1
$\begingroup$

Sorry for a very poorly worded title

My project is primarily for a stylistic display of words to layperson regarding their frequency of occurrences by way of font size. The output only needs to convey

1) Which word is the most frequent

2) Is a particular word more frequent than the other?

I am gunning for a presentable visual effect.

Here is my problem:

I am given an arbitrary list of words and their frequency. The distribution of the actual words and frequency are not known at advance. I want to to assign an appropriate font size to each word with respect to its frequency. A higher frequency word will have a larger font size. So it will be shown more prominently.

There are usually two kind of distributions:

1) A very evenly distributed list.

The frequencies are all within say 1 standard deviation of the mean

2) A very unbalanced list with extreme outliers on both end

I will use python to construct my example:

Here is the situation 1

import math


FONT_SIZE=96


def genImage1(wls, maxsize=100, color='red'):
    for wl in wls:
        print "<div style=\"font-size:%d;color:%s\">%s</div>" % (wl[1]/float(maxsize)*FONT_SIZE, color, wl[0])

wl = [['inhuman', 100], ['ironman', 90], ['hulk', 95], ['defenders', 85],
      ['punisher', 80], ['jessica jones', 83], ['daredevil', 80], ['x-men', 76],
      ['wolverine', 73], ['deadpool', 69], ['spiderman', 65], ['magneto', 60],
      ['Jean Grey', 55], ['Captain America', 53], ['Black widow', 49], ['Guardian of Galaxy', 41]];
genImage1(wl, wl[0][1])

It is very straightforward and here is the outcome. It is what I expect to see.

enter image description here

Now here is a second list, a list of uneven distribution

wl = [['inhuman', 10000], ['ironman', 3090], ['hulk', 1395], ['defenders', 1285],
      ['punisher', 1180], ['jessica jones', 1083], ['daredevil', 980], ['x-men', 976],
      ['wolverine', 873], ['deadpool', 769], ['spiderman', 665], ['magneto', 60],
      ['Jean Grey', 55], ['Captain America', 53], ['Black widow', 49], ['Guardian of Galaxy', 41]];

If I reuse the same routine above, of course it is not going to work

enter image description here

As you can see, a lot of words simply disappeared because the first word inhuman is too large by proportion.

I can use log to compress both ends and produce a better result:

 def genImage2(wls, maxsize=100, color='red'):
    for wl in wls:
        new_size = math.log(wl[1])/math.log(maxsize ) 
        print "<div style=\"font-size:%d;color:%s\">%s</div>" % (new_size*FONT_SIZE, color, wl[0])

enter image description here

However here is the problem:

The frequency of defenders 1285 and that of spiderman is 665. I would like them to appear more different in size. With log, this proportionality is simply removed (i.e. defenders should be roughly twice the size of spiderman )

My question:

Is there any mathematical function that can allow me to compress both end of outliers and yet preserve some level of proportionality for the numbers in the middle?

I would use the following chart to explain what I want to achieve:

enter image description here

Blue line represent the raw data set. Red line represents the ideal transformation of the original data series. I basically want to bring an outlier more into line while the difference between numbers in between both extreme is still visible to a viewer.

$\endgroup$
5
  • $\begingroup$ I find the phrase "compress both ends of outliers" confusing (you seem to be using the first and last words in two different senses each there, for example), and it's also not clear exactly what you mean by "preserve proportionality" in the title and near the end when you say "this proportionality is simply removed". Proportional to what exactly? It can't be to the original count as you're deliberately choosing to avoid that. Can you draw a plot of font size (as a number) against frequency for each example and add a drawing over the top (perhaps by hand) that illustrates what you seek? $\endgroup$
    – Glen_b
    Commented Nov 20, 2017 at 3:38
  • $\begingroup$ Regarding this proportionality is simply removed I mean I want to be able to show the audience defenders is actually different from spiderman (defenders is twice the size). I will try come up with a plot $\endgroup$ Commented Nov 20, 2017 at 3:58
  • $\begingroup$ I have added a plot to the question. $\endgroup$ Commented Nov 20, 2017 at 4:36
  • $\begingroup$ Have you actually looked at what the font sizes corresponding to your red curve look like? ... you may be surprised. $\endgroup$
    – Glen_b
    Commented Nov 20, 2017 at 5:59
  • $\begingroup$ Perhaps a sigmoid function would be helpful here? en.wikipedia.org/wiki/Sigmoid_function It wouldn't really "preserve proportionality" across the whole range, but it might achieve the effect you are looking for. $\endgroup$ Commented Nov 20, 2017 at 15:06

1 Answer 1

3
$\begingroup$

It kind of sounds like you're after a transformation that reduces the size of very large values but less strongly than logarithms.

There's a large number of possibilities, but perhaps you might consider something in a class of power transformations. In particular, sometimes with counts square roots may be convienient (though with the sort of data you're talking about you may want something a little stronger).

At the same time, you may want to add a linear transformation on the end so that the largest and smallest values fit in with your perception of how much the size should change overall. That is, rather than just take square-roots, rescale so that the largest and smallest font sizes fit some predetermined idea of their relative size (perhaps that the smallest point size is between 35 and 50% of the largest, or something like that).

$\endgroup$
4
  • $\begingroup$ Appreciate the answer. Is there any example of power transformation? $\endgroup$ Commented Nov 20, 2017 at 3:58
  • $\begingroup$ Unfortunately my internet has gone out. The edits I made to the above won't post and my access will be limited for a few days. Having looked at your data I think you may have a different problem to the one I originally thought from your post. $\endgroup$
    – Glen_b
    Commented Nov 20, 2017 at 5:59
  • $\begingroup$ Looks like it's back now; I may be able to reconstruct the edit $\endgroup$
    – Glen_b
    Commented Nov 21, 2017 at 1:16
  • $\begingroup$ Thanks a lot for your time and effort! Really appreciate $\endgroup$ Commented Nov 21, 2017 at 4:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.