Entity["Country", "WestBank"]["Properties"] Entity["Country", "WesternSahara"]["Properties"]

Thank you @Abrita Chakravarty. I have done suggested modifications.

I have found 7 properties beginning with small letter "w", by Common Names:

In[1]:= Select[CommonName /@ EntityProperties["Country"], StringMatchQ[#, "w" ~~ __] &]

Out[1]= {"wage and salaried workers",
          "wage and salaried workers fraction",
          "wage cost index",
          "water area",
          "water productivity",
          "waterway length",
          "wholesale price index"}

and 9, by Canonical Names:

In[2]:= Select[CanonicalName /@ EntityProperties["Country"], StringMatchQ[#, "W" ~~ __] &]

Out[2]= {"WageAndSalariedWorkers",
         "WageAndSalariedWorkersFraction",
         "WagesCostIndex",
         "WaterArea",
         "WaterArrivals",
         "WaterProductivity",
         "WaterwayLength",
         "Workforce",
         "WPI"}

2 Canonical Names have Common Names that do not start with letter "w".

It looks so that the correct result is 7, as the formulation ask for names starting with the small letter "w"..

Hi Valeriu,

The question asks for 'beginning with the letter "w"'. You need to anchor the pattern to the start of the string or it will match "w" anywhere in the name.

StartOfString ~~ "w" | "W" ~~ ___

@Abrita Chakravarty

Cases[
    EntityProperties["Country"],
    x_/;Or@@StringMatchQ[
       {CanonicalName[x],CommonName[x]},
       "W*",
       IgnoreCase->True
    ]
]

Thank you, Rohit, for your suggestion!

Daily Challenge (Day 7):

This one is from EIWL Chapter 31-- Make a histogram of where the letter “e” occurs in the words in WordList[ ]

@Valeriu Ungureanu In my experience when working with Entities it is advisable to look for both CanonicalName and CommonName. CanonicalName is the actual field name in the database so in many databases they are kept as short and cryptic to avoid naming conflicts. As an example, if you create your own SQL database and map it onto the EntityFramework using RelationalDatabase, you will see that database field names are mapped to CanonicalName. If you restrict your searching to CanonicalName only you will miss out on many relevant properties.

Regarding the question, it didn't specify what name it was referring to (that's why I did an exhaustive search using the Or in my code) but I agree one can become pedantic and take "w" as a literal for the lower case. But I don't think that was the intention of the question.

My code for today (day 7) challenge:

Last /@ Position[Characters@WordList[], "e"] // Histogram

A shorter code:

Histogram@Flatten@StringPosition[WordList[], "e"]

With such code, day 7 challenge becomes a simple problem :)

Thank you, Rohit! You are right! At the moment, I can modify the code to obtain the correct result, but it will not be as short as the precedent:

Last /@ Flatten[StringPosition["e"]@WordList[], 1] // Histogram

Another modification of the code:

Histogram@Flatten@Map[Union, StringPosition[WordList[], "e"], 2]

or

Histogram@Flatten[Union @@@ StringPosition[WordList[], "e"]]

Here is my clunky attempt at it. Probably there is a better, more humane way to do this.

Another possibility:

d = Dataset[Association[
   "Padmé Amidala" -> <|"ID" -> "19435", "Planet" -> "Naboo"|>, 
   "Bail Organa" -> <|"ID" -> "11114", "Planet" -> "Alderaan"|>, 
   "Mon Mothma" -> <|"ID" -> "10712", "Planet" -> "Chandrila"|>]]

Apply[Association, d[#, "ID"] -> <|"Name" -> #, "Planet" -> d[#, "Planet"]|> & /@ Keys[d]]

If every letter is used exactly once, then the following code solves the problem:

In[1] :=  Intersection[StringJoin /@ Permutations[{"z", "t", "c", "a", "u", "h", "p"}],
          WordList[]]
Out[1] = {"chutzpa"}

or, thanks the Rohit's idea to use the function Complements:

In[2]:= l = {"z", "t", "c", "a", "u", "h", "p"};

If[Complement[l, Characters@#] == {}, #, Nothing] & /@ WordList[]

Out[2]= {"chutzpa", "chutzpah"}

Shorter version

WordList[] // Select[Complement[letters, Characters[#]] == {} &]
(* {"chutzpa", "chutzpah"} *)

@Abrita Chakravarty I have a query regarding the question.

For the letters, {"a", "o", "l", "r", "w", "f", "m"}, can the pangram apart from containing at least one of each letter from the list also contain some of the remaining letters? Like for example "flamethrower" contains all of the letters from the list but also contains "e" twice, "h" and "t" once which are not part of the list. So is "flamethrower" also regarded as a pangram of this list in addition to "wolfram"?

@Abrita Chakravarty First unlike WordList, DictionaryLookup supports patterns. Moreover DictionaryLookup seems to have more words than the WordList (probably a bug).

WordList[]//Length(*39176*)
DictionaryLookup[]//Length(*92518*)
Complement[WordList[],DictionaryLookup[]](*{}*)

Here is my solution:

Clear[pangram];
pangram[characters_List]:=Module[{$characterSet="["<>StringJoin@characters<>"]*"},DictionaryLookup@RegularExpression[StringJoin["(?="<>$characterSet<>#<>")"&/@characters,$characterSet]]];
pangram[{"a","o","l","r","w","f","m"}](*{"wolfram"}*)
pangram[{"z","t","c","a","u","h","p"}](*{"chutzpa","chutzpah"}*)
pangram[{"e","t","l","a","b"}](*{"ablate","ballet","battle","beatable","bleat","eatable","table","tablet"}*)

Its orders of magnitude faster than doing surgery on WordList[] or DictionaryLookup[] alone.

DictionaryLookup[] contains more words, really:

In[1]:= Complement[DictionaryLookup[], WordList[]] // Length

Out[1]= 53342

So, we can construct other codes with DictionaryLookup[], such as for example:

In[1]:= DictionaryLookup[x__ /; SubsetQ[Characters@x, l]]

Out[1]= {"chutzpa", "chutzpah"}

Try your code on {"a", "o", "l", "r", "w", "f", "m"}

DictionaryLookup[x__ /; SubsetQ[Characters@x, {"a", "o", "l", "r", "w", "f", "m"}]]
{"flamethrower", "flamethrowers", "flatworm", "flatworms", "mayflower", "mayflowers", "wolfram"}

It produces incorrect result.

Yet another way:

DeleteMissing[{#, PartOfSpeech[#]} & /@ StringJoin /@ Permutations[{"z", "t", "c", "a", "u", "h", "p"}], 1, Infinity]