Trying to transform a real_32 to real_64, I'm getting
real_32: 61.55
real_64: 61.54999923706055
Am I wrong with the to_double function?
This is expected. In the particular example, the binary representation of the decimal 61.55 with single and double precision respectively is:
REAL_32: 0 10000100 11101100011001100110011
REAL_64: 0 10000000100 1110110001100110011001100110011001100110011001100110
As you can see, the trailing pattern 0011 is recurrent and should go ad infinitum to give a precise value.
When REAL_32 is assigned to REAL_64, the trailing 0011s are not added automatically, but filled with zeroes instead:
REAL_32: 0 10000100 11101100011001100110011
REAL_64: 0 10000000100 1110110001100110011001100000000000000000000000000000
In decimal notation, this corresponds to 61.54999923706055. What is essential here, 61.54999923706055 and 61.55 have exactly the same binary representation when using single precision floating numbers. You can check it yourself with print ({REAL_32} 61.55 = {REAL_32} 61.54999923706055). In other words, the results you get are correct, and the two values are the same. The only difference is that when REAL_32 is printed, it is rounded to lower number of meaningful decimal digits.
This is the reason why accounting and bookkeeping software never uses floating-point numbers, only integer and decimal.
As a workaround working for getting from JSON into typescript deserialization, the following worked:
a_real_32.out.to_real_64
I have stuck (again) and looking for smart human beings of planet earth to help me out.
Background
I have an application which distributes the amounts to some users in a given percentage. Say I have $35000 and it will distribute the amounts to 3 users (A, B and C) in some ratio. So the amount distributed will be
A - 5691.05459265518
B - 14654.473815207
C - 14654.4715921378
which totals up to $35000
The Problem
I have to provide the results on the front end in 2 decimal spaces instead of float. So I use the round function of SQL Server with the precision value of 2 to convert these to 2 decimal spaces with rounding. But the issue is that when I total these values this comes out to be $34999.9999 instead of $35000.
My Findings
I searched a bit and found
If the expression that you are rounding ends with a 5, the Round()
function will round the expression so that the last digit is an even
number. Here are some examples:
Round(34.55, 1) - Result: 34.6 (rounds up)
Round(34.65, 1) - Result: 34.6 (rounds down)
So technically the answer is correct but I am looking for a function or a way to round of the value exactly what it should have been. I found that if I start rounding off (if the value is less than 5 then leave the previous number else increment the previous digit by 1 ) from the last digit after the decimal and keep on backtracking while I am left with only 2 decimal places.
Please advise.
I have an excel spreadsheet with numbers from 000 to 999 and am trying to find repeated numbers inside a cell.
(So for example, printing 1 if the number is 022 , 555 or 115 and 0 if it isn't)
So far, I have not been able to find a solution.
Feel free to ask for more information and thanks in advance.
This will do: =IF(COUNT(SEARCH(REPT({0,1,2,3,4,5,6,7,8,9},2),A1))>0,1,0)
Note: If value in cell A1 contains 2 repeated digits it will show 1 else 0. You can customize the repetition limit by changing 2 in the part 8,9},2).
You could try this one if you wanted to find repeated digits not necessarily next to each other:-
=IF(MAX(LEN(A1)-LEN(SUBSTITUTE(A1,{0,1,2,3,4,5,6,7,8,9},"")))>1,1,0)
If the numbers are stored as 3-digit numbers and you wanted it to work for (e.g.) 001, would need:-
=IF(MAX(LEN(TEXT($A1,"000"))-LEN(SUBSTITUTE(TEXT($A1,"000"),{0,1,2,3,4,5,6,7,8,9},"")))>1,1,0)
If your data is in Range "A1:A100" and you want to locate repeated numbers in the range for instance, enter =IF(COUNTIF(A:A,A1)>1,1,0) in cell B1 and fill down. But if you want to check repetitions of specific numbers like 022, 555 or 115, enter =IF(OR(AND(A1=022,COUNTIF(A:A,A1)>1),AND(A1=555,COUNTIF(A:A,A1)>1),AND(A1=115,COUNTIF(A:A,A1)>1)),1,0) in cell B1 and fill down.
being a number, use arithmetics to break it into digits and then check if all are different.
the formula is
=INT(NOT(AND(INT(A1/100)<>INT(MOD(A1,100)/10),INT(A1/100)<>MOD(A1,10),INT(MOD(A1,100)/10)<>MOD(A1,10))))
let's analyze it step by step
first, INT(A1/100) extracts the first digit (the integer division by 100); then INT(MOD(A1,100)/10) extracts the second digit (the integer division by 10 of the modulo 100); and MOD(A1,10) extracts the last digit (the modulo 10).
next there are the three comparisons of difference <> first with second, second with third and first with third, combined with AND() and finally take the result, negate it NOT() and transforming it into an integer 0 or 1 with INT()
In my code I get an array like this:
array(['2.83100e+07', '2.74000e+07', '2.79400e+07'],dtype='|S11')
How can I "cut" my values like:
2.83100e+07 --> 2.831 ?
Best regards!
using a for loop and round(n)
In [23]: round(66.66666666666,4)
Out[23]: 66.6667
In [24]: round(1.29578293,6)
Out[24]: 1.295783
help on round():
round(number[, ndigits]) -> floating point number
Round a number to a given precision in decimal digits (default 0 digits). This always returns a floating point number. Precision may be negative
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Possible Duplicate:
How to convert integer value to Roman numeral string?
Back in my college days, when learning C and all i had come across a question of representing a year in its corresponding Roman Numeral form. There was no solution in that text, as it was among some extra questions to ponder. All i could think of was using modulus operator and a bunch of ifs. I was wondering if some one could give me a proper solution. A simple algorithm or explanation of the logic used would be appreciated.
Here's logic for numbers less than 4,000. (See below for what to do above that.) There's a basic, 4-step algorithm that is applied at each level of magnitude.
Determine the number of thousands in the number: floor(number/1000). Output that many "M"s and subtract that many thousands from the number. At this point, the number is less than 1,000.
If the number is >= 900, output "CM" and subtract 900.
If the number is >= 500, output "D" and subtract 500.
If the number is >= 400, output "CD" and subtract 400.
At this point, the number is guaranteed to be < 400. We follow a similar pattern to reduce the number to less than 40:
Determine the number of hundreds in the number, output that many "C"s and subtract that many hundreds from the number. At this point, the number is less than 100.
If the number is >= 90, output "XC" and subtract 900.
If the number is >= 50, output "L" and subtract 50.
If the number is >= 40, output "XL" and subtract 40.
At this point, the number is guaranteed to be < 40. We repeat exactly the same logic using "X", "IX", "V", and "IV". We finish by using the number (guaranteed to be < 4) as the count of how many "I"s to output.
For larger numbers, the logic is still the same, we just use the standard symbols with bars on top. Each bar represents 1,000 times the value without that bar. (So V with a bar is 5,000, etc.)