Forum Discussion
Lazy evaluation
Hi i have found this, and think it's dow to Lazy /eager evaluation; so single empty list = {}
= let alist = {} ,
tr = List.Transform( alist,(x)=> x {0} ),
p_indi = alist {0}
in tr
tr = empty , p_indi = errorBut with a nested list ;
= let alist = {{}} ,
tr = List.Transform( alist,(x)=> x {0} ),
p_indi = alist {0}
in tr
tr = error, p_indi = empty so if anyone would like to elaborate on why/ how this works, links to info etc. please reply, I have no practical use for this
Just for the brain.
Richard.
Hi Dicken
List.Transform operates on empty list {}, so transformation function (x) => x{0} is never executed resulting empty list.alist{0} directly attempts to access first element of an empty list which immediately raises an index out of range error
alist = {{}} contains one element (an empty list). List.Transform execute once passing {} as x and x{0} fails because inner list is empty, so tr becomes an error. alist{0} returns first element of the outer list which is the empty list {}, so no error
4 Replies
Hi Dicken
List.Transform operates on empty list {}, so transformation function (x) => x{0} is never executed resulting empty list.alist{0} directly attempts to access first element of an empty list which immediately raises an index out of range error
alist = {{}} contains one element (an empty list). List.Transform execute once passing {} as x and x{0} fails because inner list is empty, so tr becomes an error. alist{0} returns first element of the outer list which is the empty list {}, so no error
so on LIst.Transform( {} , (x)=> , as the list is empt theree is not list for the function to work on,
so it does nothing, stays in that original state, where as {} {0} iis executed regardless, so = error , is this along the right lines.Yes Dicken
That's how I understand it as well. With List.Transform({}, (x) => x{0}) input list is empty, so there are no elements to pass into function. Since the function is never invoked, nothing attempts to evaluate x{0} and result remains an empty list. {}{0} is a direct indexing operation on an empty list, so it is evaluated immediately and returns an index out of range error
hi,
(x) => x{0}is applied to each element of the list, not the list itself. That's why List.Transform with empty list { } returns the same empty list (nothing to transform) while List.Transform with { { } } returns ... no, not error but the list with single element {error}. And It has nothing to do with lazy / eager evaluation.
