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a. Get the list whose elements are of the form {language,number of words in dictionary)

I tried this, and it gave me the number if words in dictionary, but I don't know how to get the language part:

lang = DictionaryLookup[All];

langlength =Table[{lang[[i]], Length[DictionaryLookup[{lang[[i]], All}]]}, {i, 1,
Length[lang]}]

b. Now sort this list in descending order by the number of words and get a list of the form {position in sorted list, language, number of words}.

For part b, I have

sorted = Sort[langlength, #1[[2]] >= #2[[2]] &]
placeList[list_] := Table[{i, list[[i]]}, {i, 1, Length[list]}] 
pl = placeList[sorted]

And I got

{1, {"Finnish", 728498}, 2, {"Catalan", 602208}, 3, {"Galician", 515385}, 4, {"Portuguese", 459996}, 5, {"Hebrew", 455264}, 6, {"Danish", 376469}, 7, {"BrazilianPortuguese", 264713}, 8, {"Croatian", 260752}, 9, {"Polish", 234907}, 10, {"Hungarian", 230206}, 11, {"Dutch", 229368}, 12, {"French", 139657}, 13, {"Swedish", 121430}, 14, {"Italian", 116854}, 15, {"Faroese", 108508}, 16, {"English", 92518}, 17, {"BritishEnglish", 86135}, 18, {"Spanish", 86016}, 19, {"German", 76155}, 20, {"IrishGaelic", 65203}, 21, {"Arabic", 43768}, 22, {"Breton", 32733}, 23, {"Russian", 31801}, 24, {"Hindi", 15983}, 25, {"ScottishGaelic", 15670}, 26, {"Esperanto", 14780}, 27, {"Latin", 8777}}`

From here I don't know how to get rid of the braces around Finnish, 728498 and others.

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Try this

langlength = ({#, Length@DictionaryLookup[{#, All}]} & /@ DictionaryLookup[All]);
lst = Sort[langlength, #1[[2]] > #2[[2]] &];
TableForm[lst, TableHeadings -> {Range[Length[lst]], None}]

Edit

langlength = ({#, Length@DictionaryLookup[{#, All}]} & /@ DictionaryLookup[All]);
lst = Sort[langlength, #1[[2]] > #2[[2]] &];
placeList = Table[{i, lst[[i]]}, {i, 1, Length[lst]}]
TableForm[placeList, TableDepth -> 2]
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  • $\begingroup$ I actually figured out part a. For part b, I have placeList[list_] := Table[{i, list[[i]]}, {i, 1, Length[list]}] $\endgroup$ – user720329 Apr 18 '14 at 7:02
  • $\begingroup$ @molekyla777 Can you help me with part b. I already figured out most of the part. I just need to get rid of those braces. $\endgroup$ – user720329 Apr 18 '14 at 21:46
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Let me answer your question by first clarifying some of the fundamentals about Mathematica:

A. Mapping over list. If you have a function, you can apply it to a list of values by using Map:

Range[10]

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Map[Sqrt, Range[10]]

$\left\{1,\sqrt{2},\sqrt{3},2,\sqrt{5},\sqrt{6},\sqrt{7},2 \sqrt{2},3,\sqrt{10}\right\}$

You can do it with your own functions as well:

f[n_] := n^2 + 1 (* this function squares it's argument and adds 1 *);
Map[f, Range[10]]

{2, 5, 10, 17, 26, 37, 50, 65, 82, 101}

B. Return values. You own functions can return any kind of value, including a list:

f[n_] := {n, n^2, N[Sqrt[n]]} (* returns a list of three elements: a number, 
                                 it's square and it's square root *)
Map[f, Range[10]]

Here, you get a list of lists as a result:

{{1, 1, 1.}, {2, 4, 1.41421}, {3, 9, 1.73205}, {4, 16, 2.}, {5, 25, 2.23607}, {6, 36, 2.44949}, {7, 49, 2.64575}, {8, 64, 2.82843}, {9, 81, 3.}, {10, 100, 3.16228}}

C. Solving (a). For your first question, you seem to know already how to return a list of all languages known to Mathematica

DictionaryLookup[All]

{"Arabic", "BrazilianPortuguese", "Breton", "BritishEnglish", "Catalan", "Croatian", "Danish", "Dutch", "English", "Esperanto", "Faroese", "Finnish", "French", "Galician", "German", "Hebrew", "Hindi", "Hungarian", "IrishGaelic", "Italian", "Latin", "Polish", "Portuguese", "Russian", "ScottishGaelic", "Spanish", "Swedish"}

as well as how to get a list of the form {language,number of words in dictionary}

wordsInLanguage[lang_] := {lang, Length[DictionaryLookup[{lang, All}]]}
wordsInLanguage["Russian"]

{"Russian", 31801}

so you can just use Map to get what your need in (a):

Map[wordsInLanguage, DictionaryLookup[All]]

{{"Arabic", 43768}, {"BrazilianPortuguese", 264713}, {"Breton", 32733}, {"BritishEnglish", 86135}, {"Catalan", 602208}, {"Croatian", 260752}, {"Danish", 376469}, {"Dutch", 229368}, {"English", 92518}, {"Esperanto", 14780}, {"Faroese", 108508}, {"Finnish", 728498}, {"French", 139657}, {"Galician", 515385}, {"German", 76155}, {"Hebrew", 455264}, {"Hindi", 15983}, {"Hungarian", 230206}, {"IrishGaelic", 65203}, {"Italian", 116854}, {"Latin", 8777}, {"Polish", 234907}, {"Portuguese", 459996}, {"Russian", 31801}, {"ScottishGaelic", 15670}, {"Spanish", 86016}, {"Swedish", 121430}}

D. Flattening lists. Now, to answer your second question, you need to learn how to flatten lists. For example, given a list {1, 2}, how can you get a list {1, 2, 3}?

There are quite a few ways:

(* works if you know that the list only contains two elements *)
addThirdElement[list_, element_] := {list[[1]], list[[2]], element}
addThirdElement[{1, 2}, 3]

{1, 2, 3}

Another option is this - you can even reorder elements in the list:

addThirdElement1[{first_, second_}, third_] := {third, second, first}
addThirdElement1[{1, 2}, 3]

{3, 2, 1}

There are many more - e.g. look up Flatten in Mathematica documentation.

That should give you enough to solve the last part. Good luck!

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  • $\begingroup$ I actually have trouble with the "Flatten" command. I tried quite a few times, but it doesn't work. $\endgroup$ – user720329 Apr 18 '14 at 22:11
  • $\begingroup$ If you want to get help, you would first need to explain: 1) What have you tried (with code); 2) What do you expect; 3) What did you receive instead. You don't need to use Flatten to solve (b) - just try to read and understand the section D above. $\endgroup$ – Victor K. Apr 18 '14 at 22:14
  • $\begingroup$ I tried Flatten[pl,1] and Flatten[pl,2], but none of these worked. And if you look at my part b, you'd know what is pl, in my case. $\endgroup$ – user720329 Apr 18 '14 at 22:17
  • $\begingroup$ Let's read Flatten documentation together. What does it do? $\endgroup$ – Victor K. Apr 18 '14 at 22:18
  • $\begingroup$ Well, it flattens list to a certain level $\endgroup$ – user720329 Apr 18 '14 at 22:22

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