I have a simple function which returns a table:
F[("A";"B");(1,-1)]
I would like to apply this function passing vectors as inputs:
a:((`A;`B);(`B;`C);(`C;`D))
b:((1;-1);(1;-1);(1;-1))
I have tried:
F each a,b
F each a cross b
but this doesn't work or combines the vectors rather than keeping the 2 components separate. In addition when I do get it to work how do I row bind the resulting list of tables? I am coming from a python background.
You need to use ' each-both :
q)F:{ ([] enlist x; enlist y)} /if F is simply creating a table
q)F[("A";"B");(1,-1)]
x y
---------
"AB" 1 -1
q)a:((`A;`B);(`B;`C);(`C;`D))
q)b:((1;-1);(1;-1);(1;-1))
q)F'[a;b] /each-both
+`x`y!(,`A`B;,1 -1)
+`x`y!(,`B`C;,1 -1)
+`x`y!(,`C`D;,1 -1)
raze will format it to a table (i think row binding means appending the tables together)
q)raze F'[a;b]
x y
--------
A B 1 -1
B C 1 -1
C D 1 -1
Related
I have a table
t: flip `S`V ! ((`$"|A|B|"; `$"|B|C|D|"; `$"|B|"); 1 2 3)
and some dicts
t1: 4 10 15 20 ! 1 2 3 5;
t2: 4 10 15 20 ! 0.5 2 4 5;
Now I need to add a column with values on the the substrings in S and the function below (which is a bit pseudocode because I am stuck here).
f:{[s;v];
if[`A in "|" vs string s; t:t1;];
else if[`B in "|" vs string s; t:t2;];
k: asc key t;
:t k k binr v;
}
problems are that s and v are passed in as full column vectors when I do something like
update l:f[S,V] from t;
How can I make this an operation that works by row?
How can I make this a vectorized function?
Thanks
You will want to use the each-both adverb to apply a function over two columns by row.
In your case:
update l:f'[S;V] from t;
To help with your pseudocode function, you might want to use $, the if-else operator, e.g.
f:{[s;v]
t:$["A"in ls:"|"vs string s;t1;"B"in ls;t2;()!()];
k:asc key t;
:t k k binr v;
};
You've not mentioned a final else clause in your pseudocode but $ expects one hence the empty dictionary at the end.
Also note that in your table the columns S and V have been cast to a symbol. vs expects a string to split so I've had to use the stringoperation - this could be removed if you are able to redefine your original table.
Hope this helps!
I have a column like this in dataframe named test:
Name Client
A P
B Q
C R
D S
E T
I need to to create a new column clienttype in the same dataframe with condition that, if Client = P or Q clienttype = first. If client type = R or S clienttype = second and so on. Can anyone tell how to do this in kdb? Thanks
Could this be solved with a dictionary?
q)update ClientType:(`P`Q`R`S!`first`first`second`second)Client from tab
Name Client ClientType
----------------------
A P first
B Q first
C R second
D S second
E T
Extension to this: You can also use vector conditionals ?[;;] for these types of problems. Unfortunately in this case in would result in many nested vector conditionals:
update Clienttype:?[Client in`P`Q;`first;?[Client in`R`S;`second;`third]]from tab
Name Client Clienttype
----------------------
A P first
B Q first
C R second
D S second
E T third
If you only had two possible options (e.g. first and second) this approach could prove useful.
You did not tell us the types of the columns in your table, but if all your entries are single letter, they are best represented by q char type:
q)show t:([]Name:"ABCDE";Client:"PQRST")
Name Client
-----------
A P
B Q
C R
D S
E T
Now, the Clienttype is best stored as a number which can be computed as
q)update Clienttype:1+(("i"$Client)-"i"$"P")div 2 from t
Name Client Clienttype
----------------------
A P 1
B Q 1
C R 2
D S 2
E T 3
or if you must use symbols,
q)types:`first`second`third`fourth
q)update Clienttype:types(("i"$Client)-"i"$"P")div 2 from t
Name Client Clienttype
----------------------
A P first
B Q first
C R second
D S second
E T third
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.
I need to simplify this Boolean expression with De Morgan's laws.
¬c xor (¬b ∨ c)
Could someone help me?
(accidentally made two accounts, so just responding with this one)
Ive found the best way to visualize a logic formula you do not understand is to make a table for it.
In the case of XOR, it represents One variable or another, but not both. So, lets make a table for A XOR B
A | B | Result
T | T | F *1
T | F | T *2
F | T | T *3
F | F | F *4
To generate the smallest possible result from the above table we can first take the most complex result that takes into account each option. Converting each line into a logical statement is fairly easy.
First, throw out anything that results in a False, Then take those that result in true, and convert them into a logical statement separated by 'OR's. In this case, 1 and 4 are false, and 2 and 3 are true. This means we only need to create logical statements for 2 and 3. I think how to do so would be best explained by example
Lets say X, Y, and Z are our variables, and the table gave us the following rows as true:
T | T | F - X & Y & ¬Z
F | T | F - ¬X & Y & ¬Z
F | F | F - ¬X & ¬Y & ¬Z
then to complete, we simply 'OR' them together
(X & Y & ¬Z) V (¬X & Y & ¬Z) V (¬X & ¬Y & ¬Z)
as you can see, where the variable is true, you put the variable directly in, and where it is false, you put a '¬' before the variable. The statement above basically says...
(True when X=T,Y=T,Z=F: False otherwise) OR (True when X=F,Y=T,Z=F: False otherwise) OR (True when X=F,Y=F,Z=F: False otherwise)
So finally bringing it back to our XOR the table rows are...
*2 A & ¬B
*3 ¬A & B
and are combined to be...
(A & ¬B) V (¬A & B)
So, now that you have an explanation of what to do with xor, you can apply this example to your problem, and come up with a logical statement you can use De Morgan's laws on to simplify.
first you have to split up xor into its basic form.
XOR represents A or B where A != B. If you can do that you should have more luck using demorgans on the whole equation
So I want to simplify z:=a+I*b; Im(z) where a, b are real variables So I try:
s:= 1+2*I
Im(s) // outputs 2
z:=a+I*b
Im(z) // outputs Im(a+I*b)
So I wonder is it any how possible to simplify Im(z) so to get b as output (here we look at general case meaning z could be any complex expression from real values (like a, b, c etc and complex I))?
You didn't tell Maple that a and b were real, so the simplification doesn't work because it doesn't necessarily hold. One way to get what you want is by using the assume command to let it know:
> s:=1+2*I;
s := 1 + 2 I
> Im(s);
2
> z:=a+I*b;
z := a + b I
> Im(z);
Im(a + b I)
> assume(a,real);
> assume(b,real);
> z;
a~ + b~ I
> Im(z);
b~
The evalc command works by considering unknowns as being real.
z:=a+I*b:
Im(z);
Im(a + I b)
evalc( Im(z) );
b
See its help-page, ?evalc.