Given:
a:10 20 30
How do I get:
"102030"
?
I can only find:
`c$a
But that's definitely not it.
In q there is:
string 42
which returns:
"42"
but I cannot find the equivalent in k.
In K required statement will look like:
k) ,/$a
Where ,/ is raze equivalent and $ is string function equivalent.
There are not many good K tutorials. But I found last one created by Shakti is quite complete, though it may be not fully compatible with Kx's K
How do i passed 2 variables to a lambda function, where x is a number and y is a symbol.
I have written this, but it wouldn't process
{[x;y]
// some calculation with x and y
}
each ((til 5) ,\:/: `a`b`c`d`f)
It seems to be complaining that i am missing another arg.
Here's an example that I think does what you're looking for:
q){string[x],string y}./: raze (til 5) ,\:/: `a`b`c`d`f
The issue with your example is that you need to raze the output of ((til 5) ,\:/: `a`b`c`d`f) to get your list of 2 inputs.
Passing a list of variables into a function is accomplished using "." (dot apply) http://code.kx.com/q/ref/unclassified/#apply
.e.g
q){x+y} . 10 2
12
In my example, I've then used an "each right" to then apply to each pair. http://code.kx.com/q/ref/adverbs/#each-right
Alternatively, you could use the each instead if you wrapped the function in another lamda
q){{string[x],string y} . x} each raze (til 5) ,\:/: `a`b`c`d`f
Instead of generating a list of arguments using cross or ",/:\:" and passing each of these into your function, modify your function with each left each right ("/:\:") to give you all combination. his should take the format;
x f/:\: y
Where x and y are both lists. Reusing the example {string[x],string y};
til[5] {string[x], string y}/:\:`a`b`c`d
This will give you a matrix of all combinations of x and y. If you want to flatten that list add a 'raze'
I have written this in CoffeeScript:
expect (#controllerInstance[fn]).toHaveBeenCalled()
and it's been compiled to this:
return expect(this.controllerInstance[fn].toHaveBeenCalled());
Why has it re-arranged the method call parenthesis? And how would I make it compile to what I want?
what I need to see is:
expect(this.controllerInstance[fn]).toHaveBeenCalled()
Parentheses serve two purposes in CoffeeScript:
Expression grouping, e.g. (6 + 11) * 23 or f (-> 6), (-> 11).
Function calling, e.g. f(), g('pancakes').
Since parentheses are sometimes optional in a function call, there is some ambiguity in:
f (expr)
Are those parentheses being used to call f with expr as its argument or are the parentheses really a part of f's argument? CoffeeScript chooses the latter interpretation.
You'll see similar problems if write:
f (x) + 1
CoffeeScript sees that as:
f((x) + 1)
Similarly, if you write:
f (x, y)
you'll get an unexpected , error; CoffeeScript doesn't have a comma operator so x, y is not a valid expression.
You can remove the ambiguity by removing the whitespace before the opening parenthesis:
expect(#controllerInstance[fn]).toHaveBeenCalled()
Removing the space after expect forces CoffeeScript to view the parentheses around #controllerInstance[fn] to be seen as function-calling parentheses.
I'm trying to create a parser for program. For example,
I entered (what I want)
"(2+3)-4" it will become something like this "(minus, (plus, num 2, num 3),num 4)"
What I've done so far..
"(2+3)-4" I then split it and it becomes list Z = ["(","2","+","3",")","-","4"] then I compared if "-" is a member of Z, if true I append the element "-" into a new list ["-"]
I'm not sure if the way I'm doing is correct, I'm new to Er-lang and struggling quite a lot. If anyone is able to offer me some insight, thanks.
Consider the following, which returns a tuple-based representation of its input:
parse(Expr) ->
Elems = re:split(Expr, "([-+)(])", [{return,list}]),
parse(lists:filter(fun(E) -> E /= [] end, Elems), []).
parse([], [Result]) ->
Result;
parse([], [V2,{op,Op},V1|Tacc]) ->
parse([], [{Op,V1,V2}|Tacc]);
parse(["("|Tail], Acc) ->
parse(Tail, [open|Acc]);
parse([")"|Tail], [Op,open|TAcc]) ->
parse(Tail, [Op|TAcc]);
parse(["+"|Tail], Acc) ->
parse(Tail, [{op,plus}|Acc]);
parse(["-"|Tail], Acc) ->
parse(Tail, [{op,minus}|Acc]);
parse([V2|Tail], [{op,Op},V1|Tacc]) ->
parse(Tail, [{Op,V1,{num,list_to_integer(V2)}}|Tacc]);
parse([Val|Tail], Acc) ->
parse(Tail, [{num,list_to_integer(Val)}|Acc]).
The first function, parse/1, splits the expression along the + and - operators and parentheses, preserving these in the resulting list. It then filters that list to remove empty elements, and passes it with an empty accumulator to parse/2.
The parse/2 function has eight clauses, described below:
The first two handle the case when the parsed input list has been exhausted. The second of these handles the case where multiple elements in the accumulator need to be collapsed into a single tuple consisting of operator and operands.
The next two handle clauses parentheses. When we see an open parenthesis, we push an atom open into the accumulator. Upon seeing the matching close parenthesis, we expect to see an operation tuple and the atom open in the accumulator, and we replace them with just the tuple.
Clauses 5 and 6 handle + and - respectively. Each just pushes a {op,Operator} tuple into the accumulator, where Operator is either the atom plus or the atom minus.
The final two clauses handle values. The first one handles the case where the accumulator holds a value and an op tuple, which gets replaced with a full operation tuple consisting of the atom plus or minus followed by two num tuples each holding integer operands. The last clause just handles plain values.
Putting this in a module p, compiling it, and running it in an Erlang shell yields the following:
1> p:parse("2+3").
{plus,{num,2},{num,3}}
2> p:parse("(2+3)-4").
{minus,{plus,{num,2},{num,3}},{num,4}}
I found this quicksort implementation on a website:
q:{$[2>distinct x;x;raze q each x where each not scan x < rand x]};
I don't understand this part:
raze q each x where each not scan x < rand x
Can someone explain it to me step by step?
Lets do it step by step . I assume you have basic understanding of Quick Sort algo. Also, there is one correction in code you mentioned which I have corrected in step 5.
Example list:
q)x: 1 0 5 4 3
Take a random element from list which will act as pivot.
q) rand x
Suppose it gives us '4' from list.
Split list 'x' in 2 lists. One contains elements lesser that '4' and other greater(or equal) to '4'.
2.a) First compare all elements with pivot (4 in our case)
q) (x<rand x) / 11001b : output is boolean list
2.b) Using above boolean list we can get all elements from 'x' lesser than '4'. Here is the way:
q) x where 11001b / ( 1 0 3) : output
So we require other expression to get all elements greater(or equal) than pivot '4'. There are many ways to do it
but lets see the one used in code:
q)not scan (x<rand x) / (11001b;00110b) : output
So it gives the list which has 2 lists. First is result of (x < rand x) which is used to get elements lesser than pivot '4' and other is negation of this list which is done by 'not' and it is used to get all elements greater(or equal) that pivot '4'.
2.c) So now we can generate 2 lists using sample code from (2.b)
q) x where each (not scan (x<rand x)) / ((1 0 3);(5 4)): output list which has 2 lists
Now apply same function to each list to sort each of them
i.e. recursive call on each list of list ((1 0 3);(5 4))
q) q each x where each (not scan (x<rand x))
After all calculations , apply 'raze' to flatten all lists that are returned from each recursive call to output one single list.
End condition for recursive call is: when input list has only 1 distinct element just return it.
q) 2>count distinct x
Note: There is one correction. 'count' was missing in original code.