I'm looking for a function or solution to the following:
For the chart in SQL Reporting i need to multiply values from a Column A. For summation i would use =SUM(COLUMN_A) for the chart. But what can i use for multiplication - i was not able to find a solution so far?
Currently i am calculating the value of the stacked column as following:
=ROUND(SUM(Fields!Value_Is.Value)/SUM(Fields!StartValue.Value),3)
Instead of SUM i need something to multiply the values.
Something like that:
=ROUND(MULTIPLY(Fields!Value_Is.Value)/MULTIPLY(Fields!StartValue.Value),3)
EDIT #1
Okay tried to get this thing running.
The expression for the chart looks like this:
=Exp(Sum(Log(IIf(Fields!Menge_Ist.Value = 0, 10^-306, Fields!Menge_Ist.Value)))) / Exp(Sum(Log(IIf(Fields!Startmenge.Value = 0, 10^-306, Fields!Startmenge.Value))))
If i calculate my 'needs' manually i have to get the following result:
In my SQL Report i get the following result:
To make it easier, these are the raw values:
and you have the possibility to group the chart by CW, CQ or CY
(The values from the first pictures are aggregated Sum values from the raw values by FertStufe)
EDIT #2
Tried your expression, which results in this:
Just to make it clear:
The values in the column
=Value_IS / Start_Value
in the first picture are multiplied against each other
0,9947 x 1,0000 x 0,59401 = 0,58573
Diffusion Calenderweek 44 Sums
Startvalue: 1900,00 Value Is: 1890,00 == yield:0,99474
Waffer unbestrahlt Calenderweek 44 Sums
Startvalue: 620,00 Value Is: 620,00 == yield 1,0000
Pellet Calenderweek 44 Sums
Startvalue: 271,00 Value Is: 160,00 == yield 0,59041
yield Diffusion x yield Wafer x yield Pellet = needed Value in chart = 0,58730
EDIT #3
The raw values look like this:
The chart ist grouped - like in the image - on these fields
CY (Calendar year), CM (Calendar month), CW (Calendar week)
You can download the data as xls here:
https://www.dropbox.com/s/g0yrzo3330adgem/2013-01-17_data.xls
The expression i use (copy / past from the edit window)
=Exp(Sum(Log(Fields!Menge_Ist.Value / Fields!Startmenge.Value)))
I've exported the whole report result to excel, you can get it here:
https://www.dropbox.com/s/uogdh9ac2onuqh6/2013-01-17_report.xls
it's actually a workaround. But I am pretty sure is the only solution for this infamous problem :D
This is how I did:
Exp(∑(Log(X))), so what you should do is:
Exp(Sum(Log(Fields!YourField.Value)))
Who said math was worth nothing? =D
EDIT:
Corrected the formula.
By the way, it's tested.
Addressing Ian's concern:
Exp(Sum(Log(IIf(Fields!YourField.Value = 0, 10^-306, Fields!YourField.Value))))
The idea is change 0 with a very small number. Just an idea.
EDIT:
Based on your updated question this is what you should do:
Exp(Sum(Log(Fields!Value_IS.Value / Fields!Start_Value.Value)))
I just tested the above code and got the result you hoped for.
Related
I want to return a percentage of results from a dataset. Being a noob in Scala, tried the following
ds.filter(abs(hash(col("source"))) % 100 < percentage)
but getting abs cannot be applied to (org.apache.spark.sql.Column). I don't want to sample it, I want to return based on the hash of a column so that it's deterministic even when dataset changes.
This works just fine:
ds.filter(abs(hash(col("source"))) % 100 < percentage)
Probabely you have multiple abs in your namespace (e.g. from imports like import math._ etc. To be sure, use
ds.filter(org.apache.spark.sql.functions.abs(hash(col("source"))) % 100 < percentage)
But I think this will not garantee that you get the exact percentage, because hash values may not be equally distributed (think about a dataframe with only 1 unique value of source, hash values will all be the same.... you get either all records or none. To get the exact percentage, you would need something like :
val newDF = df
.withColumn("rnb",row_number().over(Window.orderBy($"source"))) // or order by hash if you wish
.withColumn("count",count("*").over())
.where($"rnb" < lit(fraction)*$"count")
I can get the real part of a random number to stay withing a given range but the complex part of the number doesn't stay within the range I set. see matlab / octave code below.
xmin=-.5
xmax=1
n=3
x=xmin+rand(1,n)*(xmax-xmin)+(rand(1,n)-(xmax-xmin))*1i
x=x(:)
The real part works but the complex part isn't limited to -0.5 to 1
0.2419028288441536 - 0.6579427654754871i
0.2712527227134944 - 1.451964497492678i
0.3245051849394858 - 1.107556052779179i
You have two mistakes:
x=xmin+rand(1,n)*(xmax-xmin)+(xmin + rand(1,n)*(xmax-xmin))*1i
You should add xmin to the sum and change - to * in the second part.
I've added some spaces to your code so the difference more obvious:
x = xmin+rand(1,n)*(xmax-xmin) + ( rand(1,n)-(xmax-xmin) )*1i
^^^ correct ^^^ not correct: missing `xmin+`
(and as OmG noted, also a `-` instead of a `*`)
One good way to reduce the number of bugs is by avoiding code duplication. You could for example write:
rand_sequence = #(m,xmin,xmax) xmin+rand(1,n)*(xmax-xmin);
x = rand_sequence(n,xmin,xmax) + 1i*rand_sequence(n,xmin,xmax)
(This looks like more code, but the more complicated code logic is not duplicated.)
Or like this:
x = xmin + (rand(1,n)+1i*rand(1,n)) * (xmax-xmin);
Here's the thing: I want to modify (and then return) a matrix of integers that is given in the parameters of the function. The funcion average (of the class MatrixMotionBlur) gives the average between the own pixel, upper, down and left pixels. Follows the following formula:
result(x, y) = (M1(x, y)+M1(x-1, y)+M1(x, y-1)+M1(x, y+1)) / 4
This is the code i've implemented so far
MatrixMotionBlur - Average function
MotionBlurSingleThread - run
The objetive here is to apply "average" method to alter the matrix value and return that matrix. The thing is the program gives me error when I to insert the value on the matrix.
Any ideas how to do this ?
The functional way
val updatedData = data.map{ outter =>
outter(i).map{ inner =>
mx.average(i.j)
}
}
Pay attention that Seq is immutable collection type and you can't just modify it, you can create new, modified collection only.
By the way, why you iterate starting 1, but not 0. Are you sure you want it?
I have a bunch of daily change % data. I would like to calculate cumulative change, which should just be (1+change)*previous day in a chart in Tableau.
Seems simple enough right? I can do it in a few seconds in Excel, but I've tried for hours to get it to work in Tableau and cannot do it.
My thought was that I can create a column that is (1+daily change%), then try to do a compound product. However, I can't seem to get it to work.
I can't attach any files here so I pasted the data, along with a column that is "cum change", which is what I would like the calculation to be.
Thank you much in advance!
Date Daily Change Cum Change
4/1/2015 0.47% 1
4/2/2015 0.56% 1.0056
4/3/2015 -0.72% 0.99835968
4/6/2015 -0.56% 0.992768866
4/7/2015 -0.80% 0.984826715
4/8/2015 0.44% 0.989159952
4/9/2015 -0.66% 0.982631497
4/10/2015 0.99% 0.992359549
4/13/2015 0.92% 1.001489256
4/14/2015 0.73% 1.008800128
4/15/2015 0.95% 1.018383729
4/16/2015 0.42% 1.022660941
4/17/2015 0.52% 1.027978778
4/20/2015 0.02% 1.028184373
4/21/2015 0.56% 1.033942206
4/22/2015 0.35% 1.037561004
4/23/2015 -0.34% 1.034033296
4/24/2015 0.18% 1.035894556
4/27/2015 0.61% 1.042213513
4/28/2015 0.46% 1.047007695
4/29/2015 0.94% 1.056849568
Create a calculated field:
IF INDEX() = 1
THEN 1
ELSE
(1 + AVG([Daily Change])) * PREVIOUS_VALUE(1)
END
The condition checking to see if it's the first row of the partition (INDEX() = 1) is necessary to ensure that the first value of the field is a 1. After that, you can just use the self-referential PREVIOUS_VALUE() to get the previous value of this same calculation.
As simple as in title. I have nx1 sized vector p. I'm interested in the maximum value of r = p/foo - floor(p/foo), with foo being a scalar, so I just call:
max_value = max(p/foo-floor(p/foo))
How can I get which value of p gave out max_value?
I thought about calling:
[max_value, max_index] = max(p/foo-floor(p/foo))
but soon I realised that max_index is pretty useless. I'm sorry asking this, real beginner here.
Having dropped the issue to pieces, I realized there's no unique corrispondence between values p and values in my related vector p/foo-floor(p/foo), so there's a logical issue rather than a language one.
However, given my input data, I know that the solution is unique. How can I fix this?
I ended up doing:
result = p(p/foo-floor(p/foo) == max(p/foo-floor(p/foo)))
Looks terrible, so if you know any other way...
Once you have the index, use it:
result = p(max_index)
You can create a new vector with your lets say "transformed" values:
p2 = (p/foo-floor(p/foo))
and then just use find to find the max values on p2:
max_index = find(p2 == max(p2))
that will return the index or indices of p2 with the max value of that operation, and finally just lookup the original value in p
p(max_index)
in 1 line, this is:
p(find((p/foo-floor(p/foo) == max((p/foo-floor(p/foo))))))
which is basically the same thing you did in the end :)