I have a float and would like to limit to just two decimals.
I've tried format(), and round(), and still just get 0, or 0.0
x = 8.972990688205408e-05
print ("x: ", x)
print ("x using round():", round(x))
print ("x using format():"+"{:.2f}".format(x))
output:
x: 8.972990688205408e-05
x using round(): 0
x using format():0.00
I'm expecting 8.98, or 8.97 depending on what method used. What am I missing?
You are using the scientific notation. As glhr pointed out in the comments, you are trying to round 8.972990688205408e-05 = 0.00008972990688205408. This means trying to round as type float will only print the first two 0s after the decimal points, resulting in 0.00. You will have to format via 0:.2e:
x = 8.972990688205408e-05
print("{0:.2e}".format(x))
This prints:
8.97e-05
You asked in one of your comments on how to get only the 8.97.
This is the way to do it:
y = x*1e+05
print("{0:.2f}".format(y))
output:
8.97
In python (and many other programming language), any number suffix with an e with a number, it is power of 10 with the number.
For example
8.9729e05 = 8.9729 x 10^3 = 8972.9
8.9729e-05 = 8.9729 x 10^-3 = 0.000089729
8.9729e0 = 8.9729 x 10^0 = 8.9729
8.972990688205408e-05 8.972990688205408 x 10^-5 = 0.00008972990688205408
8.9729e # invalid syntax
As pointed out by other answer, if you want to print out the exponential round up, you need to use the correct Python string format, you have many choices to choose from. i.e.
e Floating point exponential format (lowercase, precision default to 6 digit)
e Floating point exponential format (uppercase, precision default to 6 digit).
g Same as "e" if exponent is greater than -4 or less than precision, "f" otherwise
G Same as "E" if exponent is greater than -4 or less than precision, "F" otherwise
e.g.
x = 8.972990688205408e-05
print('{:e}'.format(x)) # 8.972991e-05
print('{:E}'.format(x)) # 8.972991E-05
print('{:.2e}'.format(x)) # 8.97e-05
(Update)
OP asked a way to remove the exponent "E" number. Since str.format() or "%" notation just output a string object, break the "e" notation out of the string will do the trick.
'{:.2e}'.format(x).split("e") # ['8.97', '-05']
print('{:.2e}'.format(x).split('e')[0]) # 8.97
If I understand correctly, you only want to round the mantissa/significand? If you want to keep x as a float and output a float, just specify the precision when calling round:
x = round(8.972990688205408e-05,7)
Output:
8.97e-05
However, I recommend converting x with the decimal module first, which "provides support for fast correctly-rounded decimal floating point arithmetic" (see this answer):
from decimal import Decimal
x = Decimal('8.972990688205408e-05').quantize(Decimal('1e-7')) # output: 0.0000897
print('%.2E' % x)
Output:
8.97E-05
Or use the short form of the format method, which gives the same output:
print(f"{x:.2E}")
rount() returns closest multiple of 10 to the power minus ndigits,
so there is no chance you will get 8.98 or 8.97. you can check here also.
Related
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 the following array in python
n = [565387674.45, 321772103.48,321772103.48, 214514735.66,214514735.65,
357524559.41]
if I sum all these elements, I get this:
sum(n)
1995485912.1300004
But, this sum should be:
1995485912.13
In this way, I know about floating point "error". I already used the isclose() function from numpy to check the corrected value, but
how much is this limit? Is there any way to reduce this "error"?
The main issue here is that the error propagates to other operations, for example, the below assertion must be true:
assert (sum(n) - 1995485911) ** 100 - (1995485912.13 - 1995485911) ** 100 == 0.
This is problem with floating point numbers. One solution is having them represented in string form and using decimal module:
n = ['565387674.45', '321772103.48', '321772103.48', '214514735.66', '214514735.65',
'357524559.41']
from decimal import Decimal
s = sum(Decimal(i) for i in n)
print(s)
Prints:
1995485912.13
You could use round(num, n) function which rounds the number to the desired decimal places. So in your example you would use round(sum(n), 2)
Why does the math module return the wrong result?
First test
A = 12345678917
print 'A =',A
B = sqrt(A**2)
print 'B =',int(B)
Result
A = 12345678917
B = 12345678917
Here, the result is correct.
Second test
A = 123456758365483459347856
print 'A =',A
B = sqrt(A**2)
print 'B =',int(B)
Result
A = 123456758365483459347856
B = 123456758365483467538432
Here the result is incorrect.
Why is that the case?
Because math.sqrt(..) first casts the number to a floating point and floating points have a limited mantissa: it can only represent part of the number correctly. So float(A**2) is not equal to A**2. Next it calculates the math.sqrt which is also approximately correct.
Most functions working with floating points will never be fully correct to their integer counterparts. Floating point calculations are almost inherently approximative.
If one calculates A**2 one gets:
>>> 12345678917**2
152415787921658292889L
Now if one converts it to a float(..), one gets:
>>> float(12345678917**2)
1.5241578792165828e+20
But if you now ask whether the two are equal:
>>> float(12345678917**2) == 12345678917**2
False
So information has been lost while converting it to a float.
You can read more about how floats work and why these are approximative in the Wikipedia article about IEEE-754, the formal definition on how floating points work.
The documentation for the math module states "It provides access to the mathematical functions defined by the C standard." It also states "Except when explicitly noted otherwise, all return values are floats."
Those together mean that the parameter to the square root function is a float value. In most systems that means a floating point value that fits into 8 bytes, which is called "double" in the C language. Your code converts your integer value into such a value before calculating the square root, then returns such a value.
However, the 8-byte floating point value can store at most 15 to 17 significant decimal digits. That is what you are getting in your results.
If you want better precision in your square roots, use a function that is guaranteed to give full precision for an integer argument. Just do a web search and you will find several. Those usually do a variation of the Newton-Raphson method to iterate and eventually end at the correct answer. Be aware that this is significantly slower that the math module's sqrt function.
Here is a routine that I modified from the internet. I can't cite the source right now. This version also works for non-integer arguments but just returns the integer part of the square root.
def isqrt(x):
"""Return the integer part of the square root of x, even for very
large values."""
if x < 0:
raise ValueError('square root not defined for negative numbers')
n = int(x)
if n == 0:
return 0
a, b = divmod(n.bit_length(), 2)
x = (1 << (a+b)) - 1
while True:
y = (x + n//x) // 2
if y >= x:
return x
x = y
If you want to calculate sqrt of really large numbers and you need exact results, you can use sympy:
import sympy
num = sympy.Integer(123456758365483459347856)
print(int(num) == int(sympy.sqrt(num**2)))
The way floating-point numbers are stored in memory makes calculations with them prone to slight errors that can nevertheless be significant when exact results are needed. As mentioned in one of the comments, the decimal library can help you here:
>>> A = Decimal(12345678917)
>>> A
Decimal('123456758365483459347856')
>>> B = A.sqrt()**2
>>> B
Decimal('123456758365483459347856.0000')
>>> A == B
True
>>> int(B)
123456758365483459347856
I use version 3.6, which has no hardcoded limit on the size of integers. I don't know if, in 2.7, casting B as an int would cause overflow, but decimal is incredibly useful regardless.
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