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I am trying to randomize data in-place:
#"Randomize" = Table.TransformColumns(#"Changed Type1", {{"Value", each Number.RandomBetween(0.95,1.05)*[Value]}})
However, I get the following error message:
Expression.Error: We cannot apply field access to the type Number.
Details:
Value=4567,35
Key=Value
The column [Value] is of type number, so I don't understand the problem here.
Any idea? Thanks in advance.
Solved! Go to Solution.
try this
#"Randomize" = Table.TransformColumns(#"Changed Type1", {{"Value", each Number.RandomBetween(0.95,1.05)*_}})
PS
if the answer works for you and you want to know how and why it works, just ask
It isn't in the current context. Try the following syntax:
Table.TransformColumns(
RowData,
{
{
"Value",
each _ * Number.RandomBetween(0.95,1.05),
type number
}
}
)
However, note that this isn't doing what you think. Put this in a new custom column:
Number.RandomBetween(0.5,.95)
Every value is the same random number. You can either try the method in this article, or do what I would do. Either use a Measure in DAX, or a calculated column in DAX where each row or iteration is really random.
MS should document this in their Number.Random* documentation, but they don't. I just avoid both Number.Random and Number.RandomBetween in Power Query. DAX is better here, usually.
DAX is for Analysis. Power Query is for Data Modeling
Proud to be a Super User!
MCSA: BI ReportingThank you all. Working fine with the underscore.
@edhans , the random function is also working fine if I add an index column.
Good to hear @dp32 , glad you have a solution that works.
DAX is for Analysis. Power Query is for Data Modeling
Proud to be a Super User!
MCSA: BI Reporting
I can't be sure what your intention is with the formula in question, but if you need examples of how to get some random set points, you can follow the discussion here (see in particular the last post 😁😞
How-to-turn-randbetween-results-into-a-static-column-in-Power
try this
#"Randomize" = Table.TransformColumns(#"Changed Type1", {{"Value", each Number.RandomBetween(0.95,1.05)*_}})
PS
if the answer works for you and you want to know how and why it works, just ask
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