awarded at all. If 1 million super talented sporting sages with 80% game-level accuracy enter the contest, there will only be a slightly greater than 50% chance of anyone actually winning. If we substitute in a more reasonable (but let’s face it, still wildly high) figure for participants’ accuracy of 70%, the chance becomes only 1 in 5739 (0.017%) that the top prize will even be awarded even with a 1 million strong entrant pool.
tl;dr You’re not going to win, but you’re still going to play.
If you want to reproduce the numbers and plots in this post, check out this gist.
I can think of worse places to get down and dirty with R than Kananaskis, Alberta.
Canadian Aquatic Invasive Species Networks Annual General Meeting in Kananaskis, Alberta. May 03, 3:25-5:30.
This 2-hour workshop will focus on how and why we do numerical simulation in R. Time permitting, we will also look at how to build and fit likelihood based statistical models.
We ask that you bring your laptop with both R and R-Studio installed. If you’ve never worked with R before, please have a look at the getting started with R document. You can
also check out the slides from our more introductory workshops.
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A very special package that I am rather excited about arrived in the mail recently. The package contained a set of 6-sided dice. These dice, however, don’t have the standard numbers one to six on their faces. Instead, they have assorted numbers between zero and nine. Here’s the exact configuration:
red<-c(4,4,4,4,4,9) blue<-c(2,2,2,7,7,7) olive<-c(0,5,5,5,5,5) yellow<-c(3,3,3,3,8,8) magenta<-c(1,1,6,6,6,6)
Aside from maybe making for a more interesting version of snakes and ladders, why the heck am I so excited about these wacky dice? To find out what makes them so interesting, lets start by just rolling one against another and seeing which one rolls the higher number. Simple enough. Lets roll Red against Blue. Until you get your own set, you can roll in silico.
That was fun. We can do it over and over again and we’ll find that Red beats Blue more often than not. So it seems like Red is a pretty good bet. Now lets try rolling Olive against Red. I’ll wait.
Hey, look at that, the mighty Red has fallen. Olive tends to roll a higher number than Red more often than it doesn’t. So far, we have discovered this relationship:
Olive > Red > Blue
All hail the dominant Olive! Out of these three dice, if we want the best chance of winning, we should always pick Olive right? No dice, as they say. When we roll Olive against Blue, we find that Blue wins more often!
For any one of these three dice, there is another that will roll a higher number more often than not.
Olive > Red > Blue > Olive > Red > Blue > Olive > Red > Blue..
This forms a chain of dominance relationships that is a closed cycle. This property is called intransivity, and you can use it to win riches beyond your wildest dreams, er, well, at least to impress your friends.
Neat, right? But there’s more! We can do the same trick with Yellow, Magenta, and Red (Red > Magenta > Yellow > Red > …). With all five dice, there is a chain for which the order is given by that length of the word for each colour.
Red > Blue > Olive > Yellow > Magenta > …
Awesome. But that’s not it, either! You may have noticed from our three way comparisons that there is another five way chain. This time, the chain order is given by the alphabetical order of the words for each of the colours.
Blue > Magenta > Olive > Red > Yellow > …
What are the odds?
So far I’ve just asked you to take my word for it that the dominance relationships are as I described. Working out the odds of winning for any given pairing of dice as actually quite straightforward. Start by looking at the number on each side of the first die, one at a time. Count how many sides on the opposing die are less than the current number and divide by six. Since each side on the first die has a 1/6 chance of appearing, divide by 6 again. Sum these values for all six sides and you will have the probability that the first die will roll a higher number than the second.
For example, P(Red > Blue) = 5/6 x 1/2 + 1/6, which is 7/12.
Here I’ve worked out all of the pairwise odds:
So, you can always win in this game as long as you get to be second to choose a colour. The odds are strongest in your favour when your opponent either chooses Magenta or Red, and you choose Olive or Yellow, respectively. Isn’t probability wonderful!
And if you still want more, it turns out that if you roll the Grime dice in pairs, the order of the word length chain reverses!