I understand perfectly why 0.1 + 0.2 is not equal to 0.3 due to the floating point. In most of programming languages, 0.1 + 0.2 == 0.3 is False.
But in Excel if(0.1 + 0.2 == 0.3; 1; 0) gives 1
The reason this happens in Excel is because Excel only keeps track of 15 digits of precision. Floating point math for 0.2 + 0.1 results in 0.30000000000000004, and that 4 way out there is the 17th digit. That means Excel just truncates everything after the 15th 0 and is left with 0.300000000000000 which = 0.3
See here for more info: https://en.wikipedia.org/wiki/Numeric_precision_in_Microsoft_Excel
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How is floating point stored? When does it matter?
Is floating point math broken?
Why does the following occur in the Python Interpreter?
>>> 0.1+0.1+0.1-0.3
5.551115123125783e-17
>>> 0.1+0.1
0.2
>>> 0.2+0.1
0.30000000000000004
>>> 0.3-0.3
0.0
>>> 0.2+0.1
0.30000000000000004
>>>
Why doesn't 0.2 + 0.1 = 0.3?
That's because .1 cannot be represented exactly in a binary floating point representation. If you try
>>> .1
Python will respond with .1 because it only prints up to a certain precision, but there's already a small round-off error. The same happens with .3, but when you issue
>>> .2 + .1
0.30000000000000004
then the round-off errors in .2 and .1 accumulate. Also note:
>>> .2 + .1 == .3
False
Not all floating point numbers are exactly representable on a finite machine. Neither 0.1 nor 0.2 are exactly representable in binary floating point. And nor is 0.3.
A number is exactly representable if it is of the form a/b where a and b are an integers and b is a power of 2. Obviously, the data type needs to have a large enough significand to store the number also.
I recommend Rob Kennedy's useful webpage as a nice tool to explore representability.
I don't know how to round up and round down in excel if I have variety of numbers.
For ex: If I have 9.233, then it rounds down to 9 because 0.233 < 0.5, but if it is 2.457, then it rounds up to 3 because 0.457 = 0.5 (if you round it), and then if I have 3.890, then it rounds up to 4 because 0.890 >= 0.5.
I don't know what to do. I tried:
=IF(A1-INT(A1)>0.1,ROUNDUP(A2,0),INT(A2))
but it doesn't work.
Please help
Have you tried just the ROUND(, 0) for your purpose.
This should yeild the desired result.
Num Rounded Num
-----------------------
9.233 9
2.457 2
3.89 4
I have a dataset which looks like this:
0 1 0.1
0 0 0.1
0 1 0.1
1 0 0.2
0 1 0.2
1 0 0.2
...
I now want to do the following operations on each different value in the third column of the table:
Example for 0.1:
First column values summed: 0+0+0=0
Second column values summed: 1+0+1=2
Now I want to substract these two 2-0=2 and in a last step divide them by the occurrences.
2/3 =0.667
The same for 0.2 and my plot should then plot at x=0.1, y=0.667.
I hope my problem is with the example understandable.
You can use the smooth unique option to do exactly this: sum up all y-values belonging to the same x-value and then divide the result by the number of occurences. For the second column, upon which the operation is performed, you use the difference between the second and first column:
plot 'file.txt' using 3:($2 - $1) smooth unique
However, it seems like you'll run in a strange bug then. This works only correct, if you insert an empty or commented row at the beginning of your data file:
The result with the following file.txt
#
0 1 0.1
0 0 0.1
0 1 0.1
1 0 0.2
0 1 0.2
1 0 0.2
is
I am trying to measure how much have accuracy when I convert to binary fixed point representation way.
first I tried use this 0.9375. And I got the binary 0.1111.
second I tried used this 0.9377 and I also got the binary 0.1111
There is nothing different between them.
Also how can I solve this problem?
Is there any other way? To make converting ?
For your understand, I let me know some more example,
For example, If I want to convert 3.575 to binary then 3.575 be 11.1001.
but If I back to again to decimal then 3.5625. It so quite different on original value.
From a similar question we have:
Base 2: Twos complement 4 integer, 4 bit fractional
-2^3 2^2 2^1 2^0 . 2^-1 2^-2 2^-3 2^-4
-8 4 2 1 . 0.5 0.25 0.125 0.0625
With only 4 fractional bits the represented number only has an accuracy of 0.0625
3.575 could be 11.1001 = 2+ 1+ 0.5 + 0.0625 => 3.5625 to low
or 11.1010 = 2+ 1+ 0.5 + 0.125 => 3.625 to high
This should indicate that 4 bits is just not enough to represent "3.575" exactly.
To figure out how many bit you would need multiply by a power of 2 until you get an integer: For "3.575" it is rather a lot (50 fractional bits).
3.575 * 2^2 = 14.3 (not integer)
3.575 * 2^20 = 3748659.2
3.575 * 2^30 = 3838627020.8
3.575 * 2^40 = 3930754069299.2 (not integer)
3.575 * 2^50 = 4025092166962381.0 (INTEGER) we need 50 bits!
3.575 => 11.10010011001100110011001100110011001100110011001101
Multiplying by a power of two shift the word to the left (<<) When there is no fractional bits left it means the number is fully represented, then number of shifts is the number of fractional bits required.
For fixed point you are better off thinking about the level of precision your application requires.
Have created formula for example =(E2+(G2*37))/290 which returns decimals based on data entered but can't work out how to round the answer to get the following:
If 0 to 0.39 round to 0
If 0.4 to 0.89 round to 0.5
If 0.9 to 1.39 round to 1
If 1.4 to 1.89 round to 1.5
If 1.9 to 2.39 round to 2 etc
Hope my question makes sense. Thanks
Your custom rounding is simply rounding to the nearest 0.5.....but with an "offset" of 0.15. With value to be rounded in A1 you can do that with a simple formula, i.e.
=ROUND(A1*2-0.3,0)/2
or with your calculation in place of A1 that becomes
=ROUND((E2+G2*37)/290*2-0.3,0)/2
It's a bit convoluted but this should work:
=IF(A1 + 0.1 - ROUNDDOWN(A1+0.1;0) < 0.5; ROUNDDOWN(A1+0.1;0); ROUNDDOWN(A1+0.1;0) + 0.5)
where A1 is the cell you want to round.
e.g.
N.B.
This works only for positive numbers.
Ranges are actually:
[0, 0.39999999...] [0.4 , 0.8999999...] ...
or equally:
[0, 0.4) [0.4 , 0.9) ...
You could define a VBA function to do this for you:
Public Function CustomRound(number As Double)
Dim result As Double, fraction As Double
fraction = number - Int(number)
If number <= 0.39 Then
result = 0
ElseIf fraction >= 0.4 And fraction <= 0.9 Then
result = Int(number) + 0.5
Else
result = Round(number, 0)
End If
CustomRound = result
End Function
You would call this as follows:
=CustomRound((E2+(G2*37))/290)
Example:
=CustomRound(0.23) // 0
=CustomRound(1.58) //1.5
=CustomRound(2.12) //2