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')
Related
Need some help on the below piece of code that I am working on. Why original number in "a" is different from "c" when it goes through a type conversion. Any way we can make "a" and "c" same when it goes through float -> int type conversion?
a = '46700000000987654321'
b = float(a) => 4.670000000098765e+19
c = int(b) => 46700000000987652096
a == c => False
Please read this document about Floating Point Arithmetic: Issues and Limitations :
https://docs.python.org/3/tutorial/floatingpoint.html
for your example:
from decimal import Decimal
a='46700000000987654321'
b=Decimal(a)
print(b) #46700000000987654321
c=int(b)
print(c) #46700000000987654321
Modified version of my answer to another question (reasonably) duped to this one:
This happens because 46700000000987654321 is greater than the integer representational limits of a C double (which is what a Python float is implemented in terms of).
Typically, C doubles are IEEE 754 64 bit binary floating point values, which means they have 53 bits of integer precision (the last consecutive integer values float can represent are 2 ** 53 - 1 followed by 2 ** 53; it can't represent 2 ** 53 + 1). Problem is, 46700000000987654321 requires 66 bits of integer precision to store ((46700000000987654321).bit_length() will provide this information). When a value is too large for the significand (the integer component) alone, the exponent component of the floating point value is used to scale a smaller integer value by powers of 2 to be roughly in the ballpark of the original value, but this means that the representable integers start to skip, first by 2 (as you require >53 bits), then by 4 (for >54 bits), then 8 (>55 bits), then 16 (>56 bits), etc., skipping twice as far between representable values for each additional bit of magnitude you have beyond 53 bits.
In your case, 46700000000987654321, converted to float, has an integer value of 46700000000987652096 (as you noted), having lost precision in the low digits.
If you need arbitrarily precise base-10 floating point math, replace your use of float with decimal.Decimal (conveniently, your initial value is already a string, so you don't risk loss of precision between how you type a float and the actual value stored); the default precision will handle these values, and you can increase it if you need larger values. If you do that (and convert a to an int for the comparison, since a str is never equal to any numeric type), you get the behavior you expected:
from decimal import Decimal as Dec, getcontext
a = "46700000000987654321"
b = Dec(a); print(b) # => 46700000000987654321
c = int(b); print(c) # => 46700000000987654321
print(int(a) == c) # => True
Try it online!
If you echo the Decimals in an interactive interpreter instead of using print, you'd see Decimal('46700000000987654321') instead, which is the repr form of Decimals, but it's numerically 46700000000987654321, and if converted to int, or stringified via any method that doesn't use the repr, e.g. print, it displays as just 46700000000987654321.
I need the first 4 digits after the front 0s in a decimal while also keeping the 0s in the output without getting scientific notation.
Would like to take
0.0000000000000000000000634546534
and get
0.00000000000000000000006345 but not
6.345e-23.
But would also like to take
231.00942353246
and get
231.009423.
Thank you.
Here's one way using the decimal module.
import decimal
example = 0.0000000000000000000000634546534
x = decimal.Decimal(example)
sign = x.as_tuple().sign
digits = x.as_tuple().digits
exponent = x.as_tuple().exponent
figs = 4
result = decimal.Decimal((sign, digits[:figs], len(digits)+(exponent)-figs))
precision = -1 * (len(digits) + (exponent) - figs) # for this example: -1 * (99 + (-121) - 4)
print("{:.{precision}f}".format(float(result), precision=precision))
Result:
0.00000000000000000000006345
Note that Decimal stores 99 digits because of floating point imprecision. The example variable holds a float value (due to the initializer value) that is inherently imprecise. There is no way around this, unless you are able to represent the original float value as a string, that you can use to initialize the example variable.
There are cases, where the 4th digit shown will be wrong, because in the floating point representation that digit is represented as one lesser and the next digit is a 9, for example. To fix, this, we grab one more digit as well to use for rounding. This should work in most cases, as the imprecision should be within the nearest rounding threshold.
result = decimal.Decimal((0, digits[:figs + 1], len(digits)+(exponent)-figs-1))
Lastly, to handle the case where there are numbers before the decimal, we can simply store it, remove it, and re-add it:
whole_number_part = int(example)
example -= whole_number_part
...
result += whole_number_part
Altogether, we get:
import decimal
example = 231.00942353246
whole_number_part = int(example)
example -= whole_number_part
x = decimal.Decimal(example)
sign = x.as_tuple().sign
digits = x.as_tuple().digits
exponent = x.as_tuple().exponent
figs = 4
result = decimal.Decimal((0, digits[:figs + 1], len(digits)+(exponent)-figs-1))
result += whole_number_part
precision = -1 * (len(digits) + (exponent) - figs)
print("{:.{precision}f}".format(float(result), precision=precision))
Result:
231.009423
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).
I have a problem in sas. I have a dataset where the numbers are stored as string. The problem is that the size and format of the numbers vary a lot. I will give an example:
data s;
x='123456789012345';
y=input(x,best32.);
z = '0.0001246564';
a = input(z,best32.);
put 'a='y;
put a;
keep y a;
run;
Output:
y=
1.2345679E14
a=
0.00012465644
As you can see I lose information in the large integer. How can I make my program in order to not lose information. As far as I understand this number has less than 15 digits sas large number. I really miss python where I just can set y = float(x).
No loss of information has occurred. You're just mistaking informat for format.
The informat best32. told SAS to import the string with a 32 width. All good, you have all 15 characters stored in the number.
Now if you want to see all 15 characters, you need to use a format wider than the default best12. format to see it on output:
data s;
x='123456789012345';
y=input(x,best32.);
z = '0.0001246564';
a = input(z,best32.);
put y= best32.;
put a= best32.;
keep y a;
run;
But even if you don't display it this way, the number y still is exactly equal to 123456789012345, if you were to do math with it or similar - you haven't lost any information, you just weren't displaying it right.
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