Moving average of an array in Python - python

I have an array where discreet sinewave values are recorded and stored. I want to find the max and min of the waveform. Since the sinewave data is recorded voltages using a DAQ, there will be some noise, so I want to do a weighted average. Assuming self.yArray contains my sinewave values, here is my code so far:
filterarray = []
filtersize = 2
length = len(self.yArray)
for x in range (0, length-(filtersize+1)):
for y in range (0,filtersize):
summation = sum(self.yArray[x+y])
ave = summation/filtersize
filterarray.append(ave)
My issue seems to be in the second for loop, where depending on my averaging window size (filtersize), I want to sum up the values in the window to take the average of them. I receive an error saying:
summation = sum(self.yArray[x+y])
TypeError: 'float' object is not iterable
I am an EE with very little experience in programming, so any help would be greatly appreciated!

The other answers correctly describe your error, but this type of problem really calls out for using numpy. Numpy will run faster, be more memory efficient, and is more expressive and convenient for this type of problem. Here's an example:
import numpy as np
import matplotlib.pyplot as plt
# make a sine wave with noise
times = np.arange(0, 10*np.pi, .01)
noise = .1*np.random.ranf(len(times))
wfm = np.sin(times) + noise
# smoothing it with a running average in one line using a convolution
# using a convolution, you could also easily smooth with other filters
# like a Gaussian, etc.
n_ave = 20
smoothed = np.convolve(wfm, np.ones(n_ave)/n_ave, mode='same')
plt.plot(times, wfm, times, -.5+smoothed)
plt.show()
If you don't want to use numpy, it should also be noted that there's a logical error in your program that results in the TypeError. The problem is that in the line
summation = sum(self.yArray[x+y])
you're using sum within the loop where your also calculating the sum. So either you need to use sum without the loop, or loop through the array and add up all the elements, but not both (and it's doing both, ie, applying sum to the indexed array element, that leads to the error in the first place). That is, here are two solutions:
filterarray = []
filtersize = 2
length = len(self.yArray)
for x in range (0, length-(filtersize+1)):
summation = sum(self.yArray[x:x+filtersize]) # sum over section of array
ave = summation/filtersize
filterarray.append(ave)
or
filterarray = []
filtersize = 2
length = len(self.yArray)
for x in range (0, length-(filtersize+1)):
summation = 0.
for y in range (0,filtersize):
summation = self.yArray[x+y]
ave = summation/filtersize
filterarray.append(ave)

self.yArray[x+y] is returning a single item out of the self.yArray list. If you are trying to get a subset of the yArray, you can use the slice operator instead:
summation = sum(self.yArray[x:y])
to return an iterable that the sum builtin can use.
A bit more information about python slices can be found here (scroll down to the "Sequences" section): http://docs.python.org/2/reference/datamodel.html#the-standard-type-hierarchy

You could use numpy, like:
import numpy
filtersize = 2
ysums = numpy.cumsum(numpy.array(self.yArray, dtype=float))
ylags = numpy.roll(ysums, filtersize)
ylags[0:filtersize] = 0.0
moving_avg = (ysums - ylags) / filtersize

Your original code attempts to call sum on the float value stored at yArray[x+y], where x+y is evaluating to some integer representing the index of that float value.
Try:
summation = sum(self.yArray[x:y])

Indeed numpy is the way to go. One of the nice features of python is list comprehensions, allowing you to do away with the typical nested for loop constructs. Here goes an example, for your particular problem...
import numpy as np
step=2
res=[np.sum(myarr[i:i+step],dtype=np.float)/step for i in range(len(myarr)-step+1)]

Related

Optimizing a simple Photon Detection Simulation

I am a medical physics student trying to simulate photon detection - I succeeded (below) but I want to make it better by speeding it up: it currently takes 50 seconds to run and I want it to run in some fraction of that time. I assume someone more knowledgeable in Python could optimize it to complete within less than 10 seconds (without reducing num_photons_detected values). Thank you very much for trying out this little optimization challenge.
from random import seed
from random import random
import random
import matplotlib.pyplot as plt
import numpy as np
rows, cols = (25, 25)
num_photons_detected = [10**3, 10**4, 10**5, 10**6, 10**7]
lesionPercentAboveNoiseLevel = [1, 0.20, 0.10, 0.05]
index_range = np.array([i for i in range(rows)])
for l in range(len(lesionPercentAboveNoiseLevel)):
pixels = np.array([[0.0 for i in range(cols)] for j in range(rows)])
for k in range(len(num_photons_detected)):
random.seed(a=None, version=2)
photons_random_pixel_choice = np.array([random.choice(index_range) for z in range(rows)])
counts = 0
while num_photons_detected[k] > counts:
for i in photons_random_pixel_choice:
photons_random_pixel_choice = np.array([random.choice(index_range) for z in range(rows)]) #further ensures random pixel selection
for j in photons_random_pixel_choice:
pixels[i,j] +=1
counts +=1
plt.imshow(pixels, cmap="gray") #in the resulting images/graphs, x is on the vertical and y on the horizontal
plt.show()
I think that, aside from efficiency issues, a problem with the code is that it does not select the positions of photons truly at random. Instead, it selects rows numbers, and then for each selected row, it picks column numbers where photons will be observed in that row. As a result, if a row number is not selected, there will be no photons in that row at all, and if the same row is selected several times, there will be many photons in it. This is visible in the produced plots which have a clear pattern of lighter and darker rows:
Assuming that this is unintended and that each pixel should have equal chances of being selected, here is a function generating an array of a given size, with a given number of randomly selected pixels:
import numpy as np
def generate_photons(rows, cols, num_photons):
rng = np.random.default_rng()
indices = rng.choice(rows*cols, num_photons)
np.add.at(pix:=np.zeros(rows*cols), indices, 1)
return pix.reshape(rows, cols)
You can use it to produce images with specified parameters. E.g.:
import matplotlib.pyplot as plt
pixels = generate_photons(rows=25, cols=25, num_photons=10**4)
plt.imshow(pixels, cmap="gray")
plt.show()
gives:
photons_random_pixel_choice = np.array([random.choice(index_range) for z in range(rows)])
It seems like the goal here is:
Use a pre-made sequence of integers, 0 to 24 inclusive, to select one of those values.
Repeat that process 25 times in a list comprehension, to get a Python list of 25 random values in that range.
Make a 1-d Numpy array from those results.
This is very much missing the point of using Numpy. If we want integers in a range, then we can directly ask for those. But more importantly, we should let Numpy do the looping as much as possible when using Numpy data structures. This is where it pays to read the documentation:
size: int or tuple of ints, optional
Output shape. If the given shape is, e.g., (m, n, k), then m * n * k samples are drawn. Default is None, in which case a single value is returned.
So, just make it directly: photons_random_pixel_choice = random.integers(rows, size=(rows,)).

Resampling 2-d array using Fourier transform method

I have a question on the resampling 2-d array.
Sometimes, the original size of the geoscience data should be transformed to other size. If the ratio for each axis is equal, the task is simple, in which np.reshape allow a 2-d array of 100x100 to 50x50 without data loss. The code is shown as:
## creat a original data
xc1, xc2, yc1, yc2 = 100, 110, 35, 45
XSIZE,YSIZE=100,100
lon,lat = np.linspace(xc1,xc2,XSIZE),np.linspace(yc1,yc2,YSIZE)
pop = np.random.uniform(low=1000, high=50000, size=(XSIZE*YSIZE,)).reshape(YSIZE,XSIZE)
## reshape
shape = np.array(pop.shape, dtype=float)
coarseness = 2 # the new shape is in 50 x 50
new_shape = coarseness * np.ceil(shape/coarseness).astype(int)
zp_pop = np.zeros(new_shape)
zp_pop[:int(shape[0]), :int(shape[1])] = pop
temp = zp_pop.reshape((new_shape[0] // coarseness, coarseness,
new_shape[1] // coarseness, coarseness))
coarse_pop = np.sum(temp, axis=(1,3))
print (pop.sum())
print (coarse_pop.sum())
However, when the coarse factor is different for each axis, this method can not be implemented. I turned to apply other method. Here is an example I tried to use FFT to generate a 60*80 array as output
from scipy import fftpack
pop_fft = fftpack.fft2(pop,shape = (60,80))
pop_res = fftpack.ifft2(pop_fft).real
print(pop.sum())
print(pop_res.sum())
254208134.8356425
122048754.13639387
The data loss was significant. Thus, I posted my issue here. Maybe the resampling function I used was not correct. Or there are some better approach to deal with this situation. Any advices or comments are highly appreciated!
When you set up the 'coarse array' yourself you sum over adjacent entries, instead of computing the average or interpolating.
This way the sum over all elements in the coarse and original array are identical str((coarse_pop.sum()-pop.sum())/(0.5*(pop.sum()+coarse_pop.sum()))) gives '-1.1638426077573779e-16' only a tiny numerical error.
if you compare the mean of the fftpack resampled coarse array it matches up:
print(pop.mean())
print(pop_res.mean())
25606.832220313503
25496.03271480075
alternatively you can correct for the number of elements yourself:
print(pop.sum())
print(pop_res.sum()*100*100/(60*80))
256068322.20313504
254960327.14800745
I don't know about your problem but the fftpack way of downsampling the array makes more sense to me. if it's not what you want you can apply the prefactor to the original array, like pop_fft = fftpack.fft2(pop*100*100/(60*80),shape = (60,80))

How to efficiently index a numpy array based on varying start and stop indexes per row

I have a 2D numpy array with rows being time series of a feature, based on which I'm training a neural network. For generalisation purposes, I would like to subset these time series at random points. I'd like them to have a minimum subset length as well. However, the network requires fixed length time series, so I need to pre-pad the resulting subsets with zeroes.
Currently, I'm doing it using the code below, which includes a nasty for-loop, because I don't know how I can use fancy indexing for this particular problem. As this piece of code is part of the network data generator, it needs to be fast to keep up to pace with the data-hungry GPU. Does anyone know a numpy-way of doing this without the for-loop?
import numpy as np
import matplotlib.pyplot as plt
# Amount of time series to consider
batchsize = 25
# Original length of the time series
timesteps = 150
# As an example, fill the 2D array with sine function time series
sinefunction = np.expand_dims(np.sin(np.arange(timesteps)), axis=0)
originalarray = np.repeat(sinefunction, batchsize, axis=0)
# Now the real thing, we want:
# - to start the time series at a random moment (between 0 and maxstart)
# - to end the time series at a random moment
# - however with a minimum length of the resulting subset time series (minlength)
maxstart = 50
minlength = 75
# get random starts
randomstarts = np.random.choice(np.arange(0, maxstart), size=batchsize)
# get random stops
randomstops = np.random.choice(np.arange(maxstart + minlength, timesteps), size=batchsize)
# determine the resulting random sizes of the subset time series
randomsizes = randomstops - randomstarts
# finally create a new 2D array with all the randomly subset time series, however pre-padded with zeros
# THIS IS THE FOR LOOP WE SHOULD TRY TO AVOID
cutarray = np.zeros_like(originalarray)
for i in range(batchsize):
cutarray[i, -randomsizes[i]:] = originalarray[i, randomstarts[i]:randomstops[i]]
To show what goes in and out of the function:
# Show that it worked
f, ax = plt.subplots(2, 1)
ax[0].imshow(originalarray)
ax[0].set_title('original array')
ax[1].imshow(cutarray)
ax[1].set_title('zero-padded subset array')
Approach #1 : Views-based
We can leverage np.lib.stride_tricks.as_strided based scikit-image's view_as_windows to get sliding windowed views into a zeros padded version of the input and assign into a zeros padded version of the output. All of that padding is needed for a vectorized solution on account of the ragged nature. Upside is that working on views would be efficient on memory and performance.
The implementation would look something like this -
from skimage.util.shape import view_as_windows
n = randomsizes.max()
max_extent = randomstarts.max()+n
padlen = max_extent - origalarray.shape[1]
p = np.zeros((origalarray.shape[0],padlen),dtype=origalarray.dtype)
a = np.hstack((origalarray,p))
w = view_as_windows(a,(1,n))[...,0,:]
out_vals = w[np.arange(len(randomstarts)),randomstarts]
out_starts = origalarray.shape[1]-randomsizes
out_extensions_max = out_starts.max()+n
out = np.zeros((origalarray.shape[0],out_extensions_max),dtype=origalarray.dtype)
w2 = view_as_windows(out,(1,n))[...,0,:]
w2[np.arange(len(out_starts)),out_starts] = out_vals
cutarray_out = out[:,:origalarray.shape[1]]
Approach #2 : With masking
cutarray_out = np.zeros_like(origalarray)
r = np.arange(origalarray.shape[1])
m = (randomstarts[:,None]<=r) & (randomstops[:,None]>r)
s = origalarray.shape[1]-randomsizes
m2 = s[:,None]<=r
cutarray_out[m2] = origalarray[m]

Data mean by intervals in IDL

I did the following script to integrate (average) data by intervals in python:
# N = points to mean in the array
# data = original data
# data_mean = average data each N points
data_mean = np.array([np.mean(i) for i in np.array_split(data, len(data)/N)])
How could do that in IDL?
There are a "mean" function, but a "array_split-like"?
The array_split functionality is usually done via REFORM to create a two (or higher) dimensional array from a 1-dimensional array using the same values. So for example:
n = 20
data = randomu(seed, 100)
data = reform(data, 100 / n, n)
print, mean(data, dimension=2)
The IDL mean function is equivalent to the numpy mean function, and the IDL reform can be used similarly to the numpy array_split:
data_mean = mean(reform(data, n_elements(data) / N), dimension=2)
If you don't mind data ending up with different dimensions, you can greatly speed this up using the /overwrite keyword:
data_mean = mean(reform(data, n_elements(data) / N, /overwrite), dimension=2)
Finally, if you have a version of IDL before IDL 8.0, then you won't have the dimension keyword for the mean function. Use this (less elegant) pattern instead:
data_mean = total(reform(data, n_elements(data) / N), 2) / N
Note that this version with total also accepts the /nan keyword, so that it works even when some data are missing.

Efficiently get indices of histogram bins in Python

Short Question
I have a large 10000x10000 elements image, which I bin into a few hundred different sectors/bins. I then need to perform some iterative calculation on the values contained within each bin.
How do I extract the indices of each bin to efficiently perform my calculation using the bins values?
What I am looking for is a solution which avoids the bottleneck of having to select every time ind == j from my large array. Is there a way to obtain directly, in one go, the indices of the elements belonging to every bin?
Detailed Explanation
1. Straightforward Solution
One way to achieve what I need is to use code like the following (see e.g. THIS related answer), where I digitize my values and then have a j-loop selecting digitized indices equal to j like below
import numpy as np
# This function func() is just a placemark for a much more complicated function.
# I am aware that my problem could be easily sped up in the specific case of
# of the sum() function, but I am looking for a general solution to the problem.
def func(x):
y = np.sum(x)
return y
vals = np.random.random(1e8)
nbins = 100
bins = np.linspace(0, 1, nbins+1)
ind = np.digitize(vals, bins)
result = [func(vals[ind == j]) for j in range(1, nbins)]
This is not what I want as it selects every time ind == j from my large array. This makes this solution very inefficient and slow.
2. Using binned_statistics
The above approach turns out to be the same implemented in scipy.stats.binned_statistic, for the general case of a user-defined function. Using Scipy directly an identical output can be obtained with the following
import numpy as np
from scipy.stats import binned_statistics
vals = np.random.random(1e8)
results = binned_statistic(vals, vals, statistic=func, bins=100, range=[0, 1])[0]
3. Using labeled_comprehension
Another Scipy alternative is to use scipy.ndimage.measurements.labeled_comprehension. Using that function, the above example would become
import numpy as np
from scipy.ndimage import labeled_comprehension
vals = np.random.random(1e8)
nbins = 100
bins = np.linspace(0, 1, nbins+1)
ind = np.digitize(vals, bins)
result = labeled_comprehension(vals, ind, np.arange(1, nbins), func, float, 0)
Unfortunately also this form is inefficient and in particular, it has no speed advantage over my original example.
4. Comparison with IDL language
To further clarify, what I am looking for is a functionality equivalent to the REVERSE_INDICES keyword in the HISTOGRAM function of the IDL language HERE. Can this very useful functionality be efficiently replicated in Python?
Specifically, using the IDL language the above example could be written as
vals = randomu(s, 1e8)
nbins = 100
bins = [0:1:1./nbins]
h = histogram(vals, MIN=bins[0], MAX=bins[-2], NBINS=nbins, REVERSE_INDICES=r)
result = dblarr(nbins)
for j=0, nbins-1 do begin
jbins = r[r[j]:r[j+1]-1] ; Selects indices of bin j
result[j] = func(vals[jbins])
endfor
The above IDL implementation is about 10 times faster than the Numpy one, due to the fact that the indices of the bins do not have to be selected for every bin. And the speed difference in favour of the IDL implementation increases with the number of bins.
I found that a particular sparse matrix constructor can achieve the desired result very efficiently. It's a bit obscure but we can abuse it for this purpose. The function below can be used in nearly the same way as scipy.stats.binned_statistic but can be orders of magnitude faster
import numpy as np
from scipy.sparse import csr_matrix
def binned_statistic(x, values, func, nbins, range):
'''The usage is nearly the same as scipy.stats.binned_statistic'''
N = len(values)
r0, r1 = range
digitized = (float(nbins)/(r1 - r0)*(x - r0)).astype(int)
S = csr_matrix((values, [digitized, np.arange(N)]), shape=(nbins, N))
return [func(group) for group in np.split(S.data, S.indptr[1:-1])]
I avoided np.digitize because it doesn't use the fact that all bins are equal width and hence is slow, but the method I used instead may not handle all edge cases perfectly.
I assume that the binning, done in the example with digitize, cannot be changed. This is one way to go, where you do the sorting once and for all.
vals = np.random.random(1e4)
nbins = 100
bins = np.linspace(0, 1, nbins+1)
ind = np.digitize(vals, bins)
new_order = argsort(ind)
ind = ind[new_order]
ordered_vals = vals[new_order]
# slower way of calculating first_hit (first version of this post)
# _,first_hit = unique(ind,return_index=True)
# faster way:
first_hit = searchsorted(ind,arange(1,nbins-1))
first_hit.sort()
#example of using the data:
for j in range(nbins-1):
#I am using a plotting function for your f, to show that they cluster
plot(ordered_vals[first_hit[j]:first_hit[j+1]],'o')
The figure shows that the bins are actually clusters as expected:
You can halve the computation time by sorting the array first, then use np.searchsorted.
vals = np.random.random(1e8)
vals.sort()
nbins = 100
bins = np.linspace(0, 1, nbins+1)
ind = np.digitize(vals, bins)
results = [func(vals[np.searchsorted(ind,j,side='left'):
np.searchsorted(ind,j,side='right')])
for j in range(1,nbins)]
Using 1e8 as my test case, I go from 34 seconds of computation to about 17.
One efficient solution is using the numpy_indexed package (disclaimer: I am its author):
import numpy_indexed as npi
npi.group_by(ind).split(vals)
Pandas has a very fast grouping code (I think it's written in C), so if you don't mind loading the library you could do that :
import pandas as pd
pdata=pd.DataFrame({'vals':vals,'ind':ind})
resultsp = pdata.groupby('ind').sum().values
or more generally :
pdata=pd.DataFrame({'vals':vals,'ind':ind})
resultsp = pdata.groupby('ind').agg(func).values
Although the latter is slower for standard aggregation functions
(like sum, mean, etc)

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