Forum Discussion
Function within List.Generate
Hi,
I have to grouped accumulations ;
A ; which does not work ;
= Table.Group(Source, {"Name", "Item"},
{{"Count", each
List.Generate( ()=> [ x = 0 , y = [Unit] {0} ] ,
each [x] < List.Count( [Unit] ) ,
each [ x = 0 , y = [y] + [Unit] {x} ] ,
each [y] )
} } )
and B; which does;
Table.Group(Source, {"Name", "Item"}, {{"Count", (i)=>
List.Generate( ()=> [ x = 0 , y = i[Unit] {0} ],
each [x] < List.Count( i [Unit] ) ,
each [ x = [x] + 1, y = [y] + i [Unit] {x} ] ,
each [y] )
}} )
As can be seen if defined the function (i)=> in the second, so is the first refering to an 'outside' table, while the 'each' in
generate is specifically whats going on within the List.Generate , so where both are each it gets confused and causes an error
or something like that ?
Rchard.
You have it correct.
The each keyword is a syntax sugar, which is a shorthand form of "(_) =>". In the first case, only the each keyword is used. The program will find whether there is a Unit column in the variable (identifier) named "_" in the nearest layer.
The key point is:
- The function in the first parameter has no parameters, so the Unit column can be found (there is no variable named _ at the current level, so it will be looked for outside).
- The next three parameters all use the each keyword, which uses _ as the name, so [Unit] in them cannot find the Unit column (looking for the nearest variable named _).
3 Replies
You have it correct.
Thanks, it took a while for the penny to drop when I kept getting an
error, at first I used a let statement, and named the [Units] and then I could use the name.
in generate.
The each keyword is a syntax sugar, which is a shorthand form of "(_) =>". In the first case, only the each keyword is used. The program will find whether there is a Unit column in the variable (identifier) named "_" in the nearest layer.
The key point is:
- The function in the first parameter has no parameters, so the Unit column can be found (there is no variable named _ at the current level, so it will be looked for outside).
- The next three parameters all use the each keyword, which uses _ as the name, so [Unit] in them cannot find the Unit column (looking for the nearest variable named _).
