Top rank per group: Difference between revisions
m (Fixed lang tags.) |
(added ocaml) |
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Kim Arlich, id=E10001, salary=57000 |
Kim Arlich, id=E10001, salary=57000 |
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Timothy Grove, id=E16398, salary=29900 |
Timothy Grove, id=E16398, salary=29900 |
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</pre> |
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=={{header|OCaml}}== |
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<lang ocaml>open StdLabels |
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let to_string (name,_,s,_) = (Printf.sprintf "%s (%d)" name s) |
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let take n li = |
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let rec aux i acc = function |
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| _ when i >= n -> (List.rev acc) |
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| [] -> (List.rev acc) |
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| x::xs -> aux (succ i) (x::acc) xs |
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in |
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aux 0 [] li ;; |
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let toprank data n = |
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let len = List.length data in |
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let h = Hashtbl.create len in |
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List.iter data ~f:(fun ((_,_,_,dep) as employee) -> |
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Hashtbl.add h dep employee); |
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let deps = |
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List.fold_left data ~init:[] ~f: |
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(fun ac (_,_,_,v) -> if List.mem v ac then ac else v::ac) in |
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let f dep = |
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Printf.printf "Department: %s\n " dep; |
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let l = Hashtbl.find_all h dep in |
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let l2 = List.sort (fun (_,_,v1,_) (_,_,v2,_) -> compare v2 v1) l in |
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let l3 = (take n l2) in |
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print_endline(String.concat ", " (List.map to_string l3)); |
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print_newline() |
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in |
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List.iter f deps; |
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;; |
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let data = [ |
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"Tyler Bennett", "E10297", 32000, "D101"; |
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"John Rappl", "E21437", 47000, "D050"; |
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"George Woltman", "E00127", 53500, "D101"; |
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"Adam Smith", "E63535", 18000, "D202"; |
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"Claire Buckman", "E39876", 27800, "D202"; |
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"David McClellan", "E04242", 41500, "D101"; |
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"Rich Holcomb", "E01234", 49500, "D202"; |
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"Nathan Adams", "E41298", 21900, "D050"; |
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"Richard Potter", "E43128", 15900, "D101"; |
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"David Motsinger", "E27002", 19250, "D202"; |
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"Tim Sampair", "E03033", 27000, "D101"; |
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"Kim Arlich", "E10001", 57000, "D190"; |
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"Timothy Grove", "E16398", 29900, "D190"; |
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] |
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let () = |
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toprank data 3; |
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;;</lang> |
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outputs: |
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<pre>Department: D190 |
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Kim Arlich (57000), Timothy Grove (29900) |
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Department: D202 |
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Rich Holcomb (49500), Claire Buckman (27800), David Motsinger (19250) |
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Department: D050 |
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John Rappl (47000), Nathan Adams (21900) |
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Department: D101 |
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George Woltman (53500), David McClellan (41500), Tyler Bennett (32000) |
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</pre> |
</pre> |
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Revision as of 21:15, 16 December 2009
You are encouraged to solve this task according to the task description, using any language you may know.
Find the top N salaries in each group, where N is provided as a parameter.
Use this data as a formatted internal data structure (adapt it to your language-native idioms, rather than parse at runtime), or identify your external data source: <lang csv>Employee Name,Employee ID,Salary,Department Tyler Bennett,E10297,32000,D101 John Rappl,E21437,47000,D050 George Woltman,E00127,53500,D101 Adam Smith,E63535,18000,D202 Claire Buckman,E39876,27800,D202 David McClellan,E04242,41500,D101 Rich Holcomb,E01234,49500,D202 Nathan Adams,E41298,21900,D050 Richard Potter,E43128,15900,D101 David Motsinger,E27002,19250,D202 Tim Sampair,E03033,27000,D101 Kim Arlich,E10001,57000,D190 Timothy Grove,E16398,29900,D190</lang>
C++
<lang cpp>#include <string>
- include <set>
- include <list>
- include <map>
- include <iostream>
struct Employee
{
std::string Name;
std::string ID;
unsigned long Salary;
std::string Department;
Employee(std::string _Name = "", std::string _ID = "", unsigned long _Salary = 0, std::string _Department = "")
{
Name = _Name;
ID = _ID;
Salary = _Salary;
Department = _Department;
}
void display(std::ostream& out) const { out << Name << "\t" << ID << "\t" << Salary << "\t" << Department << std::endl; } };
// We'll tell std::set to use this to sort our employees. struct CompareEarners { bool operator()(const Employee& e1, const Employee& e2) { return (e1.Salary > e2.Salary); } };
// A few typedefs to make the code easier to type, read and maintain. typedef std::list<Employee> EMPLOYEELIST;
// Notice the CompareEarners; We're telling std::set to user our specified comparison mechanism // to sort its contents. typedef std::set<Employee, CompareEarners> DEPARTMENTPAYROLL;
typedef std::map<std::string, DEPARTMENTPAYROLL> DEPARTMENTLIST;
void initialize(EMPLOYEELIST& Employees) { // Initialize our employee list data source. Employees.push_back(Employee("Tyler Bennett", "E10297", 32000, "D101")); Employees.push_back(Employee("John Rappl", "E21437", 47000, "D050")); Employees.push_back(Employee("George Woltman", "E21437", 53500, "D101")); Employees.push_back(Employee("Adam Smith", "E21437", 18000, "D202")); Employees.push_back(Employee("Claire Buckman", "E39876", 27800, "D202")); Employees.push_back(Employee("David McClellan", "E04242", 41500, "D101")); Employees.push_back(Employee("Rich Holcomb", "E01234", 49500, "D202")); Employees.push_back(Employee("Nathan Adams", "E41298", 21900, "D050")); Employees.push_back(Employee("Richard Potter", "E43128", 15900, "D101")); Employees.push_back(Employee("David Motsinger", "E27002", 19250, "D202")); Employees.push_back(Employee("Tim Sampair", "E03033", 27000, "D101")); Employees.push_back(Employee("Kim Arlich", "E10001", 57000, "D190")); Employees.push_back(Employee("Timothy Grove", "E16398", 29900, "D190")); }
void group(EMPLOYEELIST& Employees, DEPARTMENTLIST& Departments) { // Loop through all of our employees. for( EMPLOYEELIST::iterator iEmployee = Employees.begin(); Employees.end() != iEmployee; ++iEmployee ) { DEPARTMENTPAYROLL& groupSet = Departments[iEmployee->Department];
// Add our employee to this group. groupSet.insert(*iEmployee); } }
void present(DEPARTMENTLIST& Departments, unsigned int N) { // Loop through all of our departments for( DEPARTMENTLIST::iterator iDepartment = Departments.begin(); Departments.end() != iDepartment; ++iDepartment ) { std::cout << "In department " << iDepartment->first << std::endl; std::cout << "Name\t\tID\tSalary\tDepartment" << std::endl; // Get the top three employees for each employee unsigned int rank = 1; for( DEPARTMENTPAYROLL::iterator iEmployee = iDepartment->second.begin(); ( iDepartment->second.end() != iEmployee) && (rank <= N); ++iEmployee, ++rank ) { iEmployee->display(std::cout); } std::cout << std::endl; } }
int main(int argc, char* argv[]) { // Our container for our list of employees. EMPLOYEELIST Employees;
// Fill our list of employees initialize(Employees);
// Our departments. DEPARTMENTLIST Departments;
// Sort our employees into their departments. // This will also rank them. group(Employees, Departments);
// Display the top 3 earners in each department. present(Departments, 3);
return 0; }</lang>
Output:
In department D050 Name ID Salary Department John Rappl E21437 47000 D050 Nathan Adams E41298 21900 D050 In department D101 Name ID Salary Department George Woltman E21437 53500 D101 David McClellan E04242 41500 D101 Tyler Bennett E10297 32000 D101 In department D190 Name ID Salary Department Kim Arlich E10001 57000 D190 Timothy Grove E16398 29900 D190 In department D202 Name ID Salary Department Rich Holcomb E01234 49500 D202 Claire Buckman E39876 27800 D202 David Motsinger E27002 19250 D202
Common Lisp
<lang lisp>(defun top-n-by-group (n data value-key group-key predicate &key (group-test 'eql))
(let ((not-pred (complement predicate))
(group-data (make-hash-table :test group-test)))
(labels ((value (datum)
(funcall value-key datum))
(insert (x list)
(merge 'list (list x) list not-pred :key #'value))
(entry (group)
"Return the entry for the group, creating it if
necessary. An entry is a list whose first element is
k, the number of items currently associated with the
group (out of n total), and whose second element is
the list of the k current top items for the group."
(multiple-value-bind (entry presentp)
(gethash group group-data)
(if presentp entry
(setf (gethash group group-data)
(list 0 '())))))
(update-entry (entry datum)
"Update the entry using datum. If there are already n
items associated with the entry, then when datum's value
is greater than the current least item, data is merged into
the items, and the list (minus the first element) is
stored in entry. Otherwise, if there are fewer than n
items in the entry, datum is merged in, and the
entry's k is increased by 1."
(if (= n (first entry))
(when (funcall predicate (value datum) (value (first (second entry))))
(setf (second entry)
(cdr (insert datum (second entry)))))
(setf (first entry) (1+ (first entry))
(second entry) (insert datum (second entry))))))
(dolist (datum data group-data)
(update-entry (entry (funcall group-key datum)) datum)))))</lang>
Example
<lang lisp>> (defparameter *employee-data*
'(("Tyler Bennett" E10297 32000 D101)
("John Rappl" E21437 47000 D050)
("George Woltman" E00127 53500 D101)
("Adam Smith" E63535 18000 D202)
("Claire Buckman" E39876 27800 D202)
("David McClellan" E04242 41500 D101)
("Rich Holcomb" E01234 49500 D202)
("Nathan Adams" E41298 21900 D050)
("Richard Potter" E43128 15900 D101)
("David Motsinger" E27002 19250 D202)
("Tim Sampair" E03033 27000 D101)
("Kim Arlich" E10001 57000 D190)
("Timothy Grove" E16398 29900 D190))
"A list of lists of each employee's name, id, salary, and
department.")
- EMPLOYEE-DATA*
> (top-n-by-group 3 *employee-data* 'third 'fourth '>)
- <EQL Hash Table{4} 2361A0E7>
> (describe *)
- <EQL Hash Table{4} 2361A0E7> is a HASH-TABLE
D101 (3 (("Tyler Bennett" E10297 32000 D101) ("David McClellan" E04242 41500 D101) ("George Woltman" E00127 53500 D101))) D050 (2 (("Nathan Adams" E41298 21900 D050) ("John Rappl" E21437 47000 D050))) D202 (3 (("David Motsinger" E27002 19250 D202) ("Claire Buckman" E39876 27800 D202) ("Rich Holcomb" E01234 49500 D202))) D190 (2 (("Timothy Grove" E16398 29900 D190) ("Kim Arlich" E10001 57000 D190)))</lang>
E
<lang e>/** Turn a list of arrays into a list of maps with the given keys. */ def addKeys(keys, rows) {
def res := [].diverge()
for row in rows { res.push(__makeMap.fromColumns(keys, row)) }
return res.snapshot()
}
def data := addKeys(
["name", "id", "salary", "dept"], [["Tyler Bennett", "E10297", 32000, "D101"], ["John Rappl", "E21437", 47000, "D050"], ["George Woltman", "E00127", 53500, "D101"], ["Adam Smith", "E63535", 18000, "D202"], ["Claire Buckman", "E39876", 27800, "D202"], ["David McClellan", "E04242", 41500, "D101"], ["Rich Holcomb", "E01234", 49500, "D202"], ["Nathan Adams", "E41298", 21900, "D050"], ["Richard Potter", "E43128", 15900, "D101"], ["David Motsinger", "E27002", 19250, "D202"], ["Tim Sampair", "E03033", 27000, "D101"], ["Kim Arlich", "E10001", 57000, "D190"], ["Timothy Grove", "E16398", 29900, "D190"]])
def topSalaries(n, out) {
var groups := [].asMap()
for row in data {
def [=> salary, => dept] | _ := row
def top := groups.fetch(dept, fn {[]}).with([-salary, row]).sort()
groups with= (dept, top.run(0, top.size().min(n)))
}
for dept => group in groups.sortKeys() {
out.println(`Department $dept`)
out.println(`---------------`)
for [_, row] in group {
out.println(`${row["id"]} $$${row["salary"]} ${row["name"]}`)
}
out.println()
}
}</lang>
(Note: This uses an append-and-then-sort to maintain the list of top N; a sorted insert or a proper selection algorithm would be more efficient. As long as N is small, this does not matter much; the algorithm is O(n) with respect to the data set.)
<lang e>? topSalaries(3, stdout) Department D050
E21437 $47000 John Rappl E41298 $21900 Nathan Adams
Department D101
E00127 $53500 George Woltman E04242 $41500 David McClellan E10297 $32000 Tyler Bennett
Department D190
E10001 $57000 Kim Arlich E16398 $29900 Timothy Grove
Department D202
E01234 $49500 Rich Holcomb E39876 $27800 Claire Buckman E27002 $19250 David Motsinger</lang>
Haskell
<lang haskell>import Data.List import Control.Monad import Control.Arrow import Text.Printf import Data.Ord import Data.Function
groupingOn = ((==) `on`) comparingDwn = flip . comparing
type ID = Int
type DEP = String
type NAME = String
type SALARY = Double
data Employee = Employee {nr :: ID, dep :: DEP, name :: NAME, sal :: SALARY}
employees :: [Employee] employees = map (\(i,d,n,s) -> Employee i d n s)
[(1001,"AB","Janssen A.H.",41000), (101,"KA","'t Woud B.",45000),
(1013,"AB","de Bont C.A.",65000), (1101,"CC","Modaal A.M.J.",30000),
(1203,"AB","Anders H.",50000), (100,"KA","Ezelbips P.J.",52000),
(1102,"CC","Zagt A.",33000), (1103,"CC","Ternood T.R.",21000),
(1104,"CC","Lageln M.",23000), (1105,"CC","Amperwat A.",19000),
(1106,"CC","Boon T.J.",25000), (1107,"CC","Beloop L.O.",31000),
(1009,"CD","Janszoon A.",38665), (1026,"CD","Janszen H.P.",41000),
(1011,"CC","de Goeij J.",39000), (106,"KA","Pragtweik J.M.V.",42300),
(111,"KA","Bakeuro S.",31000), (105,"KA","Clubdrager C.",39800),
(104,"KA","Karendijk F.",23000), (107,"KA","Centjes R.M.",34000),
(119,"KA","Tegenstroom H.L.",39000), (1111,"CD","Telmans R.M.",55500),
(1093,"AB","de Slegte S.",46987), (1199,"CC","Uitlaat G.A.S.",44500)
]
dorank :: Int ->
(Employee -> DEP) ->
(Employee -> SALARY) ->
[Employee]-> Employee
dorank n o1 o2 = map (take n. sortBy (comparingDwn o2))
. groupBy (groupingOn o1) . sortBy (comparing o1)
toprank :: IO () toprank = do
printf "%-16s %3s %10s\n" "NAME" "DEP" "TIP" putStrLn $ replicate 31 '=' mapM_ (mapM_ (ap (ap (printf "%-16s %3s %10.2g\n" . name) dep) sal)) $ dorank 3 dep sal employees</lang>
Output: top 3 per department
*Main> toprank NAME DEP TIP =============================== de Bont C.A. AB 65000.00 Anders H. AB 50000.00 de Slegte S. AB 46987.00 Uitlaat G.A.S. CC 44500.00 de Goeij J. CC 39000.00 Zagt A. CC 33000.00 Telmans R.M. CD 55500.00 Janszen H.P. CD 41000.00 Janszoon A. CD 38665.00 Ezelbips P.J. KA 52000.00 't Woud B. KA 45000.00 Pragtweik J.M.V. KA 42300.00
J
J has a rich set of primitive functions, which combine the power of an imperative language with the expressiveness of a declarative, SQL-like language:
<lang j>NB. Dynamically generate convenience functions ('`',,;:^:_1: N=:{.Employees) =:, (_&{"1)` ([^:(_ -: ])L:0)"0 _~ i.# E =: {: Employees
NB. Show top six ranked employees in each dept N , (<@:>"1@:|:@:((6 <. #) {. ] \: SALARY)/.~ DEPT) |: <"1&> E</lang>
+-----+-----+-----------------+------+ |ID |DEPT |NAME |SALARY| +-----+-----+-----------------+------+ |1013 |AB |de Bont C.A. |65000 | |1203 |AB |Anders H. |50000 | |1093 |AB |de Slegte S. |46987 | |1001 |AB |Janssen A.H. |41000 | +-----+-----+-----------------+------+ |100 |KA |Ezelbips P.J. |52000 | |101 |KA |'t Woud B. |45000 | |106 |KA |Pragtweik J.M.V. |42300 | |105 |KA |Clubdrager C. |39800 | |119 |KA |Tegenstroom H.L. |39000 | |107 |KA |Centjes R.M. |34000 | +-----+-----+-----------------+------+ |1199 |CC |Uitlaat G.A.S. |44500 | |1011 |CC |de Goeij J. |39000 | |1102 |CC |Zagt A. |33000 | |1107 |CC |Beloop L.O. |31000 | |1101 |CC |Modaal A.M.J. |30000 | |1106 |CC |Boon T.J. |25000 | +-----+-----+-----------------+------+ |1111 |CD |Telmans R.M. |55500 | |1026 |CD |Janszen H.P. |41000 | |1009 |CD |Janszoon A. |38665 | +-----+-----+-----------------+------+
using the data set:
Employees=: (<;.1~(1 1{.~#);+./@:(;:E.S:0])@:{.)];._2 noun define
ID DEPT NAME SALARY
1001 AB Janssen A.H. 41000
101 KA 't Woud B. 45000
1013 AB de Bont C.A. 65000
1101 CC Modaal A.M.J. 30000
1203 AB Anders H. 50000
100 KA Ezelbips P.J. 52000
1102 CC Zagt A. 33000
1103 CC Ternood T.R. 21000
1104 CC Lageln M. 23000
1105 CC Amperwat A. 19000
1106 CC Boon T.J. 25000
1107 CC Beloop L.O. 31000
1009 CD Janszoon A. 38665
1026 CD Janszen H.P. 41000
1011 CC de Goeij J. 39000
106 KA Pragtweik J.M.V. 42300
111 KA Bakeuro S. 31000
105 KA Clubdrager C. 39800
104 KA Karendijk F. 23000
107 KA Centjes R.M. 34000
119 KA Tegenstroom H.L. 39000
1111 CD Telmans R.M. 55500
1093 AB de Slegte S. 46987
1199 CC Uitlaat G.A.S. 44500
)
Named as a function where the (maximum) number of employees in each department is a parameter:
<lang j>reportTopSalaries=: 3 :'N , (<@:>"1@:|:@:((y <. #) {. ] \: SALARY)/.~ DEPT) |: <"1&> E'</lang> Example of use:
reportTopSalaries 2 +-----+-----+-----------------+------+ |ID |DEPT |NAME |SALARY| +-----+-----+-----------------+------+ |1013 |AB |de Bont C.A. |65000 | |1203 |AB |Anders H. |50000 | +-----+-----+-----------------+------+ |100 |KA |Ezelbips P.J. |52000 | |101 |KA |'t Woud B. |45000 | +-----+-----+-----------------+------+ |1199 |CC |Uitlaat G.A.S. |44500 | |1011 |CC |de Goeij J. |39000 | +-----+-----+-----------------+------+ |1111 |CD |Telmans R.M. |55500 | |1026 |CD |Janszen H.P. |41000 | +-----+-----+-----------------+------+
JavaScript
<lang javascript>var data = [
{name: "Tyler Bennett", id: "E10297", salary: 32000, dept: "D101"},
{name: "John Rappl", id: "E21437", salary: 47000, dept: "D050"},
{name: "George Woltman", id: "E00127", salary: 53500, dept: "D101"},
{name: "Adam Smith", id: "E63535", salary: 18000, dept: "D202"},
{name: "Claire Buckman", id: "E39876", salary: 27800, dept: "D202"},
{name: "David McClellan", id: "E04242", salary: 41500, dept: "D101"},
{name: "Rich Holcomb", id: "E01234", salary: 49500, dept: "D202"},
{name: "Nathan Adams", id: "E41298", salary: 21900, dept: "D050"},
{name: "Richard Potter", id: "E43128", salary: 15900, dept: "D101"},
{name: "David Motsinger", id: "E27002", salary: 19250, dept: "D202"},
{name: "Tim Sampair", id: "E03033", salary: 27000, dept: "D101"},
{name: "Kim Arlich", id: "E10001", salary: 57000, dept: "D190"},
{name: "Timothy Grove", id: "E16398", salary: 29900, dept: "D190"},
];
function top_rank(n) {
var by_dept = group_by_dept(data);
for (var dept in by_dept) {
output(dept);
for (var i = 0; i < n && i < by_dept[dept].length; i++) {
var emp = by_dept[dept][i];
output(emp.name + ", id=" + emp.id + ", salary=" + emp.salary);
}
output("");
}
}
// group by dept, and sort by balary function group_by_dept(data) {
var by_dept = {};
for (var idx in data) {
var dept = data[idx].dept;
if ( ! has_property(by_dept, dept)) {
by_dept[dept] = new Array();
}
by_dept[dept].push(data[idx]);
}
for (var dept in by_dept) {
// numeric sort
by_dept[dept].sort(function (a,b){return a.salary - b.salary}).reverse();
}
return by_dept;
}
function has_property(obj, propname) {
return typeof(obj[propname]) == "undefined" ? false : true;
}
function output(str) {
try {
WScript.Echo(str); // WSH
} catch(err) {
print(str); // Rhino
}
}
top_rank(3);</lang>
outputs:
D101 George Woltman, id=E00127, salary=53500 David McClellan, id=E04242, salary=41500 Tyler Bennett, id=E10297, salary=32000 D050 John Rappl, id=E21437, salary=47000 Nathan Adams, id=E41298, salary=21900 D202 Rich Holcomb, id=E01234, salary=49500 Claire Buckman, id=E39876, salary=27800 David Motsinger, id=E27002, salary=19250 D190 Kim Arlich, id=E10001, salary=57000 Timothy Grove, id=E16398, salary=29900
OCaml
<lang ocaml>open StdLabels
let to_string (name,_,s,_) = (Printf.sprintf "%s (%d)" name s)
let take n li =
let rec aux i acc = function | _ when i >= n -> (List.rev acc) | [] -> (List.rev acc) | x::xs -> aux (succ i) (x::acc) xs in aux 0 [] li ;;
let toprank data n =
let len = List.length data in
let h = Hashtbl.create len in
List.iter data ~f:(fun ((_,_,_,dep) as employee) ->
Hashtbl.add h dep employee);
let deps =
List.fold_left data ~init:[] ~f:
(fun ac (_,_,_,v) -> if List.mem v ac then ac else v::ac) in
let f dep =
Printf.printf "Department: %s\n " dep;
let l = Hashtbl.find_all h dep in
let l2 = List.sort (fun (_,_,v1,_) (_,_,v2,_) -> compare v2 v1) l in
let l3 = (take n l2) in
print_endline(String.concat ", " (List.map to_string l3));
print_newline()
in
List.iter f deps;
let data = [
"Tyler Bennett", "E10297", 32000, "D101"; "John Rappl", "E21437", 47000, "D050"; "George Woltman", "E00127", 53500, "D101"; "Adam Smith", "E63535", 18000, "D202"; "Claire Buckman", "E39876", 27800, "D202"; "David McClellan", "E04242", 41500, "D101"; "Rich Holcomb", "E01234", 49500, "D202"; "Nathan Adams", "E41298", 21900, "D050"; "Richard Potter", "E43128", 15900, "D101"; "David Motsinger", "E27002", 19250, "D202"; "Tim Sampair", "E03033", 27000, "D101"; "Kim Arlich", "E10001", 57000, "D190"; "Timothy Grove", "E16398", 29900, "D190";
]
let () =
toprank data 3;
- </lang>
outputs:
Department: D190 Kim Arlich (57000), Timothy Grove (29900) Department: D202 Rich Holcomb (49500), Claire Buckman (27800), David Motsinger (19250) Department: D050 John Rappl (47000), Nathan Adams (21900) Department: D101 George Woltman (53500), David McClellan (41500), Tyler Bennett (32000)
Perl
<lang perl>sub zip
{my @a = @{shift()};
my @b = @{shift()};
my @l;
push @l, shift @a, shift @b while @a and @b;
return @l;}
sub uniq
{my %h;
return grep {not $h{$_}++} @_;}
my @data =
map {{ zip [qw(name id salary dept)], [split ','] }}
split "\n",
<<'EOF';
Tyler Bennett,E10297,32000,D101 John Rappl,E21437,47000,D050 George Woltman,E00127,53500,D101 Adam Smith,E63535,18000,D202 Claire Buckman,E39876,27800,D202 David McClellan,E04242,41500,D101 Rich Holcomb,E01234,49500,D202 Nathan Adams,E41298,21900,D050 Richard Potter,E43128,15900,D101 David Motsinger,E27002,19250,D202 Tim Sampair,E03033,27000,D101 Kim Arlich,E10001,57000,D190 Timothy Grove,E16398,29900,D190 EOF
@ARGV or die "Please provide a value for N.\n"; my $N = shift;
foreach my $d (sort {$a cmp $b} uniq map {$_->{dept}} @data)
{print "$d\n";
my @es =
sort {$b->{salary} <=> $a->{salary}}
grep {$_->{dept} eq $d}
@data;
foreach (1 .. $N)
{@es or last;
my $e = shift @es;
printf "%-15s | %-6s | %5d\n", @{$e}{qw(name id salary)};}
print "\n";}</lang>
PowerShell
<lang powershell>function New-Employee ($Name, $ID, $Salary, $Department) {
New-Object PSObject `
| Add-Member -PassThru NoteProperty EmployeeName $Name `
| Add-Member -PassThru NoteProperty EmployeeID $ID `
| Add-Member -PassThru NoteProperty Salary $Salary `
| Add-Member -PassThru NoteProperty Department $Department
}
$data = (New-Employee 'Tyler Bennett' E10297 32000 D101),
(New-Employee 'John Rappl' E21437 47000 D050),
(New-Employee 'George Woltman' E00127 53500 D101),
(New-Employee 'Adam Smith' E63535 18000 D202),
(New-Employee 'Claire Buckman' E39876 27800 D202),
(New-Employee 'David McClellan' E04242 41500 D101),
(New-Employee 'Rich Holcomb' E01234 49500 D202),
(New-Employee 'Nathan Adams' E41298 21900 D050),
(New-Employee 'Richard Potter' E43128 15900 D101),
(New-Employee 'David Motsinger' E27002 19250 D202),
(New-Employee 'Tim Sampair' E03033 27000 D101),
(New-Employee 'Kim Arlich' E10001 57000 D190),
(New-Employee 'Timothy Grove' E16398 29900 D190)
function Get-TopRank ($n) {
$data `
| Group-Object Department `
| ForEach-Object {
$_.Group `
| Sort-Object Salary -Descending `
| Select-Object -First $n
} `
| Format-Table -GroupBy Department
}</lang> Output:
PS> Get-TopRank 2 Department: D101 EmployeeName EmployeeID Salary Department ------------ ---------- ------ ---------- George Woltman E00127 53500 D101 David McClellan E04242 41500 D101 Department: D050 EmployeeName EmployeeID Salary Department ------------ ---------- ------ ---------- John Rappl E21437 47000 D050 Nathan Adams E41298 21900 D050 Department: D202 EmployeeName EmployeeID Salary Department ------------ ---------- ------ ---------- Rich Holcomb E01234 49500 D202 Claire Buckman E39876 27800 D202 Department: D190 EmployeeName EmployeeID Salary Department ------------ ---------- ------ ---------- Kim Arlich E10001 57000 D190 Timothy Grove E16398 29900 D190
Python
<lang python>from collections import defaultdict from heapq import nlargest
data = [('Employee Name', 'Employee ID', 'Salary', 'Department'),
('Tyler Bennett', 'E10297', 32000, 'D101'),
('John Rappl', 'E21437', 47000, 'D050'),
('George Woltman', 'E00127', 53500, 'D101'),
('Adam Smith', 'E63535', 18000, 'D202'),
('Claire Buckman', 'E39876', 27800, 'D202'),
('David McClellan', 'E04242', 41500, 'D101'),
('Rich Holcomb', 'E01234', 49500, 'D202'),
('Nathan Adams', 'E41298', 21900, 'D050'),
('Richard Potter', 'E43128', 15900, 'D101'),
('David Motsinger', 'E27002', 19250, 'D202'),
('Tim Sampair', 'E03033', 27000, 'D101'),
('Kim Arlich', 'E10001', 57000, 'D190'),
('Timothy Grove', 'E16398', 29900, 'D190')]
departments = defaultdict(list) for rec in data[1:]:
departments[rec[-1]].append(rec)
N = 3 format = "%-15s " * len(data[0]) for department, recs in departments.iteritems():
print "Department", department
print " ", format % data[0]
for rec in nlargest(N, recs, key=lambda rec: rec[-2]):
print " ", format % rec
print</lang>
Output:
Department D101 Employee Name Employee ID Salary Department George Woltman E00127 53500 D101 David McClellan E04242 41500 D101 Tyler Bennett E10297 32000 D101 Department D202 Employee Name Employee ID Salary Department Rich Holcomb E01234 49500 D202 Claire Buckman E39876 27800 D202 David Motsinger E27002 19250 D202 Department D190 Employee Name Employee ID Salary Department Kim Arlich E10001 57000 D190 Timothy Grove E16398 29900 D190 Department D050 Employee Name Employee ID Salary Department John Rappl E21437 47000 D050 Nathan Adams E41298 21900 D050
Alternative Solution
Uses namedtuples for database records, and groupby builtin to group records by Department: <lang python>from collections import namedtuple from itertools import groupby
N = 2
db = Employee Name,Employee ID,Salary,Department Tyler Bennett,E10297,32000,D101 John Rappl,E21437,47000,D050 George Woltman,E00127,53500,D101 Adam Smith,E63535,18000,D202 Claire Buckman,E39876,27800,D202 David McClellan,E04242,41500,D101 Rich Holcomb,E01234,49500,D202 Nathan Adams,E41298,21900,D050 Richard Potter,E43128,15900,D101 David Motsinger,E27002,19250,D202 Tim Sampair,E03033,27000,D101 Kim Arlich,E10001,57000,D190 Timothy Grove,E16398,29900,D190
rows = db.split('\n') DBRecord = namedtuple('DBRecord', rows[0].replace(' ', '_')) records = [ DBRecord(*row.split(',')) for row in rows[1:] ] records.sort(key = lambda record: (record.Department, -float(record.Salary))) print '\n\n'.join('\n '.join([dpt] + [str(g) for g in grp][:N])
for dpt, grp in groupby(records,
lambda record: record.Department))</lang>
Sample output
D050 DBRecord(Employee_Name='John Rappl', Employee_ID='E21437', Salary='47000', Department='D050') DBRecord(Employee_Name='Nathan Adams', Employee_ID='E41298', Salary='21900', Department='D050') D101 DBRecord(Employee_Name='George Woltman', Employee_ID='E00127', Salary='53500', Department='D101') DBRecord(Employee_Name='David McClellan', Employee_ID='E04242', Salary='41500', Department='D101') D190 DBRecord(Employee_Name='Kim Arlich', Employee_ID='E10001', Salary='57000', Department='D190') DBRecord(Employee_Name='Timothy Grove', Employee_ID='E16398', Salary='29900', Department='D190') D202 DBRecord(Employee_Name='Rich Holcomb', Employee_ID='E01234', Salary='49500', Department='D202') DBRecord(Employee_Name='Claire Buckman', Employee_ID='E39876', Salary='27800', Department='D202')
R
First, read in the data. <lang R>dfr <- read.csv(tc <- textConnection( "Employee Name,Employee ID,Salary,Department Tyler Bennett,E10297,32000,D101 John Rappl,E21437,47000,D050 George Woltman,E00127,53500,D101 Adam Smith,E63535,18000,D202 Claire Buckman,E39876,27800,D202 David McClellan,E04242,41500,D101 Rich Holcomb,E01234,49500,D202 Nathan Adams,E41298,21900,D050 Richard Potter,E43128,15900,D101 David Motsinger,E27002,19250,D202 Tim Sampair,E03033,27000,D101 Kim Arlich,E10001,57000,D190 Timothy Grove,E16398,29900,D190")); close(tc)</lang> To just return the top salary, it's very simple using tapply. <lang R>with(dfr, tapply(Salary, Department, max))</lang> To return N salaries, we replace max with our own function. <lang R>get.top.N.salaries <- function(N) {
with(dfr, tapply(Salary, Department,
function(x)
{
sort(x);
lx <- length(x)
if(N >= lx) return(x)
x[-1:(N-lx)]
}))
}
get.top.N.salaries(3)</lang>
$D050 [1] 47000 21900 $D101 [1] 41500 15900 27000 $D190 [1] 57000 29900 $D202 [1] 27800 49500 19250
To return the whole record for each of the top salaries, a different tack is required. <lang R>get.top.N.salaries2 <- function(N) {
#Sort data frame by Department, then by Salary
sorted <- dfr[with(dfr, order(Department, Salary, decreasing=TRUE)),]
#Split the dataframe up, by Department
bydept <- split(sorted, sorted$Department)
#Return the first N values (or all of them
lapply(bydept,
function(x)
{
n <- min(N, nrow(x))
x[1:n,]
})
} get.top.N.salaries2(3)</lang>
$D050
Employee.Name Employee.ID Salary Department
2 John Rappl E21437 47000 D050
8 Nathan Adams E41298 21900 D050
$D101
Employee.Name Employee.ID Salary Department
3 George Woltman E00127 53500 D101
6 David McClellan E04242 41500 D101
1 Tyler Bennett E10297 32000 D101
$D190
Employee.Name Employee.ID Salary Department
12 Kim Arlich E10001 57000 D190
13 Timothy Grove E16398 29900 D190
$D202
Employee.Name Employee.ID Salary Department
7 Rich Holcomb E01234 49500 D202
5 Claire Buckman E39876 27800 D202
10 David Motsinger E27002 19250 D202
Ruby
Without much thought to report formatting:
<lang ruby>Employee = Struct.new(:name, :employee_id, :salary, :department)
def get_employees
[
Employee.new("Tyler Bennett", "E10297", 32000, "D101"),
Employee.new("John Rappl", "E21437", 47000, "D050"),
Employee.new("George Woltman", "E00127", 53500, "D101"),
Employee.new("Adam Smith", "E63535", 18000, "D202"),
Employee.new("Claire Buckman", "E39876", 27800, "D202"),
Employee.new("David McClellan", "E04242", 41500, "D101"),
Employee.new("Rich Holcomb", "E01234", 49500, "D202"),
Employee.new("Nathan Adams", "E41298", 21900, "D050"),
Employee.new("Richard Potter", "E43128", 15900, "D101"),
Employee.new("David Motsinger", "E27002", 19250, "D202"),
Employee.new("Tim Sampair", "E03033", 27000, "D101"),
Employee.new("Kim Arlich", "E10001", 57000, "D190"),
Employee.new("Timothy Grove", "E16398", 29900, "D190"),
]
end
def show_top_salaries_per_group(groups, n)
groups.each do |dept, emps|
puts dept
# sort by salary descending
emps.sort_by {|emp| -emp.salary}[0,n].each {|e| p e}
puts
end
end
groups = get_employees.group_by {|emp| emp.department}
show_top_salaries_per_group(groups,3)</lang>
D101 #<struct Employee name="George Woltman", employee_id="E00127", salary=53500, department="D101"> #<struct Employee name="David McClellan", employee_id="E04242", salary=41500, department="D101"> #<struct Employee name="Tyler Bennett", employee_id="E10297", salary=32000, department="D101"> D190 #<struct Employee name="Kim Arlich", employee_id="E10001", salary=57000, department="D190"> #<struct Employee name="Timothy Grove", employee_id="E16398", salary=29900, department="D190"> D202 #<struct Employee name="Rich Holcomb", employee_id="E01234", salary=49500, department="D202"> #<struct Employee name="Claire Buckman", employee_id="E39876", salary=27800, department="D202"> #<struct Employee name="David Motsinger", employee_id="E27002", salary=19250, department="D202"> D050 #<struct Employee name="John Rappl", employee_id="E21437", salary=47000, department="D050"> #<struct Employee name="Nathan Adams", employee_id="E41298", salary=21900, department="D050">
SMEQL
The following SMEQL example returns the top 6 earners in each department based on this table schema:
<lang smeql>table: Employees
empID dept empName salary</lang>
Source Code:
<lang smeql>srt = orderBy(Employees, (dept, salary), order) top = group(srt, [(dept) dept2, max(order) order]) join(srt, top, a.dept=b.dept2 and b.order - a.order < 6)</lang>
Tcl
<lang tcl>package require Tcl 8.5
set text {Tyler Bennett,E10297,32000,D101 John Rappl,E21437,47000,D050 George Woltman,E00127,53500,D101 Adam Smith,E63535,18000,D202 Claire Buckman,E39876,27800,D202 David McClellan,E04242,41500,D101 Rich Holcomb,E01234,49500,D202 Nathan Adams,E41298,21900,D050 Richard Potter,E43128,15900,D101 David Motsinger,E27002,19250,D202 Tim Sampair,E03033,27000,D101 Kim Arlich,E10001,57000,D190 Timothy Grove,E16398,29900,D190}
set data [dict create] foreach line [split $text \n] {
lassign [split $line ,] name id salary dept dict lappend data $dept [list $name $id $salary]
}
proc top_n_salaries {n data} {
incr n -1
dict for {dept employees} $data {
puts "Department $dept"
foreach emp [lrange [lsort -integer -decreasing -index 2 $employees] 0 $n] {
puts [format " %-20s %-8s %8d" {*}$emp]
}
puts ""
}
}
top_n_salaries 3 $data</lang> outputs
Department D101 George Woltman E00127 53500 David McClellan E04242 41500 Tyler Bennett E10297 32000 Department D050 John Rappl E21437 47000 Nathan Adams E41298 21900 Department D202 Rich Holcomb E01234 49500 Claire Buckman E39876 27800 David Motsinger E27002 19250 Department D190 Kim Arlich E10001 57000 Timothy Grove E16398 29900
Ursala
The algorithm used by the top function is to lex the data into fields, partition by the last field, sort each partition descending by the second to last, take the first n strings in each partition, and display the list of them in a reasonably understandable form. <lang Ursala>#import std
- import nat
data =
-[ Employee Name,Employee ID,Salary,Department Tyler Bennett,E10297,32000,D101 John Rappl,E21437,47000,D050 George Woltman,E00127,53500,D101 Adam Smith,E63535,18000,D202 Claire Buckman,E39876,27800,D202 David McClellan,E04242,41500,D101 Rich Holcomb,E01234,49500,D202 Nathan Adams,E41298,21900,D050 Richard Potter,E43128,15900,D101 David Motsinger,E27002,19250,D202 Tim Sampair,E03033,27000,D101 Kim Arlich,E10001,57000,D190 Timothy Grove,E16398,29900,D190]-
top "n" = @tt sep`,*; mat0+ ^C(~&hz,mat`,*yS)*+ take/*"n"+ *zK2 (nleq+ %np~~)-<x&yzNC
- show+
main = top3 data</lang> output:
D190 Kim Arlich,E10001,57000 Timothy Grove,E16398,29900 D101 George Woltman,E00127,53500 David McClellan,E04242,41500 Tyler Bennett,E10297,32000 D202 Rich Holcomb,E01234,49500 Claire Buckman,E39876,27800 David Motsinger,E27002,19250 D050 John Rappl,E21437,47000 Nathan Adams,E41298,21900