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Very nice results! Cool stuff! Your wordvalue is fast, but looks very convoluted!

I came up with two ways:

rules = Dispatch[# -> LetterNumber[#] & /@ Alphabet[]]
wordvalue2[word_String] := Total[Characters[ToLowerCase@RemoveDiacritics@word] /. rules]
wordvalue3[word_String] := Plus @@ LetterNumber@ToLowerCase@RemoveDiacritics@word

Note that I do Plus@@ rather than Total@ in order to deal with 1 letter words...

In your version you could replace StringPartition[...,1] by Characters.

Absolutely marvelous! But could there be another way? I think many. Here is one.

Don't you think ranking letters according the order they appear in the alphabet is sort of random? I suggest we rank them according to their frequency in the full dictionary.

letterRANK = MapIndexed[Prepend[#1, First[#2]] &, 
   Sort[Tally[Select[Flatten[Characters[ToLowerCase[WordData[]]]], LetterQ]], #1[[2]] > #2[[2]] &]];

As you can see (first number) I ranked (gave most points) the highest the rarest letters. I consider them jems because I guess they make up the rarest (I hope the funniest?) words.

Multicolumn[letterRANK, 2]

And I must say the frequencies do give a weird graph:

ListLinePlot[letterRANK[[All, 3]], PlotTheme -> "Business", Filling -> Bottom, 
 FrameTicks -> { {Automatic, None}, {letterRANK[[All, {1, 2}]], None}}, GridLines -> All]

Now let's make our little indexed database:

indexRANKED = Association[Thread[letterRANK[[All, 2]] -> letterRANK[[All, 1]]]]

Here is Rodrigo's function:

wordValue[word_String] := 
 Total[Lookup[indexRANKED, #, 0] & /@ StringPartition[ToLowerCase@RemoveDiacritics@word, 1]]

and here it is - our new jem-like 100-points scoring words:

words100 = Select[DictionaryLookup[], wordValue[#] == 100 &];

But let's go a step further. Consider the most POPular words those who get 100 points the quickest --- with least amount of letters. And consider the most RARE words those who get 100 points the slowest --- with most amount of letters. We weigh them according to this logic for word cloud:

weightPOP = {#, 1/StringLength[#]} & /@ Union[ToLowerCase[words100]];
weightRARE = {#, StringLength[#]} & /@ Union[ToLowerCase[words100]];

And now behold their corresponding word clouds - I bet you will grab the dictionary. A recent quote from a friend looking at this "I'm thrilled to my very kutuzov." ;-)

WordCloud[weightPOP]
WordCloud[weightRARE, ScalingFunctions -> (#^2 &)]

Good Job!!! :-)

I am wondering how the word cloud would looks like if we set the words size according Google search statistics.

In order to get that I did a simple function:

estimatedResultCount[word_] :=
 ToExpression[
  Import[StringJoin[
     "https://www.googleapis.com/customsearch/v1?key=\
<Google KEY>&cx=<Google CX>&q=", word], "JSON"][[3, 2, 1, 2, 1, 1, 2]]]

We pass a word to the function, and it search at google and get a JSON message back, we go deep at the JSON message and get the number we are interested in.

It is important to notes that to get the Google KEY and CX you should register as a google developer, check this out:

Google Developer Console

There are some limits of requests by day and requests by second (time) at Google.

The result is:

sample = RandomSample[words100, 10];
WordCloud[#] &[{#, estimatedResultCount[#]} & /@ sample]