Translation of an excel formula to Oracle SQL

klyles95
klyles95 used Ask the Experts™
on
I am trying to convert an excel formula into oracle sql.  The excel formula is straight forward for the most part.

However, there is a portion of the formula I can not understand and so can not interpret.
Assume:
C3 = 2
D3 = 32
E3 = 28
F3 = 24
G3 = 18
H3 = 0

L3 contains the following formula: {=C$3^(D$3:G$3)}

What is this formula doing?  I thought it was 2 to the power of the sum of D3-G3, i.e. 2^102

Therefore I wrote POWER(2,102).  But based on the expected result for the entire formula, it is not.

Help will be appreciated.
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Most Valuable Expert 2012
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Commented:
>>What is this formula doing?

With your data and Excel 2007, it generates an error for me.
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Top Expert 2012

Commented:
I'm getting an error with Excel 2010
Most Valuable Expert 2012
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Commented:
Ditto for Excel 2003.
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Most Valuable Expert 2011
Top Expert 2012

Commented:
If I change the formula to

=C$3^SUM(D$3:H$3)

then I get   5,070,602,400,912,920,000,000,000,000,000.00

   which is close, but rounded to:


select power(2,102) from dual

 5070602400912917605986812821504  which checks out by my math

Author

Commented:
What I have researched so far is that this is an array formula, i.e. the curly braces. To enter array formulas you enter into the cell the following exactly:

=C$3^(D$3:G$3)

and then press CTRL+SHIFT+ENTER
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Commented:
>>and then press CTRL+SHIFT+ENTER

OK, I get a number.  Now to figure out what that number actually is...
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Commented:
doing that I get

4294967296


which is 2^32  or , with respect to your spreadsheet - 2^D3
Most Valuable Expert 2012
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Commented:
>>with respect to your spreadsheet - 2^D3

Looks like sdstuber has it.  The array formula with power seems to only take the first value in the range.

If you play around with any of the other numbers except D3, it doesn't change the value.
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Top Expert 2012

Commented:
building the array formula like this  with ctrl-shift-enter

{=C$3^SUM(D$3:G$3)}

yields  5,070,602,400,912,920,000,000,000,000,000.00


again, as expected but rounded off

so, if you want the exact value in oracle use power(2,102)
if you want oracle to round the same way then use

round(power(2,102),-16)

Author

Commented:
OK...so the array formula I am translating is this :

=TEXT(SUM(INT((27471187-INT(27471187/2^(D$3:G$3))))),"0000000000")

If input = 27471187  (I have hard coded it in above)
output = 0109884643

How do I write this in oracle SQL?
PortletPaulEE Topic Advisor
Most Valuable Expert 2014
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Commented:
I'm not convinced your Excel formula is correct in the first instance.

using CTRL+SHIFT+ENTER so that I get the array formula, the computed result is:

4294967296

which is the same as `

=C3^D3

In other words the range
 (D$3:G$3)
is only returning the first cell, D3
PortletPaulEE Topic Advisor
Most Valuable Expert 2014
Awarded 2013

Commented:
what I observe is that your Excel formula isn't working as expected... so you want some SQL that doesn't work as expected too?

This is not working as expected:
=TEXT(SUM(INT((27471187-INT(27471187/2^(D$3:G$3))))),"0000000000")

what are you expecting from this bit????
2^(D$3:G$3)

Author

Commented:
REPEAT:

the array formula I am translating is this :

{=TEXT(SUM(INT((27471187-INT(27471187/C$3^(D$3:G$3))))),"0000000000")}

If input = 27471187  (I have hard coded it in above)
output = 0109884643

How do I write this in oracle SQL?
The formula makes an array with the 4 numbers {4294967296,268435456,16777216,262144} when array entered (Ctrl+Shift+Enter).
The numbers are the result of the calculations
2^32 = 4294967296
2^28 = 268435456
2^24 = 16777216
2^18 = 262144
Go to the formula line, highlight all, press F9 to see the result, Esc to leave.
But what the purpose is I don't know, since the result in a cell, will only be the first value.

It can be used in another formula.
Sum of the 4 values =SUM(C$3^(D$3:G$3)) Array entered
With Index like this =INDEX(C$3^(D$3:G$3),2) the result will be the second value.
Could be it is made as a test before making a name with the formula.
A name AAAA with the formula in the "Refers to" =C$3^(D$3:G$3), could be used in a formula like above =INDEX(AAAA,2)
PortletPaulEE Topic Advisor
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Commented:
>>"REPEAT"

I'd like to stress that you are not listening to us. The formula in use is not good.
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Commented:
>>The formula makes an array with the 4 numbers

Learned something new about Excel today!!!

>>"REPEAT"

Let's try this approach:
Since you want to migrate from Excel to Oracle, please provide the Oracle specifics you want to replicate?

You currently have provided us with 6 cells of data and one cell with a formula.

Will this equate to 6 columns in an Oracle table and a 7th column to hold some value?

Since we now know the formula produces an array in Excel, do you want a database object column to hold the same type of array?

Author

Commented:
PortletPaul - you state the formula in use is not good but this is what is in operation in the spreadsheet so I do not know how to answer you.  A number is supplied and a new number is created.

I want to create a user defined function. The function will take a number as a parameter (example above was 27471187) and return a number

The static values in the cell (i.e. C3-H3) and the formula (L3) is the rule by which the input value will be converted into the new number.

So in the existing spreadsheet I have:

C3 = 2
D3 = 32
E3 = 28
F3 = 24
G3 = 18
H3 = 0

 INPUT PARAMETER = J3 = 27471187
OUTPUT PARAMETER = L3 (contains the array formula) = {=TEXT(SUM(INT((J3-INT(J3/C$3^(D$3:G$3))))),"0000000000")}  
      
L3 displays = 0109884643

hgholt - thank you for your response!!!  I didnt know about F9, Pressing F9 has given further clarity to what the formula is doing and now I will be able to proceed with the function.

FYI:

C$3^(D$3:G$3) == 2 to the power of the 4 numbers in the cell, e.g. 2^32, 2^28 etc.  (POWER function)
= 4294967296,268435456,16777216,262144

J3/C$3^(D$3:G$3) == divide input number 27471187 by each of the 4 numbers above
= 0.00639613415114582,0.102338146418333,1.63741034269332,104.794261932373

INT(J3/C$3^(D$3:G$3)) ==  round down the numbers to an integer (FLOOR function)
= 0,0,1,104

J3-INT(J3/C$3^(D$3:G$3)) == substract the numbers above from the input number, e.g. 27471187 - 0
= 27471187,27471187,27471186,27471083

INT((J3-INT(J3/C$3^(D$3:G$3)))) == round down the numbers to an integer (FLOOR function)
=27471187,27471187,27471186,27471083

SUM(INT((J3-INT(J3/C$3^(D$3:G$3)))))  == sum the numbers (SUM function)
=109884643

TEXT(SUM(INT((J3-INT(J3/C$3^(D$3:G$3))))),"0000000000")  == left pad the number (LPAD function)
=0109884643 = final output

Author

Commented:
The use of F9 is what I needed to crack this.  I am able to step through the process of what this formula is doing and get the output number.  

Now to write the function!

Once again thank you hgholt.  Thank you all who commented

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