Why creating a table from a tale results in a list of dictionaries?

[davidcrossey]

A non-keyed table is just a list of dictionaries, flipped:

 

q)tab:([]a:1 2 3;b:`a`b`c) 
tab 
a b 
--- 
1 a 
2 b
3 c 

q)`a`b!(1 2 3;`a`b`c) 
a| 1 2 3 
b| a b c 

q)flip `a`b!(1 2 3;`a`b`c) 
a b 
--- 
1 a 
2 b 
3 c 

q)tab~flip `a`b!(1 2 3;`a`b`c) 
1b

 

What you are doing here is creating a new table, with a column called tab. This new column has a record for each row of the original table as a dictionary

q)tab:([]a:1 2 3;b:`a`b`c) 
tab 
a b 
--- 
1 a 
2 b 
3 c 

q)([]tab) 
tab 
----------- 
`a`b!(1;`a) 
`a`b!(2;`b) 
`a`b!(3;`c)

You can also enlist the table to create a single record in the new table:

q)([] enlist tab) 
tab 
-------------------- 
+`a`b!(1 2 3;`a`b`c)

If you want to create a table from a table, typically you would utilise query filtering to select a subset of columns, or joins from another table. What is your use case here?

[cillianreilly]

First, remember that a table itself is just a list of dictionaries.

q)0N!/:tab; 
`col1`col2!(`a;1) 
`col1`col2!(`b;2) 
`col1`col2!(`c;3)

So, when you tabulate a table, each entry in the original table (a dictionary) just becomes an entry in the new table column:

q)([]tab) 
tab 
----------------- 
`col1`col2!(`a;1) 
`col1`col2!(`b;2) 
`col1`col2!(`c;3) 

q)type ([]tab) 
98h 

q)type each tab 
99 99 99h 

q)type each([]tab) 
99 99 99h