Mark Needham

Thoughts on Software Development

4 types of user

with one comment

I’ve been working with Neo4j full time for slightly more than a year now and from interacting with the community I’ve noticed that while using different features of the product people fall into 4 categories.

These are as follows:

4types

On one axis we have ‘loudness’ i.e. how vocal somebody is either on twitter, StackOverflow or by email and on the other we have ‘success’ which is how well a product feature is working for them.

The people in the top half of the diagram will get the most attention because they’re the most visible.

Of those people we’ll tend to spend more time on the people who are unhappy and vocal to try and help them solve the problems their having.

When working with the people in the top left it’s difficult to understand how representative they are for the whole user base.

It could be the case that they aren’t representative at all and actually there is a quiet majority who the product is working for and are just getting on with it with no fuss.

However, it could equally be the case that they are absolutely representative and there are a lot of users quietly suffering / giving up using the product.

I haven’t come up with a good way to come across the less vocal users but in my experience they’ll often be passive users of the user group or Stack Overflow i.e. they’ll read existing issues but not post anything themselves.

Given this uncertainty I think it makes sense to assume that the silent majority suffer the same problems as the more vocal minority.

Another interesting thing I’ve noticed about this quadrant is that the people in the top right are often the best people in the community to help those who are struggling.

It’d be interesting to know whether anyone has noticed a similar thing with the products they worked on, and if so what approach do you take to unveiling the silent majority?

Written by Mark Needham

July 29th, 2014 at 7:07 pm

R: ggplot – Plotting back to back charts using facet_wrap

without comments

Earlier in the week I showed a way to plot back to back charts using R’s ggplot library but looking back on the code it felt like it was a bit hacky to ‘glue’ two charts together using a grid.

I wanted to find a better way.

To recap, I came up with the following charts showing the RSVPs to Neo4j London meetup events using this code:

2014 07 20 17 42 40

The first thing we need to do to simplify chart generation is to return ‘yes’ and ‘no’ responses in the same cypher query, like so:

timestampToDate <- function(x) as.POSIXct(x / 1000, origin="1970-01-01", tz = "GMT")
 
query = "MATCH (e:Event)<-[:TO]-(response {response: 'yes'})
         WITH e, COLLECT(response) AS yeses
         MATCH (e)<-[:TO]-(response {response: 'no'})<-[:NEXT]-()
         WITH e, COLLECT(response) + yeses AS responses
         UNWIND responses AS response
         RETURN response.time AS time, e.time + e.utc_offset AS eventTime, response.response AS response"
allRSVPs = cypher(graph, query)
allRSVPs$time = timestampToDate(allRSVPs$time)
allRSVPs$eventTime = timestampToDate(allRSVPs$eventTime)
allRSVPs$difference = as.numeric(allRSVPs$eventTime - allRSVPs$time, units="days")

The query is a bit because we want to capture the ‘no’ responses when they initially said yes which is why we check for a ‘NEXT’ relationship when looking for the negative responses.

Let’s inspect allRSVPs:

> allRSVPs[1:10,]
                  time           eventTime response difference
1  2014-06-13 21:49:20 2014-07-22 18:30:00       no   38.86157
2  2014-07-02 22:24:06 2014-07-22 18:30:00      yes   19.83743
3  2014-05-23 23:46:02 2014-07-22 18:30:00      yes   59.78053
4  2014-06-23 21:07:11 2014-07-22 18:30:00      yes   28.89084
5  2014-06-06 15:09:29 2014-07-22 18:30:00      yes   46.13925
6  2014-05-31 13:03:09 2014-07-22 18:30:00      yes   52.22698
7  2014-05-23 23:46:02 2014-07-22 18:30:00      yes   59.78053
8  2014-07-02 12:28:22 2014-07-22 18:30:00      yes   20.25113
9  2014-06-30 23:44:39 2014-07-22 18:30:00      yes   21.78149
10 2014-06-06 15:35:53 2014-07-22 18:30:00      yes   46.12091

We’ve returned the actual response with each row so that we can distinguish between responses. It will also come in useful for pivoting our single chart later on.

The next step is to get ggplot to generate our side by side charts. I started off by plotting both types of response on the same chart:

ggplot(allRSVPs, aes(x = difference, fill=response)) + 
  geom_bar(binwidth=1)

2014 07 25 22 14 28

This one stacks the ‘yes’ and ‘no’ responses on top of each other which isn’t what we want as it’s difficult to compare the two.

What we need is the facet_wrap function which allows us to generate multiple charts grouped by key. We’ll group by ‘response’:

ggplot(allRSVPs, aes(x = difference, fill=response)) + 
  geom_bar(binwidth=1) + 
  facet_wrap(~ response, nrow=2, ncol=1)

2014 07 25 22 34 46

The only thing we’re missing now is the red and green colours which is where the scale_fill_manual function comes in handy:

ggplot(allRSVPs, aes(x = difference, fill=response)) + 
  scale_fill_manual(values=c("#FF0000", "#00FF00")) + 
  geom_bar(binwidth=1) +
  facet_wrap(~ response, nrow=2, ncol=1)

2014 07 25 22 39 56

If we want to show the ‘yes’ chart on top we can pass in an extra parameter to facet_wrap to change where it places the highest value:

ggplot(allRSVPs, aes(x = difference, fill=response)) + 
  scale_fill_manual(values=c("#FF0000", "#00FF00")) + 
  geom_bar(binwidth=1) +
  facet_wrap(~ response, nrow=2, ncol=1, as.table = FALSE)

2014 07 25 22 43 29

We could go one step further and group by response and day. First let’s add a ‘day’ column to our data frame:

allRSVPs$dayOfWeek = format(allRSVPs$eventTime, "%A")

And now let’s plot the charts using both columns:

ggplot(allRSVPs, aes(x = difference, fill=response)) + 
  scale_fill_manual(values=c("#FF0000", "#00FF00")) + 
  geom_bar(binwidth=1) +
  facet_wrap(~ response + dayOfWeek, as.table = FALSE)

2014 07 25 22 49 57

The distribution of dropouts looks fairly similar for all the days – Thursday is just at an order of magnitude below the other days because we haven’t run many events on Thursdays so far.

At a glance it doesn’t appear that so many people sign up for Thursday events on the day or one day before.

One potential hypothesis is that people have things planned for Thursday whereas they decide more last minute what to do on the other days.

We’ll have to run some more events on Thursdays to see whether that trend holds out.

The code is on github if you want to play with it

Written by Mark Needham

July 25th, 2014 at 9:57 pm

Posted in R

Tagged with

Java: Determining the status of data import using kill signals

with 2 comments

A few weeks ago I was working on the initial import of ~ 60 million bits of data into Neo4j and we kept running into a problem where the import process just seemed to freeze and nothing else was imported.

It was very difficult to tell what was happening inside the process – taking a thread dump merely informed us that it was attempting to process one line of a CSV line and was somehow unable to do so.

One way to help debug this would have been to print out every single line of the CSV as we processed it and then watch where it got stuck but this seemed a bit over kill. Ideally we wanted to only print out the line we were processing on demand.

As luck would have it we can do exactly this by sending a kill signal to our import process and have it print out where it had got up to. We had to make sure we picked a signal which wasn’t already being handled by the JVM and decided to go with ‘SIGTRAP’ i.e. kill -5 [pid]

We came across a neat blog post that explained how to wire everything up and then created our own version:

class Kill3Handler implements SignalHandler
{
    private AtomicInteger linesProcessed;
    private AtomicReference<Map<String, Object>> lastRowProcessed;
 
    public Kill3Handler( AtomicInteger linesProcessed, AtomicReference<Map<String, Object>> lastRowProcessed )
    {
        this.linesProcessed = linesProcessed;
        this.lastRowProcessed = lastRowProcessed;
    }
 
    @Override
    public void handle( Signal signal )
    {
        System.out.println("Last Line Processed: " + linesProcessed.get() + " " + lastRowProcessed.get());
    }
}

We then wired that up like so:

AtomicInteger linesProcessed = new AtomicInteger( 0 );
AtomicReference<Map<String, Object>> lastRowProcessed = new AtomicReference<>(  );
Kill3Handler kill3Handler = new Kill3Handler( linesProcessed, lastRowProcessed );
Signal.handle(new Signal("TRAP"), kill3Handler);
 
// as we iterate each line we update those variables
 
linesProcessed.incrementAndGet();
lastRowProcessed.getAndSet( properties ); // properties = a representation of the row we're processing

This worked really well for us and we were able to work out that we had a slight problem with some of the data in our CSV file which was causing it to be processed incorrectly.

We hadn’t been able to see this by visual inspection since the CSV files were a few GB in size. We’d therefore only skimmed a few lines as a sanity check.

I didn’t even know you could do this but it’s a neat trick to keep in mind – I’m sure it shall come in useful again.

Written by Mark Needham

July 23rd, 2014 at 10:20 pm

Posted in Java

Tagged with

R: ggplot – Plotting back to back bar charts

without comments

I’ve been playing around with R’s ggplot library to explore the Neo4j London meetup and the next thing I wanted to do was plot back to back bar charts showing ‘yes’ and ‘no’ RSVPs.

I’d already done the ‘yes’ bar chart using the following code:

query = "MATCH (e:Event)<-[:TO]-(response {response: 'yes'})
         RETURN response.time AS time, e.time + e.utc_offset AS eventTime"
allYesRSVPs = cypher(graph, query)
allYesRSVPs$time = timestampToDate(allYesRSVPs$time)
allYesRSVPs$eventTime = timestampToDate(allYesRSVPs$eventTime)
allYesRSVPs$difference = as.numeric(allYesRSVPs$eventTime - allYesRSVPs$time, units="days")
 
ggplot(allYesRSVPs, aes(x=difference)) + geom_histogram(binwidth=1, fill="green")
2014 07 20 01 15 39

The next step was to create a similar thing for people who’d RSVP’d ‘no’ having originally RSVP’d ‘yes’ i.e. people who dropped out:

query = "MATCH (e:Event)<-[:TO]-(response {response: 'no'})<-[:NEXT]-()
         RETURN response.time AS time, e.time + e.utc_offset AS eventTime"
allNoRSVPs = cypher(graph, query)
allNoRSVPs$time = timestampToDate(allNoRSVPs$time)
allNoRSVPs$eventTime = timestampToDate(allNoRSVPs$eventTime)
allNoRSVPs$difference = as.numeric(allNoRSVPs$eventTime - allNoRSVPs$time, units="days")
 
ggplot(allNoRSVPs, aes(x=difference)) + geom_histogram(binwidth=1, fill="red")
2014 07 20 17 25 03

As expected if people are going to drop out they do so a day or two before the event happens. By including the need for a ‘NEXT’ relationship we only capture the people who replied ‘yes’ and changed it to ‘no’. We don’t capture the people who said ‘no’ straight away.

I thought it’d be cool to be able to have the two charts back to back using the same scale so I could compare them against each other which led to my first attempt:

yes = ggplot(allYesRSVPs, aes(x=difference)) + geom_histogram(binwidth=1, fill="green")
no = ggplot(allNoRSVPs, aes(x=difference)) + geom_histogram(binwidth=1, fill="red") + scale_y_reverse()
library(gridExtra)
grid.arrange(yes,no,ncol=1,widths=c(1,1))

scale_y_reverse() flips the y axis so we’d see the ‘no’ chart upside down. The last line plots the two charts in a grid containing 1 column which forces them to go next to each other vertically.

2014 07 20 17 29 27

When we compare them next to each other we can see that the ‘yes’ replies are much more spread out whereas if people are going to drop out it nearly always happens a week or so before the event happens. This is what we thought was happening but it’s cool to have it confirmed by the data.

One annoying thing about that visualisation is that the two charts aren’t on the same scale. The ‘no’ chart only goes up to 100 days whereas the ‘yes’ one goes up to 120 days. In addition, the top end of the ‘yes’ chart is around 200 whereas the ‘no’ is around 400.

Luckily we can solve that problem by fixing the axes for both plots:

yes = ggplot(allYesRSVPs, aes(x=difference)) + 
  geom_histogram(binwidth=1, fill="green") +
  xlim(0,120) + 
  ylim(0, 400)
 
no = ggplot(allNoRSVPs, aes(x=difference)) +
  geom_histogram(binwidth=1, fill="red") +
  xlim(0,120) + 
  ylim(0, 400) +
  scale_y_reverse()

Now if we re-render it looks much better:

2014 07 20 17 42 40

From having comparable axes we can see that a lot more people drop out of an event (500) as it approaches than new people sign up (300). This is quite helpful for working out how many people are likely to show up.

We’ve found that the number of people RSVP’d ‘yes’ to an event will drop by 15-20% overall from 2 days before an event up until the evening of the event and the data seems to confirm this.

The only annoying thing about this approach is that the axes are repeated due to them being completely separate charts.

I expect it would look better if I can work out how to combine the two data frames together and then pull out back to back charts based on a variable in the combined data frame.

I’m still working on that so suggestions are most welcome. The code is on github if you want to play with it.

Written by Mark Needham

July 20th, 2014 at 4:50 pm

Posted in R

Tagged with

Neo4j 2.1.2: Finding where I am in a linked list

without comments

I was recently asked how to calculate the position of a node in a linked list and realised that as the list increases in size this is one of the occasions when we should write an unmanaged extension rather than using cypher.

I wrote a quick bit of code to create a linked list with 10,000 elements in it:

public class Chains 
{
    public static void main(String[] args)
    {
        String simpleChains = "/tmp/longchains";
        populate( simpleChains, 10000 );
    }
 
    private static void populate( String path, int chainSize )
    {
        GraphDatabaseService db = new GraphDatabaseFactory().newEmbeddedDatabase( path );
        try(Transaction tx = db.beginTx()) {
            Node currentNode = null;
            for ( int i = 0; i < chainSize; i++ )
            {
                Node node = db.createNode();
 
                if(currentNode != null) {
                    currentNode.createRelationshipTo( node, NEXT );
                }
                currentNode = node;
            }
            tx.success();
        }
 
 
        db.shutdown();
    }
}

To find our distance from the end of the linked list we could write the following cypher query:

match n  where id(n) = {nodeId}  with n 
match path = (n)-[:NEXT*]->() 
RETURN id(n) AS nodeId, length(path) AS length 
ORDER BY length DESC 
LIMIT 1;

For simplicity we’re finding a node by it’s internal node id and then finding the ‘NEXT’ relationships going out from this node recursively. We then filter the results so that we only get the longest path back which will be our distance to the end of the list.

I noticed that this query would sometimes take 10s of seconds so I wrote a version using the Java Traversal API to see whether I could get it any quicker.

This is the Java version:

try(Transaction tx = db.beginTx()) {
    Node startNode = db.getNodeById( nodeId );
    TraversalDescription traversal = db.traversalDescription();
    Traverser traverse = traversal
            .depthFirst()
            .relationships( NEXT, Direction.OUTGOING )
            .sort( new Comparator<Path>()
            {
                @Override
                public int compare( Path o1, Path o2 )
                {
                    return Integer.valueOf( o2.length() ).compareTo( o1 .length() );
                }
            } )
            .traverse( startNode );
 
    Collection<Path> paths = IteratorUtil.asCollection( traverse );
 
    int maxLength = traverse.iterator().next().length();
    System.out.print( maxLength );
 
    tx.failure();
}

This is a bit more verbose than the cypher version but computes the same result. We’ve sorted the paths by length using a comparator to ensure we get the longest path back first.

I created a little program to warm up the caches and kick off a few iterations where I queried from different nodes and returned the length and time taken. These were the results:

--------
(Traversal API) Node:    1, Length: 9998, Time (ms):  15
       (Cypher) Node:    1, Length: 9998, Time (ms): 26225
(Traversal API) Node:  456, Length: 9543, Time (ms):  10
       (Cypher) Node:  456, Length: 9543, Time (ms): 24881
(Traversal API) Node:  761, Length: 9238, Time (ms):   9
       (Cypher) Node:  761, Length: 9238, Time (ms): 9941
--------
(Traversal API) Node:    1, Length: 9998, Time (ms):   9
       (Cypher) Node:    1, Length: 9998, Time (ms): 12537
(Traversal API) Node:  456, Length: 9543, Time (ms):   8
       (Cypher) Node:  456, Length: 9543, Time (ms): 15690
(Traversal API) Node:  761, Length: 9238, Time (ms):   7
       (Cypher) Node:  761, Length: 9238, Time (ms): 9202
--------
(Traversal API) Node:    1, Length: 9998, Time (ms):   8
       (Cypher) Node:    1, Length: 9998, Time (ms): 11905
(Traversal API) Node:  456, Length: 9543, Time (ms):   7
       (Cypher) Node:  456, Length: 9543, Time (ms): 22296
(Traversal API) Node:  761, Length: 9238, Time (ms):   8
       (Cypher) Node:  761, Length: 9238, Time (ms): 8739
--------

Interestingly when I reduced the size of the linked list to 1000 the difference wasn’t so pronounced:

--------
(Traversal API) Node:    1, Length: 998, Time (ms):   5
       (Cypher) Node:    1, Length: 998, Time (ms): 174
(Traversal API) Node:  456, Length: 543, Time (ms):   2
       (Cypher) Node:  456, Length: 543, Time (ms):  71
(Traversal API) Node:  761, Length: 238, Time (ms):   1
       (Cypher) Node:  761, Length: 238, Time (ms):  13
--------
(Traversal API) Node:    1, Length: 998, Time (ms):   2
       (Cypher) Node:    1, Length: 998, Time (ms): 111
(Traversal API) Node:  456, Length: 543, Time (ms):   1
       (Cypher) Node:  456, Length: 543, Time (ms):  40
(Traversal API) Node:  761, Length: 238, Time (ms):   1
       (Cypher) Node:  761, Length: 238, Time (ms):  12
--------
(Traversal API) Node:    1, Length: 998, Time (ms):   3
       (Cypher) Node:    1, Length: 998, Time (ms): 129
(Traversal API) Node:  456, Length: 543, Time (ms):   2
       (Cypher) Node:  456, Length: 543, Time (ms):  48
(Traversal API) Node:  761, Length: 238, Time (ms):   0
       (Cypher) Node:  761, Length: 238, Time (ms):  12
--------

which is good news as most linked lists that we’ll create will be in the 10s – 100s range rather than 10,000 which was what I was faced with.

I’m sure cypher will reach parity for this type of query in future which will be great as I like writing cypher much more than I do Java. For now though it’s good to know we have a backup option to call on when necessary.

The code is available as a gist if you want to play around with it further.

Written by Mark Needham

July 20th, 2014 at 3:13 pm

Posted in neo4j

Tagged with

R: ggplot – Don’t know how to automatically pick scale for object of type difftime – Discrete value supplied to continuous scale

without comments

While reading ‘Why The R Programming Language Is Good For Business‘ I came across Udacity’s ‘Data Analysis with R‘ courses – part of which focuses exploring data sets using visualisations, something I haven’t done much of yet.

I thought it’d be interesting to create some visualisations around the times that people RSVP ‘yes’ to the various Neo4j events that we run in London.

I started off with the following query which returns the date time that people replied ‘Yes’ to an event and the date time of the event:

library(Rneo4j)
query = "MATCH (e:Event)<-[:TO]-(response {response: 'yes'})
         RETURN response.time AS time, e.time + e.utc_offset AS eventTime"
allYesRSVPs = cypher(graph, query)
allYesRSVPs$time = timestampToDate(allYesRSVPs$time)
allYesRSVPs$eventTime = timestampToDate(allYesRSVPs$eventTime)
 
> allYesRSVPs[1:10,]
                  time           eventTime
1  2011-06-05 12:12:27 2011-06-29 18:30:00
2  2011-06-05 14:49:04 2011-06-29 18:30:00
3  2011-06-10 11:22:47 2011-06-29 18:30:00
4  2011-06-07 15:27:07 2011-06-29 18:30:00
5  2011-06-06 20:21:45 2011-06-29 18:30:00
6  2011-07-04 19:49:04 2011-07-27 19:00:00
7  2011-07-05 16:40:10 2011-07-27 19:00:00
8  2011-08-19 07:41:10 2011-08-31 18:30:00
9  2011-08-24 12:47:40 2011-08-31 18:30:00
10 2011-08-18 09:56:53 2011-08-31 18:30:00

I wanted to create a bar chart showing the amount of time in advance of a meetup that people RSVP’d ‘yes’ so I added the following column to my data frame:

allYesRSVPs$difference = allYesRSVPs$eventTime - allYesRSVPs$time
 
> allYesRSVPs[1:10,]
                  time           eventTime    difference
1  2011-06-05 12:12:27 2011-06-29 18:30:00 34937.55 mins
2  2011-06-05 14:49:04 2011-06-29 18:30:00 34780.93 mins
3  2011-06-10 11:22:47 2011-06-29 18:30:00 27787.22 mins
4  2011-06-07 15:27:07 2011-06-29 18:30:00 31862.88 mins
5  2011-06-06 20:21:45 2011-06-29 18:30:00 33008.25 mins
6  2011-07-04 19:49:04 2011-07-27 19:00:00 33070.93 mins
7  2011-07-05 16:40:10 2011-07-27 19:00:00 31819.83 mins
8  2011-08-19 07:41:10 2011-08-31 18:30:00 17928.83 mins
9  2011-08-24 12:47:40 2011-08-31 18:30:00 10422.33 mins
10 2011-08-18 09:56:53 2011-08-31 18:30:00 19233.12 mins

I then tried to use ggplot to create a bar chart of that data:

> ggplot(allYesRSVPs, aes(x=difference)) + geom_histogram(binwidth=1, fill="green")

Unfortunately that resulted in this error:

Don't know how to automatically pick scale for object of type difftime. Defaulting to continuous
Error: Discrete value supplied to continuous scale

I couldn’t find anyone who had come across this problem before in my search but I did find the as.numeric function which seemed like it would put the difference into an appropriate format:

allYesRSVPs$difference = as.numeric(allYesRSVPs$eventTime - allYesRSVPs$time, units="days")
> ggplot(allYesRSVPs, aes(x=difference)) + geom_histogram(binwidth=1, fill="green")

that resulted in the following chart:

2014 07 20 01 15 39

We can see there is quite a heavy concentration of people RSVPing yes in the few days before the event and then the rest are scattered across the first 30 days.

We usually announce events 3/4 weeks in advance so I don’t know that it tells us anything interesting other than that it seems like people sign up for events when an email is sent out about them.

The date the meetup was announced (by email) isn’t currently exposed by the API but hopefully one day it will be.

The code is on github if you want to have a play – any suggestions welcome.

Written by Mark Needham

July 20th, 2014 at 12:21 am

Posted in R

Tagged with

R: Apply a custom function across multiple lists

without comments

In my continued playing around with R I wanted to map a custom function over two lists comparing each item with its corresponding items.

If we just want to use a built in function such as subtraction between two lists it’s quite easy to do:

> c(10,9,8,7,6,5,4,3,2,1) - c(5,4,3,4,3,2,2,1,2,1)
 [1] 5 5 5 3 3 3 2 2 0 0

I wanted to do a slight variation on that where instead of returning the difference I wanted to return a text value representing the difference e.g. ’5 or more’, ’3 to 5′ etc.

I spent a long time trying to figure out how to do that before finding an excellent blog post which describes all the different ‘apply’ functions available in R.

As far as I understand ‘apply’ is the equivalent of ‘map’ in Clojure or other functional languages.

In this case we want the mapply variant which we can use like so:

> mapply(function(x, y) { 
    if((x-y) >= 5) {
        "5 or more"
    } else if((x-y) >= 3) {
        "3 to 5"
    } else {
        "less than 5"
    }    
  }, c(10,9,8,7,6,5,4,3,2,1),c(5,4,3,4,3,2,2,1,2,1))
 [1] "5 or more"   "5 or more"   "5 or more"   "3 to 5"      "3 to 5"      "3 to 5"      "less than 5"
 [8] "less than 5" "less than 5" "less than 5"

We could then pull that out into a function if we wanted:

summarisedDifference <- function(one, two) {
  mapply(function(x, y) { 
    if((x-y) >= 5) {
      "5 or more"
    } else if((x-y) >= 3) {
      "3 to 5"
    } else {
      "less than 5"
    }    
  }, one, two)
}

which we could call like so:

> summarisedDifference(c(10,9,8,7,6,5,4,3,2,1),c(5,4,3,4,3,2,2,1,2,1))
 [1] "5 or more"   "5 or more"   "5 or more"   "3 to 5"      "3 to 5"      "3 to 5"      "less than 5"
 [8] "less than 5" "less than 5" "less than 5"

I also wanted to be able to compare a list of items to a single item which was much easier than I expected:

> summarisedDifference(c(10,9,8,7,6,5,4,3,2,1), 1)
 [1] "5 or more"   "5 or more"   "5 or more"   "5 or more"   "5 or more"   "3 to 5"      "3 to 5"     
 [8] "less than 5" "less than 5" "less than 5"

If we wanted to get a summary of the differences between the lists we could plug them into ddply like so:

> library(plyr)
> df = data.frame(x=c(10,9,8,7,6,5,4,3,2,1), y=c(5,4,3,4,3,2,2,1,2,1))
> ddply(df, .(difference=summarisedDifference(x,y)), summarise, count=length(x))
   difference count
1      3 to 5     3
2   5 or more     3
3 less than 5     4

Written by Mark Needham

July 16th, 2014 at 5:04 am

Posted in R

Tagged with

Neo4j: LOAD CSV – Processing hidden arrays in your CSV documents

without comments

I was recently asked how to process an ‘array’ of values inside a column in a CSV file using Neo4j’s LOAD CSV tool and although I initially thought this wouldn’t be possible as every cell is treated as a String, Michael showed me a way of working around this which I thought was pretty neat.

Let’s say we have a CSV file representing people and their friends. It might look like this:

name,friends
"Mark","Michael,Peter"
"Michael","Peter,Kenny"
"Kenny","Anders,Michael"

And what we want is to have the following nodes:

  • Mark
  • Michael
  • Peter
  • Kenny
  • Anders

And the following friends relationships:

  • Mark -> Michael
  • Mark -> Peter
  • Michael -> Peter
  • Michael -> Kenny
  • Kenny -> Anders
  • Kenny -> Michael

We’ll start by loading the CSV file and returning each row:

$ load csv with headers from "file:/Users/markneedham/Desktop/friends.csv" AS row RETURN row;
+------------------------------------------------+
| row                                            |
+------------------------------------------------+
| {name -> "Mark", friends -> "Michael,Peter"}   |
| {name -> "Michael", friends -> "Peter,Kenny"}  |
| {name -> "Kenny", friends -> "Anders,Michael"} |
+------------------------------------------------+
3 rows

As expected the ‘friends’ column is being treated as a String which means we can use the split function to get an array of people that we want to be friends with:

$ load csv with headers from "file:/Users/markneedham/Desktop/friends.csv" AS row RETURN row, split(row.friends, ",") AS friends;
+-----------------------------------------------------------------------+
| row                                            | friends              |
+-----------------------------------------------------------------------+
| {name -> "Mark", friends -> "Michael,Peter"}   | ["Michael","Peter"]  |
| {name -> "Michael", friends -> "Peter,Kenny"}  | ["Peter","Kenny"]    |
| {name -> "Kenny", friends -> "Anders,Michael"} | ["Anders","Michael"] |
+-----------------------------------------------------------------------+
3 rows

Now that we’ve got them as an array we can use UNWIND to get pairs of friends that we want to create:

$ load csv with headers from "file:/Users/markneedham/Desktop/friends.csv" AS row 
  WITH row, split(row.friends, ",") AS friends 
  UNWIND friends AS friend 
  RETURN row.name, friend;
+-----------------------+
| row.name  | friend    |
+-----------------------+
| "Mark"    | "Michael" |
| "Mark"    | "Peter"   |
| "Michael" | "Peter"   |
| "Michael" | "Kenny"   |
| "Kenny"   | "Anders"  |
| "Kenny"   | "Michael" |
+-----------------------+
6 rows

And now we’ll introduce some MERGE statements to create the appropriate nodes and relationships:

$ load csv with headers from "file:/Users/markneedham/Desktop/friends.csv" AS row 
  WITH row, split(row.friends, ",") AS friends 
  UNWIND friends AS friend  
  MERGE (p1:Person {name: row.name}) 
  MERGE (p2:Person {name: friend}) 
  MERGE (p1)-[:FRIENDS_WITH]->(p2);
+-------------------+
| No data returned. |
+-------------------+
Nodes created: 5
Relationships created: 6
Properties set: 5
Labels added: 5
373 ms

And now if we query the database to get back all the nodes + relationships…

$ match (p1:Person)-[r]->(p2) RETURN p1,r, p2;
+------------------------------------------------------------------------+
| p1                      | r                  | p2                      |
+------------------------------------------------------------------------+
| Node[0]{name:"Mark"}    | :FRIENDS_WITH[0]{} | Node[1]{name:"Michael"} |
| Node[0]{name:"Mark"}    | :FRIENDS_WITH[1]{} | Node[2]{name:"Peter"}   |
| Node[1]{name:"Michael"} | :FRIENDS_WITH[2]{} | Node[2]{name:"Peter"}   |
| Node[1]{name:"Michael"} | :FRIENDS_WITH[3]{} | Node[3]{name:"Kenny"}   |
| Node[3]{name:"Kenny"}   | :FRIENDS_WITH[4]{} | Node[4]{name:"Anders"}  |
| Node[3]{name:"Kenny"}   | :FRIENDS_WITH[5]{} | Node[1]{name:"Michael"} |
+------------------------------------------------------------------------+
6 rows

…you’ll see that we have everything.

If instead of a comma separated list of people we have a literal array in the cell…

name,friends
"Mark", "[Michael,Peter]"
"Michael", "[Peter,Kenny]"
"Kenny", "[Anders,Michael]"

…we’d need to tweak the part of the query which extracts our friends to strip off the first and last characters:

$ load csv with headers from "file:/Users/markneedham/Desktop/friendsa.csv" AS row 
  RETURN row, split(substring(row.friends, 1, length(row.friends) -2), ",") AS friends;
+-------------------------------------------------------------------------+
| row                                              | friends              |
+-------------------------------------------------------------------------+
| {name -> "Mark", friends -> "[Michael,Peter]"}   | ["Michael","Peter"]  |
| {name -> "Michael", friends -> "[Peter,Kenny]"}  | ["Peter","Kenny"]    |
| {name -> "Kenny", friends -> "[Anders,Michael]"} | ["Anders","Michael"] |
+-------------------------------------------------------------------------+
3 rows

And then if we put the whole query together we end up with this:

$ load csv with headers from "file:/Users/markneedham/Desktop/friendsa.csv" AS row 
  WITH row, split(substring(row.friends, 1, length(row.friends) -2), ",") AS friends 
  UNWIND friends AS friend  
  MERGE (p1:Person {name: row.name}) 
  MERGE (p2:Person {name: friend}) 
  MERGE (p1)-[:FRIENDS_WITH]->(p2);;
+-------------------+
| No data returned. |
+-------------------+
Nodes created: 5
Relationships created: 6
Properties set: 5
Labels added: 5

Written by Mark Needham

July 10th, 2014 at 2:54 pm

Posted in neo4j

Tagged with

R/plyr: ddply – Error in vector(type, length) : vector: cannot make a vector of mode ‘closure’.

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In my continued playing around with plyr’s ddply function I was trying to group a data frame by one of its columns and return a count of the number of rows with specific values and ran into a strange (to me) error message.

I had a data frame:

n = c(2, 3, 5) 
s = c("aa", "bb", "cc") 
b = c(TRUE, FALSE, TRUE) 
df = data.frame(n, s, b)

And wanted to group and count on column ‘b’ so I’d get back a count of 2 for TRUE and 1 for FALSE. I wrote this code:

ddply(df, "b", function(x) { 
  countr <- length(x$n) 
  data.frame(count = count) 
})

which when evaluated gave the following error:

Error in vector(type, length) : 
  vector: cannot make a vector of mode 'closure'.

It took me quite a while to realise that I’d just made a typo in assigned the count to a variable called ‘countr’ instead of ‘count’.

As a result of that typo I think the R compiler was trying to find a variable called ‘count’ somwhere else in the lexical scope but was unable to. If I’d defined the variable ‘count’ outside the call to ddply function then my typo wouldn’t have resulted in an error but rather an unexpected resulte.g.

> count = 10
> ddply(df, "b", function(x) { 
+   countr <- length(x$n) 
+   data.frame(count = count) 
+ })
      b count
1 FALSE     4
2  TRUE     4

Once I spotted the typo and fixed it things worked as expected:

> ddply(df, "b", function(x) { 
+   count <- length(x$n) 
+   data.frame(count = count) 
+ })
      b count
1 FALSE     1
2  TRUE     2

Written by Mark Needham

July 7th, 2014 at 6:07 am

Posted in R

Tagged with

R/plyr: ddply – Renaming the grouping/generated column when grouping by date

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On Nicole’s recommendation I’ve been having a look at R’s plyr package to see if I could simplify my meetup analysis and I started by translating my code that grouped meetup join dates by day of the week.

To refresh, the code without plyr looked like this:

library(Rneo4j)
timestampToDate <- function(x) as.POSIXct(x / 1000, origin="1970-01-01")
 
query = "MATCH (:Person)-[:HAS_MEETUP_PROFILE]->()-[:HAS_MEMBERSHIP]->(membership)-[:OF_GROUP]->(g:Group {name: \"Neo4j - London User Group\"})
         RETURN membership.joined AS joinDate"
meetupMembers = cypher(graph, query)
meetupMembers$joined <- timestampToDate(meetupMembers$joinDate)
 
dd = aggregate(meetupMembers$joined, by=list(format(meetupMembers$joined, "%A")), function(x) length(x))
colnames(dd) = c("dayOfWeek", "count")

which returns the following:

> dd
  dayOfWeek count
1    Friday   135
2    Monday   287
3  Saturday    80
4    Sunday   102
5  Thursday   187
6   Tuesday   286
7 Wednesday   211

We need to use plyr’s ddply function which takes a data frame and transforms it into another one.

To refresh, this is what the initial data frame looks like:

> meetupMembers[1:10,]
       joinDate              joined
1  1.376572e+12 2013-08-15 14:13:40
2  1.379491e+12 2013-09-18 08:55:11
3  1.349454e+12 2012-10-05 17:28:04
4  1.383127e+12 2013-10-30 09:59:03
5  1.372239e+12 2013-06-26 10:27:40
6  1.330295e+12 2012-02-26 22:27:00
7  1.379676e+12 2013-09-20 12:22:39
8  1.398462e+12 2014-04-25 22:41:19
9  1.331734e+12 2012-03-14 14:11:43
10 1.396874e+12 2014-04-07 13:32:26

Most of the examples of using ddply show how to group by a specific ‘column’ e.g. joined but I want to group by part of the value in that column and eventually came across an example which showed how to do it:

> ddply(meetupMembers, .(format(joined, "%A")), function(x) {
    count <- length(x$joined)
    data.frame(count = count)
  })
  format(joined, "%A") count
1               Friday   135
2               Monday   287
3             Saturday    80
4               Sunday   102
5             Thursday   187
6              Tuesday   286
7            Wednesday   211

Unfortunately the generated column heading for the group by key isn’t very readable and it took me way longer than it should have to work out how to name it as I wanted! This is how you do it:

> ddply(meetupMembers, .(dayOfWeek=format(joined, "%A")), function(x) {
    count <- length(x$joined)
    data.frame(count = count)
  })
  dayOfWeek count
1    Friday   135
2    Monday   287
3  Saturday    80
4    Sunday   102
5  Thursday   187
6   Tuesday   286
7 Wednesday   211

If we want to sort that in descending order by ‘count’ we can wrap that ddply in another one:

> ddply(ddply(meetupMembers, .(dayOfWeek=format(joined, "%A")), function(x) {
    count <- length(x$joined)
    data.frame(count = count)
  }), .(count = count* -1))
  dayOfWeek count
1    Monday   287
2   Tuesday   286
3 Wednesday   211
4  Thursday   187
5    Friday   135
6    Sunday   102
7  Saturday    80

From reading a bit about ddply I gather that its slower than using some other approaches e.g. data.table but I’m not dealing with much data so it’s not an issue yet.

Once I got the hang of how it worked ddply was quite nice to work with so I think I’ll have a go at translating some of my other code to use it now.

Written by Mark Needham

July 2nd, 2014 at 6:30 am

Posted in R

Tagged with