In my last post I talked about some of the challenges of reproducing the analysis, offered in Stephen Ramsay’s Reading Machines of the distinctiveness of characters’ vocabulary in Woolf’s The Waves. Here I follow up a bit more on why I was unable to get the same results Ramsay reports.

R, tm, and weightTfIdf

To get R to generate tf—idf scores, one can do the following:

# Generate a Document Term Matrix
dtm <- DocumentTermMatrix(corpus, control=list(weighting=weightTfIdf))

This was how I got my scores last time; and those scores didn’t match the one’s reported by Ramsay. So, one explanation is that I bollocks’d the data; the other is that there may be something going on under the hood.

To explore this latter possibility, I had a look at the weightTfIdf function.Doing so is easy; if you’re in an R console, with the tm package loaded, just type weightTfIdf and you’ll get all 26 lines of the code. It is worth noting that by default, the function normalizes its scores according to text. That is, rather than calculating the tf-idf score using a raw count of the number of times a particular term appears in a single document (in our case, the number of times a particular character uses a word), it divides the raw term frequency by the total number of words in that document (in our case, the total number of words said by a character). This normalizing process explains why the scores Ramsay reports for Louis (12) start at 5.9, while my own scores were always small numbers (0.006…; 0.003…, etc).

Short digression on misunderstanding the code

When I first dug into weightTfIdf code, I thought I discovered a difference in implementation:In order to work with the data, this function transposes the Document Term Matrix into a Term Document Matrix; this means that rather than the columns being terms and the rows texts, the rows are terms and the columns documents. I don’t know why the function does this; but it does, and that explains why we’re looking at row sums in what follows rather than column sums.

lnrs <- log2(nDocs(m)/rs)

Here we’re calculating part of the tf-idf score—the part that corresponds to Ramsay’s log(N/df). Ignore the log2 for a moment and look at that divisor, rs. That should represent the number of documents which contain a particular term. How is that calculated? Well, the code says:

rs <- row_sums(m > 0)

That row_sums function comes the slam package; and when I first saw this, I thought I had my explanation. Ah ha! If the divisor in the logarithm is the sum of a row, then it is dividing the number of documents not by the number of documents which contain the specified term, but by the sum of the row—that is, by the total occurrences of that term! Here is our explanation!

“Well, but what is that comparison, m > 0, doing in the function call,” you ask? Good question! Well, I assumed that it was just a way to pass non-zero values to the row_sums function, and went about trying to rewrite the function properly. This heady delirium lead to questions on StackOverflow and wasted time. Because, of course, I totally misunderstood how this part of the code works.

The trick is that a relational operator (like >) on a matrix, returns a matrix with boolean values, evaluating that expression for each item in the matrix. So, each cell in the term document matrix is now either TRUE or FALSE based on whether the value of that term was greater than 0. So it might look something like this:

             Docs
Terms         bernard.txt jinny.txt louis.txt neville.txt rhoda.txt susan.txt
  absorption         TRUE     FALSE     FALSE       FALSE     FALSE     FALSE
  abstract           TRUE     FALSE     FALSE       FALSE     FALSE     FALSE
  abstraction       FALSE     FALSE      TRUE       FALSE     FALSE     FALSE
  absurd             TRUE     FALSE     FALSE        TRUE     FALSE     FALSE
  absurdity          TRUE     FALSE     FALSE        TRUE     FALSE     FALSE
  absurdly          FALSE     FALSE     FALSE        TRUE     FALSE     FALSE

And, since TRUE is treated as numerically equivalent to 1, and FALSE to 0, if we were to sum these rows, we would get… exactly what we were looking for; i.e., the number of documents in which the term appears. So, that was wasted time.

Another thing to note here is that weightTfIdf uses log2(), the binary logarithmWhich returns the exponent to which you would raise 2 in order to get the specified term; i.e. log2(16)=4, b/c 2^4=16.. The logarithm here works to essentially scale the term’s frequency based on how specific the term is to any particular document; terms which occur in all documents will have N nearly equal to df, or N/df nearly equal to 1. And log(1) == log2(1) == 0. That is, it will push the weight to (or towards) 0. Whereas a term which occurs in only one document will maximize N/1. A logarithmic function is used simply to dampen that effect, preventing scores linearly increasing in the case of terms highly specific to a single document.

Does it make a difference whether one uses log2() or the natural logarithm or the base 10 “common” logarithm I learned back in school? I don’t really know; I doubt it. The particular logarithmic function one chooses will change the score, but it wouldn’t change the relationship among the terms (which had the highest score).

Hand Simulations

After all that, I tried to tinker a bit to bring results into line. I stopped normalizing my data and tried different log functions in an attempt to match Ramsay’s scores. Well, okay, let’s take one last look.

Despairing, I returned to the clearest data Ramsay gives us—the list of terms and scores which appears on page 12 and picked a few, to see if I could manage this by hand.My high school computer science teacher (we switched from Pascal to C++ midway through my high school career) used to make us “hand simulate” algorithms; print out the source code, jot down the variable names, and debug by hand. Smart guy. (Though the phrase “hand simulation” may have been unfortunate.) (I don’t really mean by hand of course. Just… more slowly.) Here are the terms Ramsay lists for Louis, which have a score of 5.0021615: australian, beast, grained, though, wilt.

So, I returned to my data.

> rawdtm <- DocumentTermMatrix(characters)  # A basic Document Term Matrix of raw frequency counts
> terms <- c('australian','beast','grained','thou','wilt')
> inspect(rawdtm[,terms])
  A document-term matrix (6 documents, 5 terms)

    Non-/sparse entries: 6/24
    Sparsity           : 80%
    Maximal term length: 10 
    Weighting          : term frequency (tf)

                 Terms
    Docs          australian beast grained thou wilt
      bernard.txt          1     0       0    0    0
      jinny.txt            0     0       0    0    0
      louis.txt            6     6       6    6    6
      neville.txt          0     0       0    0    0
      rhoda.txt            0     0       0    0    0
      susan.txt            0     0       0    0    0

This is already puzzling. These five terms all had the same score. Yet, australian occurs six times in Louis, but across two texts; while the other terms occurs six time in Louis, and only in Louis.

So, back to the data. I open up the original gutenberg and search through it for australian. And, indeed, looks like 7 total occurrences; one of which is from Bernard:

I thought how Louis would mount those steps in his neat suit with his cane in his hand and his angular, rather detached gait. With his Australian accent (“My father, a banker at Brisbane”) he would come, I thought, with greater respect to these old ceremonies than I do, who have heard the same lullabies for a thousand years.

Let’s check grained. I found six occurrences; all in Louis’s speech and, as I was doing this by hand, I noticed that all those occurrences of grained were in the phrase grained oak:

Why does grained occur in Ramsay & Steger’s list, but not oak? Well, maybe everyone talks about oak trees, but only Louis talks about grained oak. Back to our data:

>inspect(rawdtm[,c('grained','oak')])
A document-term matrix (6 documents, 2 terms)

Non-/sparse entries: 3/9
Sparsity           : 75%
Maximal term length: 7 
Weighting          : term frequency (tf)
                 Terms
Docs          grained oak
bernard.txt       0   2
jinny.txt         0   0
louis.txt         6  10
neville.txt       0   0
rhoda.txt         0   0
susan.txt         0   0

Hmm… well, let’s do the math for these three terms (the first two of which got the same score in Ramsay’s analysis; the last of which didn’t appear at all). For these three terms, australian, oak, and grained for Louis, we have:

             tf    df    N    1+tf*(log2(N/df))   1+tf*(log10(N/df))    1+tf*(log(N/df))
australian    6     2    6         10.50978          3.862728               7.59164
grained       6     1    6         16.50978          5.668909              11.75056 
oak           10    2    6         16.84963          5.771213              11.98612

So oak should have the highest score of the three.

Something is clearly wrong.

I’ve returned to the raw counts, double checked them against the original file; and computed these things individually. I’ve tried to imagine all possible ways, but have been unable to produce any of the scores listed on pg 12.

I suspect that there may be a problem in the Ramsay and Steger’s data. There is significant overlap in Ramsay and Steger’s results and my own. It is for Louis that our data is most consistent; we share 19 of the top 24 terms. It is least similar for Bernard (sharing only 4/24 terms) and Rhoda (11/24). Did the Ramsay & Steger’s analysis, perhaps, discard the final chapter where only Bernard talks (perhaps deliberately)? That would obviously greatly change the Bernard data set; and, by removing so much text, it could easily affect the df term for other characters, accounting for the other discrepancies. (But it wouldn’t explain oak.)

It may very well be my data that is the problem, but if so I’ve been unable to locate the error. I’m abandoning trying to reconcile these approaches to a method that should, in principle, be eminently reproducible. I will though summarize the top 24 terms with highest tf-idf scores, using the best data I’ve got, and not normalizing.

    > louis[1:24]
       western        oak      beast    grained       thou       wilt     accent 
     23.264663  15.849625  15.509775  15.509775  15.509775  15.509775  14.000000 
      boasting       nile    average     clerks     stamps australian       boys 
     12.679700  12.679700  10.339850  10.339850  10.000000   9.509775   8.000000 
      pitchers      steel     beaten      boast    bobbing    custard eatingshop 
      7.924813   7.924813   7.754888   7.754888   7.754888   7.754888   7.754888 
       england      eyres        ham 
      7.754888   7.754888   7.754888 
    > bernard[1:24]
    curiosity   hampton    letter   phrases     byron   elderly    heaven   married 
     25.84963  23.77444  23.26466  22.22858  22.18948  20.67970  20.67970  20.67970 
     observed    dinner    phrase    willow       fin    simple  describe      self 
     20.67970  20.60451  20.00000  19.01955  18.09474  18.09474  17.43459  17.43459 
        stick     sense     story    nature  pictures  thinking    canopy     enemy 
     17.43459  16.37895  16.00000  15.84963  15.84963  15.84963  15.50978  15.50978 
    > neville[1:24]
          story      doomed immitigable      papers       byron    catullus 
      12.000000   10.339850   10.339850   10.339850    7.924813    7.924813 
          cheep  perfection       camel      detect         don   hosepipes 
       7.924813    7.924813    7.754888    7.754888    7.754888    7.754888 
         hubbub       loads      mallet      marvel   squirting       waits 
       7.754888    7.754888    7.754888    7.754888    7.754888    7.754888 
           boys     founder      knives      pocket       scene shakespeare 
       7.000000    6.339850    6.339850    6.339850    6.339850    6.339850 
    > jinny[1:24]
        tunnel   prepared  billowing       game     native      peers    quicker 
      9.509775   7.924813   7.754888   7.754888   7.754888   7.754888   7.754888 
    melancholy     bodies       band    cabinet      coach       crag     dazzle 
      6.339850   5.264663   5.169925   5.169925   5.169925   5.169925   5.169925 
        deftly   equipped   eyebrows     felled  haymarket       jump    lockets 
      5.169925   5.169925   5.169925   5.169925   5.169925   5.169925   5.169925 
      matthews   murmured    prepare 
      5.169925   5.169925   5.169925 
    > rhoda[1:24]
       oblong      dips     tiger    fuller   swallow    fallen     steep suspended 
    18.094738 12.924813 11.094738 10.339850  9.509775  8.000000  7.924813  7.924813 
       cliffs   minnows      pond    terror     bunch      foam     party    puddle 
     7.754888  7.754888  7.754888  7.000000  6.339850  6.339850  6.339850  6.339850 
        pools   violets       bow   caverns     chirp     choke    column   columns 
     6.000000  6.000000  5.169925  5.169925  5.169925  5.169925  5.169925  5.169925 
    > susan[1:24]
       kitchen     setter    washing       bury       cart       gate      apron 
     15.849625  10.339850  10.339850   8.000000   7.924813   7.924813   7.754888 
       seasons   squirrel windowpane       beds     butter      clean        wet 
      7.754888   7.754888   7.754888   6.339850   6.339850   6.339850   6.339850 
         blown     winter       baby     bitten       boil   cabbages   carbolic 
      6.000000   5.264663   5.169925   5.169925   5.169925   5.169925   5.169925 
         clara     cradle       eggs 
      5.169925   5.169925   5.169925