I am working with an MPU-6050 3-axis accelerometer and using this code to read the current Z axis value with 1/10 second between readings:
az=round(imu.accel.z,2) + 0.04 (0.04 is the calibration value)
print(str(az))
Most times the value displayed with the print statement is correct (i.e., 0.84). But sometimes the value printed is the full seven-decimal place value (0.8400001). Is there a way to correct this so the two-decimal place value is displayed consistently?
Simply, perform math with calibration value and round after
az=round(float(imu.accel.z) + 0.04,2)
print(str(az))
Related
Why is why is np.exp(x) not equal to np.exp(1)**x?
For example:
np.exp(400)
>>>5.221469689764144e+173
np.exp(1)**400
>>>5.221469689764033e+173
np.exp(400)-np.exp(1)**400
>>>1.1093513018771065e+160
This is optimisation of numpy that raise this diff.
Indeed, you have to understand how is calculated the Euler number in math:
e = (1/n)**n with n == inf.
I think numpy stop at a certain order:
You have in the numpy exp documentation here that is not very clear about how the Euler number is calculated.
Because of this order that is not equal to infinity, you have this small difference in the two calculations.
Indeed the value np.exp(400) is calculated using this: (1 + 400/n)**n
>>> (1 + 400/n)**n
5.221642085428121e+173
>>> numpy.exp(400)
5.221469689764144e+173
Here you have n = 1000000000000 wich is very small and raise this difference at 10e-5.
Indeed there is no exact value of the Euler number. Like Pi, you can only have an approched value.
It looks like a rounding issue. In the first case it's internally using a very precise value of e, while in the second you get a less precise value, which when multiplied 400 times the precision issues become more apparent.
The actual result when using the Windows calculator is 5.2214696897641439505887630066496e+173, so you can see your first outcome is fine, while the second is not.
5.2214696897641439505887630066496e+173 // calculator
5.221469689764144e+173 // exp(400)
5.221469689764033e+173 // exp(1)**400
Starting from your result, it looks it's using a value with 15 digits of precision.
2.7182818284590452353602874713527 // e
2.7182818284590450909589085441968 // 400th root of the 2nd result
This is my problem:
The first input is the observed data of MUSE, which is an astronomical instrument provides cubes, i.e. an image for each wavelength with a certain range. This means that, taken all the wavelengths corresponding to the pixel i,j, I can extract the spectrum for this pixel. Since these images are observed, for each pixel I have an error.
The second input is a spectrum template, i.e. a model of a spectrum. This template is assumed to be without error. I map this spectra at various redshift (this means multiply the wavelenghts for a factor 1+z, where z belong to a certain range).
The core of my code is the cross-correlation between the cube, i.e. the spectra extracted from each pixel, and the template mapped at different redshift. The result is a cross-correlation function for each pixel for each z, let's call this computed function as f(z). Taking, for each pixel, the argmax of f(z), I get the best redshift.
This is a common and widely-used process, indeed, it actually works well.
My question:
Since my input, i.e. the MUSE cube, has an error, I have propagated this error through the cross-correlation, obtaining an error on f(z), i.e. each f_i has a error sigma_i. So, how can I compute the error on z_max, which is the value of z corresponding to the maximum of f?
Maybe a solution could be the implementation of bootstrap method: I can extract, within the error of f, a certain number of function, for each of them I computed the argamx, so i can have an idea about the scatter of z_max.
By the way, I'm using python (3.x) and tensorflow has been used to compute the cross-correlation function.
Thanks!
EDIT
Following #TF_Support suggestion I'm trying to add some code and some figures to better understand the problem. But, before this, maybe it's better a little of math.
With this expression I had computed the cross-correlation:
where S is the spectra, T is the template and N is the normalization coefficient. Since S has an error, I had propagated these errors through the previous relation founding:
where SST_k is the the sum of the template squared and sigma_ij is the error on on S_ij (actually, I should have written sigma_S_ij).
The follow function (implemented with tensorflow 2.1) makes the cross-correlation between one template and the spectra of batch pixels, and computes the error on the cross-correlation function:
#tf.function
def make_xcorr_err1(T, S, sigma_S):
sum_spectra_sq = tf.reduce_sum(tf.square(S), 1) #shape (batch,)
sum_template_sq = tf.reduce_sum(tf.square(T), 0) #shape (Nz, )
norm = tf.sqrt(tf.reshape(sum_spectra_sq, (-1,1))*tf.reshape(sum_template_sq, (1,-1))) #shape (batch, Nz)
xcorr = tf.matmul(S, T, transpose_a = False, transpose_b= False)/norm
foo1 = tf.matmul(sigma_S**2, T**2, transpose_a = False, transpose_b= False)/norm**2
foo2 = xcorr**2 * tf.reshape(sum_template_sq**2, (1,-1)) * tf.reshape(tf.reduce_sum((S*sigma_S)**2, 1), (-1,1))/norm**4
foo3 = - 2 * xcorr * tf.reshape(sum_template_sq, (1,-1)) * tf.matmul(S*(sigma_S)**2, T, transpose_a = False, transpose_b= False)/norm**3
sigma_xcorr = tf.sqrt(tf.maximum(foo1+foo2+foo3, 0.))
Maybe, in order to understand my problem, more important than code is an image representing an output. This is the cross-correlation function for a single pixel, in red the maximum value, let's call z_best, i.e. the best cross-correlated value. The figure also shows the 3 sigma errors (the grey limits are +3sigma -3sigma).
If i zoom-in near the peak, I get this:
As you can see the maximum (as any other value) oscillates within a certain range. I would like to find a way to map this fluctuations of maximum (or the fluctuations around the maximum, or the fluctuations of the whole function) to an error on the value corresponding the maximum, i.e. an error on z_best.
I'm trying to construct a neural network for the Mnist database. When computing the softmax function I receive an error to the same ends as "you can't store a float that size"
code is as follows:
def softmax(vector): # REQUIRES a unidimensional numpy array
adjustedVals = [0] * len(vector)
totalExp = np.exp(vector)
print("totalExp equals")
print(totalExp)
totalSum = totalExp.sum()
for i in range(len(vector)):
adjustedVals[i] = (np.exp(vector[i])) / totalSum
return adjustedVals # this throws back an error sometimes?!?!
After inspection, most recommend using the decimal module. However when I've messed around with the values being used in the command line with this module, that is:
from decimal import Decimal
import math
test = Decimal(math.exp(720))
I receive a similar error for any values which are math.exp(>709).
OverflowError: (34, 'Numerical result out of range')
My conclusion is that even decimal cannot handle this number. Does anyone know of another method I could use to represent these very large floats.
There is a technique which makes the softmax function more feasible computationally for a certain kind of value distribution in your vector. Namely, you can subtract the maximum value in the vector (let's call it x_max) from each of its elements. If you recall the softmax formula, such operation doesn't affect the outcome as it reduced to multiplication of the result by e^(x_max) / e^(x_max) = 1. This way the highest intermediate value you get is e^(x_max - x_max) = 1 so you avoid the overflow.
For additional explanation I recommend the following article: https://nolanbconaway.github.io/blog/2017/softmax-numpy
With a value above 709 the function 'math.exp' exceeds the floating point range and throws this overflow error.
If, instead of math.exp, you use numpy.exp for such large exponents you will see that it evaluates to the special value inf (infinity).
All this apart, I wonder why you would want to produce such a big number (not sure you are aware how big it is. Just to give you an idea, the number of atoms in the universe is estimated to be in the range of 10 to the power of 80. The number you are trying to produce is MUCH larger than that).
I have a situation that: I have one vector A, say 10000x1, and another vector B 10000x1, both are numerical arrays with floating point numbers in it. Now I want to write the data into one line of string as below:
A(1):B(1) A(2):B(2) ....A(10000):B(10000)
Is there an efficient way to do this? Right now, i am just using a for loop, change the floating number to string first, than add the ':', and then concatenate them together. This is very slow. Could anybody help? Thanks a lot.
For dimension nx1 (Column Matrix)
tic
A=rand(10000,1);
B=rand(10000,1);
finalString=sprintf(' %f:%f',[A.'; B.']);
finalString(1)=[];
toc
Elapsed time is 0.036697 seconds.
For dimension 1xn (Row Matrix)
tic
A=rand(1,10000);
B=rand(1,10000);
finalString=sprintf(' %f:%f',[A; B]);
finalString(1)=[];
toc
Elapsed time is 0.036879 seconds.
Value Type
%f --> Floating-point number(Fixed-point notation)
%d --> Integer, signed(Base 10)
For more value types http://in.mathworks.com/help/matlab/ref/sprintf.html has a table for conversion characters to format numeric and character data as text or you can search sprintf in matlab help.
This should do it relatively quickly. I included a tic-toc to provide a reference execution time if someone provides an alternative implementation.
tic
a=rand(10000,1);
b=rand(10000,1);
c=zeros(20000,1);
c(1:2:end)=a;
c(2:2:end)=b;
c_string=mat2str(c);
idx=find(c_string==';');
c_string(idx(1:2:end))=':';
c_string(idx(2:2:end))=' ';
toc
%Elapsed time is 0.365694 seconds.
I'm trying to get the XY coordinates of a moving sprite in SmileBASIC, and I can't figure it out. I have the single variable returned from SPCHK, but when I print it, I get a single number '4' constantly as the sprite moves. How do I get each bit?
From the documentation:
Return Values for SPCHK
|b00| XY-coordinates (1), #CHKXY
|b01| Z-coordinates (2), #CHKZ
|b02| UV-coordinates (4), #CHKUV
|b03| Definition number (8), #CHKI
|b04| Rotation (16), #CHKR
|b05| Magnification XY (32), #CHKS
|b06| Display color (64), #CHKC
|b07| Variable (128), #CHKV
For each bit, a target is assigned (If 0 is assigned for all bits, animation is being stopped)
SPCHK only tells you which properties are currently being animated, not their values.
To get the actual position, you can use SPOFS id OUT x,y
Example:
SPSET 0,17
SPANIM 0,"XY",-10,100,100
WAIT 5
SPOFS 0 OUT X,Y
?X,Y 'should be 50,50