Say for example I have a table:
q)([] aa: til 5; ab: til 5; bb: til 5)
aa ab bb
--------
0 0 0
1 1 1
2 2 2
3 3 3
4 4 4
Is there a simple way of querying for just the columns that begin with a?
You can use functional select
?[t;();0b;{x!x}cols[t] where cols[t] like "a*"]
https://code.kx.com/q/basics/funsql/
If your table is in-memory and unkeyed then take (#) can also do it for you:
{where[c!(c:cols x)like"a*"]#x}t
But Matts solution is more general and thus more useful!
Related
I have a method of achieving this which also explains my question.
a:1 2 3 4;
b:5 6 7;
cond:1101001b;
comb:(count cond) # 0N;
comb[where cond]:a;
comb[where not cond]:b
But q has so many utilities for manipulating lists, I am wondering if there is a more direct way of doing this.
rank is what you need.
q)comb
1 2 5 3 6 7 4
q)(b,a)rank cond
1 2 5 3 6 7 4
You could write the expression in a single line
comb:#[;where not cond;:;b] #[;where cond;:;a] (count cond)#0N
Alternatively, assuming the 1s and 0s of cond matches the lengths of a and b:
(a,b) iasc where[cond],where not cond
I have a list of items and need to merge them into a single column
using the list
list:(1 2;3 4 5 7;0 1 3)
index value
0 1 2
1 3 4 5 7
2 0 1 3
my goal is
select from list2
value
1
2
3
4
5
7
0
1
3
'raze' function flattens out 1 level of the list.
q) raze (1 2;3 4 5 7;0 1 3)
q) 1 2 3 4 5 7 0 1 3
If you have list with multi level indexing then use 'over' adverb with raze:
q) (raze/)(1 2 3;(11 12;33 44);5 6)
To convert that to table column:
q) t:([]c:raze list)
ungroup would also work provided your table doesn't have multiple columns with different nesting (or strings)
q)ungroup ([]list)
list
----
1
2
3
4
5
7
0
1
3
If you just wanted your list to appear like that I would do the following.
1 cut raze list
I see that you have used a select statement, however if you want your column defined as this in your table do the following
a:raze list
tab:([] b:a)
Your output from this should look like this
q)tab
b
-
1
2
3
4
5
7
0
1
3
Overall, a more concise way to achieve what you want to do would be
select from ([]raze list)
To avoid any errors you should not call the column header 'value' as this is a protected keyword in kdb+ and when you try to reassign it as a column header kdb will through an assign error
`assign
Hope this helps
I'm currently learning kdb+/q.
I have a table of data. I want to take 2 columns of data (just numbers), compare them and create a new Boolean column that will display whether the value in column 1 is greater than or equal to the value in column 2.
I am comfortable using the update command to create a new column, but I don't know how to ensure that it is Boolean, how to compare the values and a method to display the "greater-than-or-equal-to-ness" - is it possible to do a simple Y/N output for that?
Thanks.
/ dummy data
q) show t:([] a:1 2 3; b: 0 2 4)
a b
---
1 0
2 2
3 4
/ add column name 'ge' with value from b>=a
q) update ge:b>=a from t
a b ge
------
1 0 0
2 2 1
3 4 1
Use a vector conditional:
http://code.kx.com/q/ref/lists/#vector-conditional
q)t:([]c1:1 10 7 5 9;c2:8 5 3 4 9)
q)r:update goe:?[c1>=c2;1b;0b] from t
c1 c2 goe
-------------
1 8 0
10 5 1
7 3 1
5 4 1
9 9 1
Use meta to confirm the goe column is of boolean type:
q)meta r
c | t f a
-------| -----
c1 | j
c2 | j
goe | b
The operation <= works well with vectors, but in some cases when a function needs atoms as input for performing an operation, you might want to use ' (each-both operator).
e.g. To compare the length of symbol string with another column value
q)f:{x<=count string y}
q)f[3;`ab]
0b
q)t:([] l:1 2 3; s: `a`bc`de)
q)update r:f'[l;s] from t
l s r
------
1 a 1
2 bc 1
3 de 0
I have a table rCom which has various columns. I would like to sum across each row..
for example:
Date TypeA TypeB TypeC TypeD
date1 40.5 23.1 45.1 65.2
date2 23.3 32.2 56.1 30.1
How can I write a q query to add a fourth column 'Total' that sums across each row?
why not just:
update Total: TypeA+TypeB+TypeC+TypeD from rCom
?
Sum will work just fine:
q)flip`a`b`c!3 3#til 9
a b c
-----
0 3 6
1 4 7
2 5 8
q)update d:sum(a;b;c) from flip`a`b`c!3 3#til 9
a b c d
--------
0 3 6 9
1 4 7 12
2 5 8 15
Sum has map reduce which will be better for a huge table.
One quick point regarding summing across rows. You should be careful about nulls in 1 column resulting in a null result for the sum. Borrowing #WooiKent Lee's example.
We put a null into the first position of the a column. Notice how our sum now becomes null
q)wn:.[flip`a`b`c!3 3#til 9;(0;`a);first 0#] //with null
q)update d:sum (a;b;c) from wn
a b c d
--------
3 6
1 4 7 12
2 5 8 15
This is a direct effect of the way nulls in q are treated. If you sum across a simple list, the nulls are ignored
q)sum 1 2 3 0N
6
However, a sum across a general list will not display this behavior
q)sum (),/:1 2 3 0N
,0N
So, for your table situation, you might want to fill in with a zero beforehand
q)update d:sum 0^(a;b;c) from wn
a b c d
--------
3 6 9
1 4 7 12
2 5 8 15
Or alternatively, make it s.t. you are actually summing across simple lists rather than general lists.
q)update d:sum each flip (a;b;c) from wn
a b c d
--------
3 6 9
1 4 7 12
2 5 8 15
For a more complete reference on null treatment please see the reference website
This is what worked:
select Answer:{[x;y;z;a] x+y+z+a }'[TypeA;TypeB;TypeC;TypeD] from
([] dt:2014.01.01 2014.01.02 2014.01.03; TypeA:4 5 6; TypeB:1 2 3; TypeC:8 9 10; TypeD:3 4 5)
I'd like to calculate a new column which is a function of several columns using select.
My actual application will involve a grouping in the select so the columns entries which I will pass to the function will contain lists. But this simple example illustrates my question
t:([] a:1 2 3; b:10 20 30; c:5 6 7)
/ Pass one argument, using projection (set first two arguments to 1)
select s:{[x;y;z] x+y+z}[1;1;] each a from t
/ Pass two arguments using each-both (set first arg to 1)
select s:a {[x;y;z] x+y+z}[1;;]'b from t
Now, how can I pass three or more arguments?
Each' will work in general but it's best to use vector operations where possible. Here I use the . operator to apply our function, \t to time both methods. I store their results to r1/r2 to show they are the same:
q)t:([]a:til n;b:til n;c:til n:1200300)
q)\t r1:update d:{x+y+z}'[a;b;c] from t
289
q)\t r2:update d:{x+y+z} . (a;b;c) from t
20
q)r1~r2
1b
q)r2
a b c d
-----------
0 0 0 0
1 1 1 3
2 2 2 6
3 3 3 9
4 4 4 12
5 5 5 15
..
Cheers,
Ryan
The following form works in general
q)t:([]a:til 10;b:til 10;c:til 10)
q)select d:{x+y+z}'[a;b;c] from t
d
--
0
3
6
9
..