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Histograms with strings, grouped by repeat count (w/ data)

6 messages · Matthew Trunnell, jim holtman, Deepayan Sarkar

Original
Matthew Trunnell
# 
Hello R gurus,

I just spent my first weekend wrestling with R, but so far have come
up empty handed.

I have a dataset that represents file downloads; it has 4 dimensions:
date, filename, email, and country.  (sample data below)

My first goal is to get an idea of the frequency of repeated
downloads.  Let me explain that.  Some people tend to download
multiple times, e.g. if the download fails they keep trying over and
over.  I'm trying to build a histogram that shows the repeat count
along the x-axis, that is, how many people downloaded once, twice,
three times, etc.  I plan to compare the median of that before and
after we switched ISPs.

To accomplish this, I'm assuming that I'll first need to combine the
email and filename columns so as to represent a single download
attempt by an individual.  Does that sound right?  Later, it would be
nice to limit the histogram to a single filename, country, or company.
 I can probably figure that out myself after I understand how to write
this funky histogram expression.

With the help of Verzani's introductory text, I've learned how to read
in the CSV data and do some simple tables, like this:

hist(table(d$filename))
hist(table(d$filename[substring(d$filename, 1, 5)=="file1"]))
hist(sort(table(d$filename[substring(d$filename, 1, 5)=="file1"])))

Obviously, these commands count the frequency of the files.  What I'd
like to see are the repeats grouped along the x-axis;  I'd like to
find, for all files, the distribution of retries.  I hope that makes
sense. :)

Can someone point me in the right direction?  I'm very new to R and to
statistics, but I write code for a living.  At this point I'd almost
be better off writing a program do this kind of simple counting... but
I have a feeling R would be so useful if I could just get past the
initial learning curve.

Thank you in advance,
Matt

Here's some real data, with the private info replaced :)

 d<-read.table(file="C:\\users\\trunnellm\\downloads\\statistics\\downloads.csv",
sep=",", quote="\"", header=TRUE)

filename,last_modified,email_addr,country_residence
file1,3/4/2006 13:54,email1,Korea (South)
file2,3/4/2006 14:33,email2,United States
file2,3/4/2006 16:03,email2,United States
file2,3/4/2006 16:17,email3,United States
file2,3/4/2006 16:28,email3,United States
file3,3/4/2006 19:13,email4,United States
file2,3/4/2006 21:22,email5,India
file4,3/4/2006 21:46,email6,United States
file1,3/4/2006 22:04,email7,Japan
file2,3/4/2006 22:09,email8,Croatia
file1,3/4/2006 22:22,email7,Japan
file1,3/4/2006 22:29,email9,India
file1,3/4/2006 23:06,email6,United States
file1,3/4/2006 23:33,email6,United States
file5,3/4/2006 23:44,email10,China
file1,3/5/2006 0:13,email9,India
file2,3/5/2006 0:52,email8,Croatia
file2,3/5/2006 0:54,email8,Croatia
file2,3/5/2006 1:10,email5,India
file6,3/5/2006 2:17,email9,India
file2,3/5/2006 2:24,email11,Italy
file7,3/5/2006 2:36,email12,Italy
file8,3/5/2006 2:52,email12,Italy
file2,3/5/2006 3:09,email13,United Kingdom
file2,3/5/2006 4:02,email14,India
file2,3/5/2006 4:07,email14,India
file2,3/5/2006 4:14,email14,India
file2,3/5/2006 4:37,email5,India
file2,3/5/2006 4:44,email15,Belgium
file1,3/5/2006 5:02,email9,India
file1,3/5/2006 5:24,email16,Taiwan
file2,3/5/2006 6:06,email17,Saudi Arabia
file2,3/5/2006 7:32,email17,Saudi Arabia
file2,3/5/2006 8:12,email18,Brazil
file2,3/5/2006 8:26,email18,Brazil
file2,3/5/2006 9:49,email19,United Kingdom
file1,3/5/2006 10:49,email11,Italy
file1,3/5/2006 11:16,email13,United Kingdom
file1,3/5/2006 11:16,email13,United Kingdom
file1,3/5/2006 11:45,email13,United Kingdom
file1,3/5/2006 14:34,email20,Australia
file9,3/5/2006 14:56,email20,Australia
file9,3/5/2006 14:56,email20,Australia
file5,3/5/2006 16:43,email21,United States
file1,3/5/2006 17:17,email7,Japan
file2,3/5/2006 17:26,email22,Japan
file2,3/5/2006 17:27,email22,Japan
file2,3/5/2006 17:33,email23,China
file1,3/5/2006 17:45,email22,Japan
file2,3/5/2006 17:45,email22,Japan
file2,3/5/2006 17:59,email23,China
file1,3/5/2006 18:27,email24,Japan
file1,3/5/2006 18:47,email25,Taiwan
file2,3/5/2006 18:48,email26,New Zealand
file2,3/5/2006 19:15,email27,Canada
file2,3/5/2006 19:23,email28,Canada
file2,3/5/2006 19:24,email28,Canada
file10,3/5/2006 19:49,email29,Japan
file10,3/5/2006 19:52,email29,Japan
file10,3/5/2006 19:57,email29,Japan
file2,3/5/2006 20:01,email29,Japan
file2,3/5/2006 20:02,email29,Japan
file2,3/5/2006 20:06,email29,Japan
Matthew Trunnell
# 
Jim,
Thanks for the quick reply!  When I run your code, I end up with a
single barplot of one datapoint, file9 vs email20 == 2.0.  I see the
call to barplot is inside a for loop... maybe it's zooming through the
display of many barplots, but all I see is the last one?

In any case, I need to figure out the distribution of the retries, such as
No. Retries   Count
1                 6
2                 13
3                 5
4                 3
5                 2
6                 1

That is, 6 people retried the download once; 13 people retried the
download twice, etc.  So it would be counting the frequency of the
email-filename combination, and grouping those together by the number
of retries.  Does that make sense?

When I look at the counts object from your code, I can see that it's
close to what I need.  How do I access the properties of the counts
object-- it's a table, right?  If I look at counts[1,1], that returns
1.  But how do I get at the row/col name of that cell?  Is that cell
an object?  rownames(counts[1,1]) returns null.

Thanks,
Matt
On 6/18/07, jim holtman <jholtman at gmail.com> wrote:
Matthew Trunnell
# 
Aha!  So to expand that from the original expression,
0   1   2   3
253  20   8   9

I think that is exactly what I'm looking for.  I knew it must be
simple!!!  What does the 0 column represent?

Also, does this tell me the same thing, filtered by Japan?
0   1   2   3
958   5   2   1

How does that differ logically from this?
0  1  2  3
51  4  2  1

I don't understand why that produces different results.  The first one
adds a third dimension to the table, but limits that third dimension
to a single element, Japan.  Shouldn't it be the same?  And again,
what's that zero column?

Thank you,
Matt
On 6/18/07, jim holtman <jholtman at gmail.com> wrote:
Deepayan Sarkar
#
On 6/18/07, Matthew Trunnell <trunnell at cognix.net> wrote:
Number of unique filename:email_addr combinations that don't occur in the data.
No it doesn't.
[1] 4350
[1] 63

You are using an index that is meaningless.


There's an alternative tabulation function that uses a formula
interface similar to that used in modeling functions; this might be
more transparent for your case:
+     xtabs(~filename + email_addr, data = d,
+           subset = country_residence == "Japan")
count
  0   1   3
284   2   4
This is also using meaningless indexing.

Note, incidentally, that you are indexing a matrix of dimension 10x29
as if it were a vector of length 290, which is probably not what you
meant to do anyway:
'table' int [1:10, 1:29] 1 0 0 0 0 0 0 0 0 0 ...
 - attr(*, "dimnames")=List of 2
  ..$ : chr [1:10] "file1" "file10" "file2" "file3" ...
  ..$ : chr [1:29] "email1" "email10" "email11" "email12" ...

You need to read help(Extract) carefully and play around with some
simple examples.
As before, they are the empty combinations.

-Deepayan