How do you iterate a function of multivalent rank (>1), e.g. f:{[x;y] ...} where the function inputs in the next iteration step depend on the last iteration step? Examples in the reference manual only iterate unary functions.
I was able to achieve this indirectly (and verbosely) by passing a dictionary of arguments (state) into unary function:
f:{[arg] key[arg]!(min arg;arg[`y]-2)}
f/[{0<x`x};`x`y!6 3]
Note that projection, e.g. f[x;]/[whilecond;y] would only work in the scenario where the x in the next iteration step does not depend on the result of the last iteration (i.e. when x is path-independent).
In relation to Rahul's answer, you could use one of the following (slightly less verbose) methods to achieve the same result:
q)g:{(min x,y;y-2)}
q)(g .)/[{0<x 0};6 3]
-1 -3
q).[g]/[{0<x 0};6 3]
-1 -3
Alternatively, you could use the .z.s self function, which recursively calls the function g and takes the output of the last iteration as its arguments. For example,
q)g:{[x;y] x: min x,y; y:y-2; $[x<0; (x;y); .z.s[x;y]]}
q)g[6;3]
-1 -3
Function that is used with '/' and '\' can only accept result from last iteration as a single item which means only 1 function parameter is reserved for the result. It is unary in that sense.
For function whose multiple input parameters depends on last iteration result, one solution is to wrap that function inside a unary function and use apply operator to execute that function on the last iteration result.
Ex:
q) g:{(min x,y;y-2)} / function with rank 2
q) f:{x . y}[g;] / function g wrapped inside unary function to iterate
q) f/[{0<x 0};6 3]
Over time I stumbled upon even shorter way which does not require parentheses or brackets:
q)g:{(min x,y;y-2)}
q){0<x 0} g//6 3
-1 -3
Why does double over (//) work ? The / adverb can sometimes be used in place of the . (apply) operator:
q)(*) . 2 3
6
q)(*/) 2 3
6
I have a String stored in a table with two logic gates which looks like 1*((CUBL>1)*((DIFL>1)*1)).
I have been able to replace the text with values from a different table i.e. 1*((1500>1)*((0>1)*1)). The issue i am having is being able to evaluate the equations between the brackets. The following is a list of steps that i believe are needed.
1*((CUBL>1)*((DIFL>1)*1)) -> 1*((1500>1)*((0>1)*1))
1*((1500>1)*((0>1)*1)) -> 1*(1*(0*1))
1*(1*(0*1)) -> 1*(1*0)
1*(1*0) -> 1*0 = 0
I have this function f
f:{{z+x*y}[x]/[y]}
I am able to call f without a 3rd parameter and I get that, but how is the inner {z+x*y} able to complete without a third parameter?
kdb will assume, if given a single list to a function which takes two parameters, that you want the first one to be x and the remainder to be y (within the context of over and scan, not in general). For example:
q){x+y}/[1;2 3 4]
10
can also be achieved by:
q){x+y}/[1 2 3 4]
10
This is likely what's happening in your example.
EDIT:
In particular, you would use this function like
q){{z+x*y}[x]/[y]}[2;3 4 5 6]
56
which is equivalent to (due to the projection of x):
q){y+2*x}/[3 4 5 6]
56
which is equivalent to (due to my original point above):
q){y+2*x}/[3;4 5 6]
56
Which explains why the "third" parameter wasn't needed
You need to understand 2 things: 'over' behavior with dyadic functions and projection.
1. Understand how over/scan works on dyadic function:
http://code.kx.com/q/ref/adverbs/#over
If you have a list like (x1,x2,x3) and funtion 'f' then
f/(x1,x2,x3) ~ f[ f[x1;x2];x3]
So in every iteration it takes one element from list which will be 'y' and result from last iteration will be 'x'. Except in first iteration where first element will be 'x' and second 'y'.
Ex:
q) f:{x*y} / call with -> f/ (4 5 6)
first iteration : x=4, y=5, result=20
second iteration: x=20, y=6, result=120
2. Projection:
Lets take an example funtion f3 which takes 3 parameters:
q) f3:{[a;b;c] a+b+c}
now we can project it to f2 by fixing (passing) one parameter
q) f2:f3[4] / which means=> f2={[b;c] 4+b+c}
so f2 is dyadic now- it accepts only 2 parameters.
So now coming to your example and applying above 2 concepts, inner function will eventually become dyadic because of projection and then finally 'over' function works on this new dyadic function.
We can rewrite the function as :
f:{
f3:{z+x*y};
f2:f3[x];
f2/y
}
I am given three variables having finite values ( all are integers) m,n, r.
Now I need to do m<-r and n<-r ( assign m and n the value of r ) and I have read in "The Art of Computer Programming vol. 1 " that the operations can be combined as
m<-n<-r
But will the above statement not mean "assign m the value of n and then n the value of r".
Thanks in advance.
The order of assignment is from right to left. Thus, m<-n<-r will be interpreted as: n<-r and then m<-n.
Since n equals r after the first assignment, m<-n and m<-r are identical.
Assignment = operator is like assigning the right side value to left side. For eg
int a = 1 + 2;
Here first 1+2 is evaluated and assigned to a because it follows right to left associativity.
Now if you have something like this
int a=b=2;
It again follows right to left associativity. From right first b=2 is evaluated and assign 2 to b then b is assigned to a. It works like this a=(b=2)
Know in your question you have m<-n<-r . This will work like this m<-(n<-r)
You can see reference Operator Associativity
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}}