I exported a XML file from Qlikview and the dates are in this 16-letter/digits form (i.e. 40E5A40D641FDB97). I have tried multiple ways to convert it to floating decimals and then dates but all methods have failed (incl. Excel HEX2DEC).
Anyone has dealt with this issue before? Would greatly appreciate any help!
Here is a Power Query routine that will convert that Hex number into its Date Equivalent:
I generate the binary equivalent of the Hex number using a lookup table and concatenating the results.
The algorithm should be clear in the coding, and it follows the rules set out in IEEE-754.
For the dates you mention in your question, it provides the same results.
Note that this routine assumes a valid value encoded as you describe your date representations from Qlikview. It is not a general purpose routine.
let
//don't really need the Decimal column
hexConvTable = Table.FromRecords({
[Hex="0", Dec=0, Bin = "0000"],
[Hex="1", Dec=1, Bin = "0001"],
[Hex="2", Dec=2, Bin = "0010"],
[Hex="3", Dec=3, Bin = "0011"],
[Hex="4", Dec=4, Bin = "0100"],
[Hex="5", Dec=5, Bin = "0101"],
[Hex="6", Dec=6, Bin = "0110"],
[Hex="7", Dec=7, Bin = "0111"],
[Hex="8", Dec=8, Bin = "1000"],
[Hex="9", Dec=9, Bin = "1001"],
[Hex="A", Dec=10, Bin = "1010"],
[Hex="B", Dec=11, Bin = "1011"],
[Hex="C", Dec=12, Bin = "1100"],
[Hex="D", Dec=13, Bin = "1101"],
[Hex="E", Dec=14, Bin = "1110"],
[Hex="F", Dec=15, Bin = "1111"]},
type table[Hex = Text.Type, Dec = Int64.Type, Bin = Text.Type]),
hexUp = Text.Upper(hexNum),
hexSplit = Table.FromList(Text.ToList(hexUp),Splitter.SplitByNothing(),{"hexNum"}),
//To sort back to original order
addIndex = Table.AddIndexColumn(hexSplit,"Index",0,1,Int64.Type),
//combine with conversion table
binConv = Table.Sort(
Table.Join(
addIndex,"hexNum",hexConvTable,"Hex",JoinKind.LeftOuter),
{"Index", Order.Ascending}),
//equivalent binary
binText = Text.Combine(binConv[Bin]),
sign = Text.Start(binText,1),
//change exponent binary parts to numbers
expBin = List.Transform(Text.ToList(Text.Middle(binText,1,11)),Number.FromText),
//exponent bias will vary depending on the precision being used
expBias = 1023, //Number.Power(2,10-List.PositionOf(expBin,1))-1,
expPwr= List.Reverse({0..10}),
exp = List.Accumulate({0..10},0,(state, current) =>
state + (expBin){current} * Number.Power(2,expPwr{current})) - expBias,
mantBin = List.Transform(Text.ToList(Text.Middle(binText,11,52)),Number.FromText),
mantPwr = {0..51},
mant = List.Accumulate({0..51},0,(state, current) =>
state + (mantBin){current} / Number.Power(2,mantPwr{current})) +1,
dt = mant * Number.Power(2,exp)
in
DateTime.From(dt)
you can use standard windows formatting with Num# (convert text to number) and Num to convert from hex to bin in Qlikview :
# example data from inline table in loading script
[our_hex_numbers]:
LOAD
Num(Num#(hex,'(HEX)'),'(BIN)') as bin
Inline
[hex,
'A',
'B',
'C'];
here is result:
This reference shows how floating point numbers are represented. In double precision (using a total of 64 bits) there is a sign bit, 11-bit exponent and 53-bit significand or mantissa. Observant readers will notice that gives a total of 65 bits: this is because the most significant bit in the mantissa is a hidden bit which by convention is always set to 1 and does not have to be stored.
Taking the first example:
we have
Exponent
The exponent is the first three hexadecimal digits (sign bit plus 11 digits - but the sign bit will always be zero for dates since they are positive numbers). It can be converted using any suitable standard method e.g. in Excel 365:
=LET(L,LEN(A2),seq,SEQUENCE(L),SUM((FIND(MID(A2,seq,1),"0123456789ABCDEF")-1)*16^(L-seq)))
The correct result is obtained by subtracting 1023 (the offset) from the converted value e.g.
40E -> 1038
1038 - 1023 -> 15
So the multiplier is 2^15.
Significand
We need to take the right-hand 13 hexadecimal digits (52 bits) of the string and convert it to a fraction using whatever is your favourite conversion method e.g. in Excel 365:
=LET(L,LEN(A2),seq,SEQUENCE(L),SUM((FIND(MID(A2,seq,1),"0123456789ABCDEF")-1)*16^(-seq)))
Then you need to add 1 (this is the hidden bit which is always set to 1).
Putting this together:
I made a report on QlikView licenses for myself using the file CalData.pgo.xml
and I ran into a non - critical problem of converting hex to date .. but without this transformation, the report would not be complete. (LastUsed, ToBeDeleted)
In general, I was looking for it, but I didn't find anything useful right away, except for converting 13x hex to excel.
But in the file CalData.pgo.xml the date is set in 16 digits, not 13.. I did not understand how to adapt the excel formula for 16 digits, but I realized that it is possible to trim a 16-bit hex to 13 digits. and it seems that nothing significant is lost at the same time .
it works fine for me
=date((num(Num#(right([PerDocumentCalData/NamedCalsAllocated/CalAllocEntry.LastUsed],13),'(HEX)') )*pow(16,-13)+1)*Pow(2,15),'DD.MM.YYYY hh:mm')
I am writing a program where I need to delete duplicate points stored in a matrix. The problem is that when it comes to check whether those points are in the matrix, MATLAB can't recognize them in the matrix although they exist.
In the following code, intersections function gets the intersection points:
[points(:,1), points(:,2)] = intersections(...
obj.modifiedVGVertices(1,:), obj.modifiedVGVertices(2,:), ...
[vertex1(1) vertex2(1)], [vertex1(2) vertex2(2)]);
The result:
>> points
points =
12.0000 15.0000
33.0000 24.0000
33.0000 24.0000
>> vertex1
vertex1 =
12
15
>> vertex2
vertex2 =
33
24
Two points (vertex1 and vertex2) should be eliminated from the result. It should be done by the below commands:
points = points((points(:,1) ~= vertex1(1)) | (points(:,2) ~= vertex1(2)), :);
points = points((points(:,1) ~= vertex2(1)) | (points(:,2) ~= vertex2(2)), :);
After doing that, we have this unexpected outcome:
>> points
points =
33.0000 24.0000
The outcome should be an empty matrix. As you can see, the first (or second?) pair of [33.0000 24.0000] has been eliminated, but not the second one.
Then I checked these two expressions:
>> points(1) ~= vertex2(1)
ans =
0
>> points(2) ~= vertex2(2)
ans =
1 % <-- It means 24.0000 is not equal to 24.0000?
What is the problem?
More surprisingly, I made a new script that has only these commands:
points = [12.0000 15.0000
33.0000 24.0000
33.0000 24.0000];
vertex1 = [12 ; 15];
vertex2 = [33 ; 24];
points = points((points(:,1) ~= vertex1(1)) | (points(:,2) ~= vertex1(2)), :);
points = points((points(:,1) ~= vertex2(1)) | (points(:,2) ~= vertex2(2)), :);
The result as expected:
>> points
points =
Empty matrix: 0-by-2
The problem you're having relates to how floating-point numbers are represented on a computer. A more detailed discussion of floating-point representations appears towards the end of my answer (The "Floating-point representation" section). The TL;DR version: because computers have finite amounts of memory, numbers can only be represented with finite precision. Thus, the accuracy of floating-point numbers is limited to a certain number of decimal places (about 16 significant digits for double-precision values, the default used in MATLAB).
Actual vs. displayed precision
Now to address the specific example in the question... while 24.0000 and 24.0000 are displayed in the same manner, it turns out that they actually differ by very small decimal amounts in this case. You don't see it because MATLAB only displays 4 significant digits by default, keeping the overall display neat and tidy. If you want to see the full precision, you should either issue the format long command or view a hexadecimal representation of the number:
>> pi
ans =
3.1416
>> format long
>> pi
ans =
3.141592653589793
>> num2hex(pi)
ans =
400921fb54442d18
Initialized values vs. computed values
Since there are only a finite number of values that can be represented for a floating-point number, it's possible for a computation to result in a value that falls between two of these representations. In such a case, the result has to be rounded off to one of them. This introduces a small machine-precision error. This also means that initializing a value directly or by some computation can give slightly different results. For example, the value 0.1 doesn't have an exact floating-point representation (i.e. it gets slightly rounded off), and so you end up with counter-intuitive results like this due to the way round-off errors accumulate:
>> a=sum([0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1]); % Sum 10 0.1s
>> b=1; % Initialize to 1
>> a == b
ans =
logical
0 % They are unequal!
>> num2hex(a) % Let's check their hex representation to confirm
ans =
3fefffffffffffff
>> num2hex(b)
ans =
3ff0000000000000
How to correctly handle floating-point comparisons
Since floating-point values can differ by very small amounts, any comparisons should be done by checking that the values are within some range (i.e. tolerance) of one another, as opposed to exactly equal to each other. For example:
a = 24;
b = 24.000001;
tolerance = 0.001;
if abs(a-b) < tolerance, disp('Equal!'); end
will display "Equal!".
You could then change your code to something like:
points = points((abs(points(:,1)-vertex1(1)) > tolerance) | ...
(abs(points(:,2)-vertex1(2)) > tolerance),:)
Floating-point representation
A good overview of floating-point numbers (and specifically the IEEE 754 standard for floating-point arithmetic) is What Every Computer Scientist Should Know About Floating-Point Arithmetic by David Goldberg.
A binary floating-point number is actually represented by three integers: a sign bit s, a significand (or coefficient/fraction) b, and an exponent e. For double-precision floating-point format, each number is represented by 64 bits laid out in memory as follows:
The real value can then be found with the following formula:
This format allows for number representations in the range 10^-308 to 10^308. For MATLAB you can get these limits from realmin and realmax:
>> realmin
ans =
2.225073858507201e-308
>> realmax
ans =
1.797693134862316e+308
Since there are a finite number of bits used to represent a floating-point number, there are only so many finite numbers that can be represented within the above given range. Computations will often result in a value that doesn't exactly match one of these finite representations, so the values must be rounded off. These machine-precision errors make themselves evident in different ways, as discussed in the above examples.
In order to better understand these round-off errors it's useful to look at the relative floating-point accuracy provided by the function eps, which quantifies the distance from a given number to the next largest floating-point representation:
>> eps(1)
ans =
2.220446049250313e-16
>> eps(1000)
ans =
1.136868377216160e-13
Notice that the precision is relative to the size of a given number being represented; larger numbers will have larger distances between floating-point representations, and will thus have fewer digits of precision following the decimal point. This can be an important consideration with some calculations. Consider the following example:
>> format long % Display full precision
>> x = rand(1, 10); % Get 10 random values between 0 and 1
>> a = mean(x) % Take the mean
a =
0.587307428244141
>> b = mean(x+10000)-10000 % Take the mean at a different scale, then shift back
b =
0.587307428244458
Note that when we shift the values of x from the range [0 1] to the range [10000 10001], compute a mean, then subtract the mean offset for comparison, we get a value that differs for the last 3 significant digits. This illustrates how an offset or scaling of data can change the accuracy of calculations performed on it, which is something that has to be accounted for with certain problems.
Look at this article: The Perils of Floating Point. Though its examples are in FORTRAN it has sense for virtually any modern programming language, including MATLAB. Your problem (and solution for it) is described in "Safe Comparisons" section.
type
format long g
This command will show the FULL value of the number. It's likely to be something like 24.00000021321 != 24.00000123124
Try writing
0.1 + 0.1 + 0.1 == 0.3.
Warning: You might be surprised about the result!
Maybe the two numbers are really 24.0 and 24.000000001 but you're not seeing all the decimal places.
Check out the Matlab EPS function.
Matlab uses floating point math up to 16 digits of precision (only 5 are displayed).
a = 28.85
b = 2000
print(a*b)
Result 57700.0
select name from fake limit 57700.0 ,10
This sentence is incorrect.
Multiplying a float with an int naturally gives you a float as an answer.
So, as #Rakesh suggested, truncate it with int(a*b).
Beware that you will lose everything after the dot...
You can keep the result as float if you use format in your print statement and cut all numbers behind the dot.
print("{:0.0f}".format(a * b))
So, I've learned quite a few ways to control the precision when I'm dealing with floats.
Here is an example of 3 different techniques:
somefloat=0.0123456789
print("{0:.10f}".format(somefloat))
print("%.5f" % somefloat)
print(Decimal(somefloat).quantize(Decimal(".01")))
This will print:
0.0123456789
0.01235
0.01
In all of the above examples, the precision itself is a fixed value, but how could I turn the precision itself a variable that could be
be entered by the end-user?
I mean, the fixed precision values are now inside quatations marks, and I can't seem to find a way to add any variable there. Is there a way, anyway?
I'm on Python 3.
Using format:
somefloat=0.0123456789
precision = 5
print("{0:.{1}f}".format(somefloat, precision))
# 0.01235
Using old-style string interpolation:
print("%.*f" % (precision, somefloat))
# 0.01235
Using decimal:
import decimal
D = decimal.Decimal
q = D(10) ** -precision
print(D(somefloat).quantize(q))
# 0.01235
In an IOS program i am trying to divide some float value but the result is incorrect
float a = 179.891891;
float b = 8.994595;
NSLog(#"Result %f",a/b);
On dividing the two (a/b) the output i get is 20.0000 instead of 19.9999989993991 . I have tried using double instead of float but still the same issue . The value of "b" keeps on varying as i obtain it from some calculations . I need the result to be exactly precise as it further is required for some calculations and 20.0000 instead of 19.9999989993991 make a lot of difference in the final output i get .
Any help on this would be really great :).
I dont know in Objective C, but in C, you should do casting: (float)(a/b) .Otherwise it is integer division.