functional modify of all dictionary keys (or column names) in

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1

looking for a function to take an arbitrary k object and modify all dictionary key values (or table columns) with a function

symbol and char vector key types would be most useful, but a general solution is welcome

2

Hi,

For example:


q) t:flip m:ab`c!(1 2;3 4;5 6)

q) f:10+

q) f each m

a| 11 12
b| 13 14
c| 15 16

q) flip f each flip t

a b c
--------
11 13 15
12 14 16

q) t @[;;f]/cols t
a b c
--------
11 13 15
12 14 16

q) @[;;]/[t;cols t;(10+;20*;30%)]
a b c
-------
11 60 6
12 80 5

q) (10+;20*;30%) @’ m
a| 11 12
b| 60 80
c| 6 5

Regards,

András

3

Hi, could you please provide an example. 

4

given some arbitrary k-object:

d:`a`b`c!1 2 3d0:`D`E`F!10 20 30d1:(`$("a_1";"b-2";"c|3"))!100 200 300t:flip`g`h`i!(1 2 3;`a`b`c;1.0 1.1 1.2)a:()!()a[`]:(::)a[`test]:da[`test2]:(1;"a";d0)a[`test3]:(2;"b";`fff;a[`test2];d1)a[`test4]:(a`test3;t;`g xkey t)I want to write a function that will rename the keys in dictionaries(or key tables) and cols in tables(or key tables), at all depthse.g. lowercase key,cols ... remove special chars ... etc
5

inspired but ridiculous, sloppy JSON files being fed into kdb

6

Something like this? (using your example, doing upper case)

 

q)a

     | ::

test | ab`c!1 2 3

test2| (1;“a”;DE`F!10 20 30)

test3| (2;“b”;fff;(1;"a";DEF!10 20 30);a_1b-2`c|3! 100 200 300)

test4| ((2;“b”;fff;(1;"a";DEF!10 20 30);a_1b-2c|3! [100 200 300]("tel:100%20200%20300"));+ghi!(1 2 3;abc;1 1.1 1.2);(+(,g)!,1 2 3)!+hi!(ab`c;1 1.1 1.2))

 

q){[f;x] $[.Q.qt x; f[cols x] xcol x;99=type x;f[key x]!.z.s[f] each value x;0h=type x; .z.s[f] each x;x]}[upper;a]

     | ::

TEST | AB`C!1 2 3

TEST2| (1;“a”;DE`F!10 20 30)

TEST3| (2;“b”;fff;(1;"a";DEF!10 20 30);A_1B-2`C|3! 100 200 300)

TEST4| ((2;“b”;fff;(1;"a";DEF!10 20 30);A_1B-2C|3! [100 200 300]("tel:100%20200%20300"));+GHI!(1 2 3;abc;1 1.1 1.2);(+(,G)!,1 2 3)!+HI!(ab`c;1 1.1 1.2))

 

 

If you expected dictionaries inside  rows of your table, you can modify the first cond to something like “f[cols x] xcol .z.s[f] each x”