Excel BI Challenge Workout 192

LinkedIn Post by:
John Jairo V.

Hi to all!

One option could be:
:white_check_mark:

=LET(y,ROW(1900:2100),FILTER(y,y-MMULT(-MID(y,{1,2,3,4},1),{1;1;1;1})=YEAR(NOW())))

Blessings!

LinkedIn Post by:
Bo Rydobon

M-Code

let
Source = let n = Date.Year(DateTime.LocalNow()) in [List.Select](http://list.select/)({1900..n}, each List.Sum(List.Transform(Text.ToList(Text.From(_)),Number.From))+_ = n)
in
Source

LinkedIn Post by:
Oscar Mendez Roca Farell

hi everyone, my proposal solution:

=LET(_s,SEQUENCE(8100)+1899, FILTER(_s, MAP(_s, LAMBDA(a, a+SUM(--MID(a, SEQUENCE(4), 1))))=2023))

LinkedIn post by:
Kris Jaganah

=LET(a,SEQUENCE(9999-1900+1,,1900),b,BYROW(HSTACK(a,--MID(J2#,{1,2,3,4},1)),LAMBDA(x,SUM(x))),FILTER(a,b=2023))

LinkedIn Post by:
Hussein SATOUR

Thank you Excel BI

=LET(a, SEQUENCE(8100,,1900), b, MAP(a, LAMBDA(x, SUM(--MID(x, SEQUENCE(4), 1), x))), FILTER(a, b=YEAR(NOW())))

LinkedIn Post by:
Amr Tawfik, CMA P1

=FILTER(SEQUENCE((9999-1900)+1,,1900,1),BYROW(--MID(SEQUENCE((9999-1900)+1,,1900,1),SEQUENCE(,4,1,1),1),LAMBDA(x,SUM(x)))+SEQUENCE((9999-1900)+1,,1900,1)=2023)

LinkedIn post by:
Daniel G.

=LET(y,SEQUENCE(8100,,1900),m,MAP(y,LAMBDA(x,LET(m,MID(x,SEQUENCE(LEN(x)),1),SUM(--m)))),FILTER(y,m+y=2023))

LinkedIn Post by:
Quadri Olayinka Atharu

=LET(_yrs,SEQUENCE(9999-1900+1,,1900),
FILTER(_yrs,MAP(_yrs,LAMBDA(x,
x+SUM(--MID(x,SEQUENCE(LEN(x)),1))))=YEAR(TODAY())))

LinkedIn Post by:
Caroline Blake

=LET(y,SEQUENCE((9999-1900),,1900),a,VALUE(MID(y,{1,2,3,4},1)),
b,BYROW(a,LAMBDA(x,SUM(x))),FILTER(y,y+b=2023))

LinkedIn Post by:
Rafael González B.View Rafael González B.’s profile

Hi everyone, Here’s my M Code:

let
 Ldates = Table.FromList({1900..9999}, Splitter.SplitByNothing(), {"Years"}, null, ExtraValues.Error),
 Result = 
 Table.SelectRows(
 Table.AddColumn(
 Ldates, 
 "YearsDigits", 
 each if
 [Years] + List.Sum(
 List.Transform(Text.ToList( Text.From([Years])), 
 each Number.From(_))) = Date.Year( Date.From( DateTime.LocalNow())) 
 then 1 else 0) , 
 each [YearsDigits] = 1)[[Years]]
 
in
 Result

LinkedIn post by:
Paolo Pozzoli

Two quite similar solutions in the attached picture.