In F#, assume I have a person record as:
type person =
{ LastName: string option
BirthDate: System.DateTime option }
Now, I want to create a list of 100 persons (this fails. Both name and The System.DateTime(...) is incorrect):
let people = [for a in 1 .. 100
do yield {
LastName= Some "LastName"+a
BirthDate = System.DateTime(2012,11,27)
}]
How is this done?
TIA
There are two separate issues with the code, but your general approach is good!
First, Some "LastName"+a is interpereted as (Some "LastName")+a, which is not the right parenthesization. Also a is an int which cannot be automatically turned into a string, so you need to explicitly convert it. The correct version is Some("LastName" + string a).
Second, System.DateTime(2012,11,27) is DateTime, but you need an option. You can fix this just by adding Some and the right parentheses, i.e. Some(System.DateTime(2012,11,27)).
As a bonus, you can reduce do yield to -> (this is just a syntactic sugar to make this kind of thing shorter). I would write:
open System
let people =
[ for a in 1 .. 100 ->
{ LastName= Some ("LastName"+string a)
BirthDate = Some(DateTime(2012,11,27)) } ]
Related
I have an app that is reading an ingredients list. At this point I've already retrieved a list of the 2500 most common ingredients. So I've got a list of, say 10 ingredients as strings, and a list of 2500 ingredients, with names as well as other properties. If an ingredient in this list of strings matches the name of an ingredient in the list of ingredients, I'd like to add it to another list third list, of ingredients that exist. The only way I know how to do that is with basically a for loop.
I'd do it as
fun compareLists(listOfIng: List<String>): List<ListIngredientsQuery.Item> {
var returnList = mutableListOf<ListIngredientsQuery.Item>()
for (ing in listOfIng) {
for (serverIngredient in MyApp.metaIngredientList!!) {
if (serverIngredient.name() == ing) {
returnList!!.add(serverIngredient)
}
}
}
return returnList
}
Which would technically work, but I have to imagine there's a better, faster way than iterating over 2500 items, as many times as there are Ingredients in an Ingredient list. What is the like, proper, preferred by real developers, way of doing this.
As each ingredient name is unique, you can use hash map for storing your 2500 ingredients with its name as the key. This way you do not need to loop over that huge collection any more, but just look thing up by the name and let the hash map deal with it.
To put some code to what Marcin said, here is what I would do:
fun compareLists(listOfIng: List<String>) =
MyApp.metaIngredientList!!
.associateBy { it.name() }
.let { metaIngredientMap -> listOfIng.mapNotNull { metaIngredientMap[it] }}
Or if we wanna avoid using !!
fun compareLists(listOfIng: List<String) =
MyApp.metaIngredientList
?.associateBy { it.name() }
?.let { metaIngredientMap -> listOfIng.mapNotNull { metaIngredientMap[it] }}
?: emptyList<ListIngredientQuery.Item>()
Of course, ideally, you would want that MyApp.metaIngredientList to be already a Map and not convert it into a Map for each operation
I have a dataset from Google Analytics I have a hard time mapping to something useful that I can use in production.
The raw data is a list of case class objects looking like this:
case class Stats(id: String, date: Int, pViews: Int, uViews: Int)
I have gone about making the list of objects like this:
val skuMatch = "[A-Z][0-9]{10}".r
val mappedRaw = rawData
.map(row => {
val idCheck = idMatch.findFirstIn(row.getDimensions()(1))
val id = if (idCheck.isDefined) idCheck.get else ""
val date = row.getDimensions()(0).toInt
val pViews = row.getMetrics()(0).getValues()(0).toInt
val uViews = row.getMetrics()(0).getValues()(1).toInt
Stats(id, date, pViews, uViews)
}).filter(!_.id.isEmpty)
}
So now I have a list of objects that are "clean" so no empty dirty, values. An object might look like this:
Stats("A0000000008", 201608, 10, 1)
Now, the list contains multiple objects with the same id value because they then have a different date value or they might be valid duplicates where the pViews and uViews data is still relevant.
Stats("A0000000008", 201609, 12, 5)
Stats("A0000000008", 201609, 8, 2)
My question therefore is, how do I make a structure that looks like this:
case class DataMonth(monthYear: Int, uViews: Int, pViews: Int)
case class NewStats(id: String, months: List(DataMonth))
Example object:
List(
NewStats(
"A0000000008",
List(
DataMonth(
201608,
10,
1
),
DataMonth(
201609,
20,
7
)
)
)
)
So that it has taken all the duplicates grouped the data by months aka. the date value.
I have a really hard time wrapping my head around these kinds of problems so I would be grateful for any kind of help really.
Thanks in advance.
Please see I want list employee objects. So I have below two options.
List<Tuple<int, string, string, string>>
List<Employee>where Employee is class contains 4 properties.
My doubt is what should I use(tuple or list of employee object?
If it is List<Employee> then in which scenario I should use List<Tuple<int, string, ...>>.
You should not use tuples unless you are doing some sort of arithmetic operation where tuple would be an acceptable and widely understood method of supplying values. Tuples make it a maintenance nightmare for anyone who is not familiar with your process as you built it.
Edit: Think about the difference between seeing:
var employeeList = DAL.getEmployees();
var activeEmployees = employeeList.Where(employee => employee.IsActive);
vs
var employeeTuple = DAL.getEmployees();
var activeEmployees = employeeTuple.Where(employee => employee.Item3);
In the second example, I know THAT you created an active employee list, but I don't know HOW you did it.
That's rather obvious. If you already have the Employee class then using List<Employee> is straightforward:
List<Employee> list = new List<Employee>();
list.Add( e );
...
Employee e = list.Where( i => i.Name == "John" ).FirstOrDefault();
whereas using List<Tuple<...>> is at least cumbersome:
List<Tuple<....>> list = new List<Tuple<....>>();
list.Add( new Tuple<...>( e.Name, e.Surname, e.Whateverelse, e.YetAnother ) );
...
// retrieve the tuple
var tuple = list.Where( i => i.Item1 == "John" );
// make Employee out of it
Employee e = new Employee( e.Item1, e.Item2, e.Item3, e.Item4 );
I have 2 immutable case classes A(source, key, value) and B(source, key, value)
I want to add A over B in such a way when 'source' and 'key' doesn't exist, to be added from A to the B and when 'source' and 'key' exist to replace the value from B with the one from A. The same way 'merge_array' function from php works on a multidimensional array.
I tried with 'A.union(B).groupBy(.key)' and then 'groupBy(.source)' and get the 1st value. But then I realized that I can never be sure that first value will always be the value of A.
I'm quite new to scala and I really ran out of ideas how I could do this from a functional immutable point of view.
Anyone has any idea how I could do this?
Thank you
Edit:
case class TranslationValue(source: String, key: String, value: String)
def main(args:Array[String]):Unit = {
println(merge(data1.toSet, data2.toSet))
}
def merge(a: Set[TranslationValue], b: Set[TranslationValue]) = {
a.union(b).groupBy(_.key).flatMap{ case (s, v) =>
v.groupBy(_.source).flatMap{case (s1, v1) => {
for (res <- 0 to 0) yield v1.head
}
}
}
}
Example
data1 has this data
Set(
TranslationValue(messages,No,No),
TranslationValue(messages,OrdRef,Order Reference),
TranslationValue(messages,OrdId,Order Id)
)
data2 has this data
Set(
TranslationValue(messages,No,No),
TranslationValue(messages,OrdRef,OrderRef)
TranslationValue(messages,Name,Name)
)
putting data1 over data2 I want to obtain
List(
TranslationValue(messages,No,No),
TranslationValue(messages,OrdRef,Order Reference),
TranslationValue(messages,OrdId,Order Id)
TranslationValue(messages,Name,Name)
)
I know that what I do can be done better, but like I said, I'm learning :)
you can group in one go:
def merge(a: Seq[TranslationValue], b: Seq[TranslationValue]) = {
a.union(b).groupBy(t=>(t.key,t.source)).map(c=>c._2.head)
}
i think you could also override the equals method for TranslationValue so that two translation values are equal when source and key are the same(the hashcode method has also to be overridden). Then a.union(b) would be enough.
edit:
It seems Set doesnt guarantee order of items(Scala: Can I rely on the order of items in a Set?), but a seq should.
For example i have erlang record:
-record(state, {clients
}).
Can i make from clients field list?
That I could keep in client filed as in normal list? And how can i add some values in this list?
Thank you.
Maybe you mean something like:
-module(reclist).
-export([empty_state/0, some_state/0,
add_client/1, del_client/1,
get_clients/1]).
-record(state,
{
clients = [] ::[pos_integer()],
dbname ::char()
}).
empty_state() ->
#state{}.
some_state() ->
#state{
clients = [1,2,3],
dbname = "QA"}.
del_client(Client) ->
S = some_state(),
C = S#state.clients,
S#state{clients = lists:delete(Client, C)}.
add_client(Client) ->
S = some_state(),
C = S#state.clients,
S#state{clients = [Client|C]}.
get_clients(#state{clients = C, dbname = _D}) ->
C.
Test:
1> reclist:empty_state().
{state,[],undefined}
2> reclist:some_state().
{state,[1,2,3],"QA"}
3> reclist:add_client(4).
{state,[4,1,2,3],"QA"}
4> reclist:del_client(2).
{state,[1,3],"QA"}
::[pos_integer()] means that the type of the field is a list of positive integer values, starting from 1; it's the hint for the analysis tool dialyzer, when it performs type checking.
Erlang also allows you use pattern matching on records:
5> reclist:get_clients(reclist:some_state()).
[1,2,3]
Further reading:
Records
Types and Function Specifications
dialyzer(1)
#JUST MY correct OPINION's answer made me remember that I love how Haskell goes about getting the values of the fields in the data type.
Here's a definition of a data type, stolen from Learn You a Haskell for Great Good!, which leverages record syntax:
data Car = Car {company :: String
,model :: String
,year :: Int
} deriving (Show)
It creates functions company, model and year, that lookup fields in the data type. We first make a new car:
ghci> Car "Toyota" "Supra" 2005
Car {company = "Toyota", model = "Supra", year = 2005}
Or, using record syntax (the order of fields doesn't matter):
ghci> Car {model = "Supra", year = 2005, company = "Toyota"}
Car {company = "Toyota", model = "Supra", year = 2005}
ghci> let supra = Car {model = "Supra", year = 2005, company = "Toyota"}
ghci> year supra
2005
We can even use pattern matching:
ghci> let (Car {company = c, model = m, year = y}) = supra
ghci> "This " ++ c ++ " " ++ m ++ " was made in " ++ show y
"This Toyota Supra was made in 2005"
I remember there were attempts to implement something similar to Haskell's record syntax in Erlang, but not sure if they were successful.
Some posts, concerning these attempts:
In Response to "What Sucks About Erlang"
Geeking out with Lisp Flavoured Erlang. However I would ignore parameterized modules here.
It seems that LFE uses macros, which are similar to what provides Scheme (Racket, for instance), when you want to create a new value of some structure:
> (define-struct car (company model year))
> (define supra (make-car "Toyota" "Supra" 2005))
> (car-model supra)
"Supra"
I hope we'll have something close to Haskell record syntax in the future, that would be really practically useful and handy.
Yasir's answer is the correct one, but I'm going to show you WHY it works the way it works so you can understand records a bit better.
Records in Erlang are a hack (and a pretty ugly one). Using the record definition from Yasir's answer...
-record(state,
{
clients = [] ::[pos_integer()],
dbname ::char()
}).
...when you instantiate this with #state{} (as Yasir did in empty_state/0 function), what you really get back is this:
{state, [], undefined}
That is to say your "record" is just a tuple tagged with the name of the record (state in this case) followed by the record's contents. Inside BEAM itself there is no record. It's just another tuple with Erlang data types contained within it. This is the key to understanding how things work (and the limitations of records to boot).
Now when Yasir did this...
add_client(Client) ->
S = some_state(),
C = S#state.clients,
S#state{clients = [Client|C]}.
...the S#state.clients bit translates into code internally that looks like element(2,S). You're using, in other words, standard tuple manipulation functions. S#state.clients is just a symbolic way of saying the same thing, but in a way that lets you know what element 2 actually is. It's syntactic saccharine that's an improvement over keeping track of individual fields in your tuples in an error-prone way.
Now for that last S#state{clients = [Client|C]} bit, I'm not absolutely positive as to what code is generated behind the scenes, but it is likely just straightforward stuff that does the equivalent of {state, [Client|C], element(3,S)}. It:
tags a new tuple with the name of the record (provided as #state),
copies the elements from S (dictated by the S# portion),
except for the clients piece overridden by {clients = [Client|C]}.
All of this magic is done via a preprocessing hack behind the scenes.
Understanding how records work behind the scenes is beneficial both for understanding code written using records as well as for understanding how to use them yourself (not to mention understanding why things that seem to "make sense" don't work with records -- because they don't actually exist down in the abstract machine...yet).
If you are only adding or removing single items from the clients list in the state you could cut down on typing with a macro.
-record(state, {clients = [] }).
-define(AddClientToState(Client,State),
State#state{clients = lists:append([Client], State#state.clients) } ).
-define(RemoveClientFromState(Client,State),
State#state{clients = lists:delete(Client, State#state.clients) } ).
Here is a test escript that demonstrates:
#!/usr/bin/env escript
-record(state, {clients = [] }).
-define(AddClientToState(Client,State),
State#state{clients = lists:append([Client], State#state.clients)} ).
-define(RemoveClientFromState(Client,State),
State#state{clients = lists:delete(Client, State#state.clients)} ).
main(_) ->
%Start with a state with a empty list of clients.
State0 = #state{},
io:format("Empty State: ~p~n",[State0]),
%Add foo to the list
State1 = ?AddClientToState(foo,State0),
io:format("State after adding foo: ~p~n",[State1]),
%Add bar to the list.
State2 = ?AddClientToState(bar,State1),
io:format("State after adding bar: ~p~n",[State2]),
%Add baz to the list.
State3 = ?AddClientToState(baz,State2),
io:format("State after adding baz: ~p~n",[State3]),
%Remove bar from the list.
State4 = ?RemoveClientFromState(bar,State3),
io:format("State after removing bar: ~p~n",[State4]).
Result:
Empty State: {state,[]}
State after adding foo: {state,[foo]}
State after adding bar: {state,[bar,foo]}
State after adding baz: {state,[baz,bar,foo]}
State after removing bar: {state,[baz,foo]}