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Why do we put a 0# here: @[`.;t;@[;`sym;`g#]0#]
I know roughly what apply does:
In the other apply we are applying to tables in the namespace the inner apply.
The inner apply is grouping on the sym column and it is a projection hence the empty slot.
But what exactly is 0# doing here?
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1 Reply
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TLDR: the expression initialises a list
tof tables in the default namespace, setting the grouped attribute on the symbol column and removing all the table rows.How that? The function called is not Apply but Amend At
@in its ternary form.The first argument is the default namespace
`., which, as a namespace, is also a dictionary.The second argument
twill be a symbol vector of names of tables in the default namespace; i.e. globals.The third argument is the function to be applied to each table. Youve already spotted that the unary projection
@[;`sym;`g#]sets the grouped attribute on thesymcolumn. It is composed (by juxtaposition) with the unary projection0#. Applied to a table,0#removes all the rows.So the function applied to each table named in
tis the composition@[;`sym;`g#] 0#. That is, remove the rows and set grouped on thesymcolumn.
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