Forum Discussion
Issue with groups
Hi Ayupchap,
That case, would you please make the sample data more clear and more readable? Would you please show me the sample table as what I have posted above?
If I understand corrently, value like "Grade 6 ELA #1 16 Book 1" is stored in a signle column, right? I think it won't matter. You can split it to multiple columns.
Column2 = LEFT(Table3[Column1],17) Column3 = RIGHT(Table3[Column1],LEN(Table3[Column1])-LEN(Table3[Column2]))
Best regards,
Yuliana Gu
Hi Ayupchap,
You can add a calculated column as a new group:
NewGroup = RIGHT('Table 1'[Column1],2)
Alternatively, you could create a summarized table to group original records:
table 1_1 =
SUMMARIZE (
'Table 1',
'Table 1'[Column1],
"Column2", AVERAGE ( 'Table 1'[Column2] ),
"Book", CONCATENATEX ( 'Table 1', 'Table 1'[Column3], "," )
)
Also, you can try this:
table 1_1 =
SUMMARIZE (
'Table 1',
'Table 1'[Column1],
"Column2", AVERAGE ( 'Table 1'[Column2] ),
"CountBook", COUNT ( 'Table 1'[Column3] )
)
Best regards,
Yuliana Gu
Oh wow thanks so much for this
Onyl issue is the assesment name and the book name is combined in one field unlike the way its displayed on your amazing example, do you think this will mater?
Hi Ayupchap,
That case, would you please make the sample data more clear and more readable? Would you please show me the sample table as what I have posted above?
If I understand corrently, value like "Grade 6 ELA #1 16 Book 1" is stored in a signle column, right? I think it won't matter. You can split it to multiple columns.
Column2 = LEFT(Table3[Column1],17) Column3 = RIGHT(Table3[Column1],LEN(Table3[Column1])-LEN(Table3[Column2]))
Best regards,
Yuliana Gu
Apologies for not making it clear, the example you use here is exactly how I have it, this is really amazing thanks so much! This totally solves my problem