How to efficiently calculate 160146 by 160146 matrix inverse in python? - python-3.x

My research is into structural dynamics and i am dealing with large symmetric sparse matrix calculation. Recently, i have to calculate the stiffness matrix (160146 by 160146) inverse with 4813762 non zero elements. I did calculate a smaller stiffness matrix inverse for a 15000 by 15000 size and it came out to almost or full dense. Initially i tried with almost all scipy.sparse.linalg functions to calculate inverse through Ax=b form. Currently, i am using superlu to calculate the L and U matrices then using that i calculate the inverse using Solve(). Since the matrix inverse is dense and i could not store in RAM memory i opted for pytables.
Unfortunately, writing time of one column of inverse matrix takes about 16 minutes(time for each step is shown below after the code) and a total of 160146 columns exist for the stiffness matrix. I would like to know how i can boost the writing speed so that this inverse task will finish in couple of days. The code is as follows,
LU= scipy.sparse.linalg.splu(interior_stiff)
interior_dof_row_ptr=160146
#---PyTables creation Code for interior_stiff_inverse begins--#
if(os.path.isfile("HDF5_Interior.h5")==False):
f=tables.open_file("HDF5_Interior.h5", 'w')
# compression-level and compression library
filters=tables.Filters(complevel=0, complib='blosc')
# f.root-> your default group in the HDF5 file "firstdata"->name of the dataset
# tables.Float32Atom()->WHat si your atomic data object?
if(f.__contains__("/DS_interior_stiff_inverse")==False):
print("DS_Interior_stiff_inverse DOESN'T EXIST!!!!!")
out=f.create_carray(f.root, "DS_interior_stiff_inverse", tables.Float32Atom(), shape=(interior_dof_row_ptr,interior_dof_row_ptr), filters=filters)
#out=f.create_earray(f.root, "DS_interior_stiff_inverse", tables.Float32Atom(), shape=(interior_dof_row_ptr,0), filters=filters, expectedrows=interior_dof_row_ptr)
else:
print("DS_interior_stiff_inverse EXISTS!!!!!")
out=f.get_node("/", "DS_interior_stiff_inverse")
#interior_stiff_inverse=numpy.zeros((interior_dof_row_ptr,interior_dof_row_ptr))
for i in range(0,interior_dof_row_ptr):
I=numpy.zeros((interior_dof_row_ptr,1))
I[i,0]=1
#-- COmmented by Libni - interior_stiff_inverse[:,i]=LU.solve(I[:,0]) #In pytables how we define the variables. So interior_stiff_inverse_1 only needs to be stored in pytables.
print("stating solve() calculation for inverse: ", datetime.datetime.now())
tmpResult=LU.solve(I[:,0])
print("solve() calculation for inverse DONE: ", datetime.datetime.now())
out[:,i]=tmpResult
print("Written to hdf5 (pytable) :", datetime.datetime.now())
#out.append(LU.solve(I[:,0]))
print(str(i) + "th iteration of " + str(interior_dof_row_ptr) + " Interior Inv done")
f.flush()
print("After FLUSH line: ", datetime.datetime.now())
f.close()
#--***PyTables creation Code for interior_stiff_inverse begins-***
Time taken for Solve () calculation and writing to hdf5 is as follows,
stating solve() calculation for inverse: 2017-08-26 01:04:20.424447
solve() calculation for inverse DONE: 2017-08-26 01:04:20.596045
Written to hdf5 (pytable) :2017-08-26 01:20:57.228322
After FLUSH line: 01:20:57.555922
which clearly indicate that writing one column of inverse matrix to hdf5 takes 16 minutes. As per this if i need to calculate the entire matrix inverse it will take me 1779 days. I am sure the writing time can be boosted up. I dont know how i can achieve this. Please help me in boosting the writing speed to hdf5 so that the matrix inverse run can be finished within couple of days.
I have used 0 compression in hdf5 creation thinking that this will be helping in reading and writing fast.
My computer spec include i7 with 4 cores and 16 RAM.
Any help will be appreciated.
Thank You,
Paul Thomas

Related

Comparing 2 image content using python [duplicate]

I'm trying to compare images to each other to find out whether they are different. First I tried to make a Pearson correleation of the RGB values, which works also quite good unless the pictures are a litte bit shifted. So if a have a 100% identical images but one is a little bit moved, I get a bad correlation value.
Any suggestions for a better algorithm?
BTW, I'm talking about to compare thousand of imgages...
Edit:
Here is an example of my pictures (microscopic):
im1:
im2:
im3:
im1 and im2 are the same but a little bit shifted/cutted, im3 should be recognized as completly different...
Edit:
Problem is solved with the suggestions of Peter Hansen! Works very well! Thanks to all answers! Some results can be found here
http://labtools.ipk-gatersleben.de/image%20comparison/image%20comparision.pdf
A similar question was asked a year ago and has numerous responses, including one regarding pixelizing the images, which I was going to suggest as at least a pre-qualification step (as it would exclude very non-similar images quite quickly).
There are also links there to still-earlier questions which have even more references and good answers.
Here's an implementation using some of the ideas with Scipy, using your above three images (saved as im1.jpg, im2.jpg, im3.jpg, respectively). The final output shows im1 compared with itself, as a baseline, and then each image compared with the others.
>>> import scipy as sp
>>> from scipy.misc import imread
>>> from scipy.signal.signaltools import correlate2d as c2d
>>>
>>> def get(i):
... # get JPG image as Scipy array, RGB (3 layer)
... data = imread('im%s.jpg' % i)
... # convert to grey-scale using W3C luminance calc
... data = sp.inner(data, [299, 587, 114]) / 1000.0
... # normalize per http://en.wikipedia.org/wiki/Cross-correlation
... return (data - data.mean()) / data.std()
...
>>> im1 = get(1)
>>> im2 = get(2)
>>> im3 = get(3)
>>> im1.shape
(105, 401)
>>> im2.shape
(109, 373)
>>> im3.shape
(121, 457)
>>> c11 = c2d(im1, im1, mode='same') # baseline
>>> c12 = c2d(im1, im2, mode='same')
>>> c13 = c2d(im1, im3, mode='same')
>>> c23 = c2d(im2, im3, mode='same')
>>> c11.max(), c12.max(), c13.max(), c23.max()
(42105.00000000259, 39898.103896795357, 16482.883608327804, 15873.465425120798)
So note that im1 compared with itself gives a score of 42105, im2 compared with im1 is not far off that, but im3 compared with either of the others gives well under half that value. You'd have to experiment with other images to see how well this might perform and how you might improve it.
Run time is long... several minutes on my machine. I would try some pre-filtering to avoid wasting time comparing very dissimilar images, maybe with the "compare jpg file size" trick mentioned in responses to the other question, or with pixelization. The fact that you have images of different sizes complicates things, but you didn't give enough information about the extent of butchering one might expect, so it's hard to give a specific answer that takes that into account.
I have one done this with an image histogram comparison. My basic algorithm was this:
Split image into red, green and blue
Create normalized histograms for red, green and blue channel and concatenate them into a vector (r0...rn, g0...gn, b0...bn) where n is the number of "buckets", 256 should be enough
subtract this histogram from the histogram of another image and calculate the distance
here is some code with numpy and pil
r = numpy.asarray(im.convert( "RGB", (1,0,0,0, 1,0,0,0, 1,0,0,0) ))
g = numpy.asarray(im.convert( "RGB", (0,1,0,0, 0,1,0,0, 0,1,0,0) ))
b = numpy.asarray(im.convert( "RGB", (0,0,1,0, 0,0,1,0, 0,0,1,0) ))
hr, h_bins = numpy.histogram(r, bins=256, new=True, normed=True)
hg, h_bins = numpy.histogram(g, bins=256, new=True, normed=True)
hb, h_bins = numpy.histogram(b, bins=256, new=True, normed=True)
hist = numpy.array([hr, hg, hb]).ravel()
if you have two histograms, you can get the distance like this:
diff = hist1 - hist2
distance = numpy.sqrt(numpy.dot(diff, diff))
If the two images are identical, the distance is 0, the more they diverge, the greater the distance.
It worked quite well for photos for me but failed on graphics like texts and logos.
You really need to specify the question better, but, looking at those 5 images, the organisms all seem to be oriented the same way. If this is always the case, you can try doing a normalized cross-correlation between the two images and taking the peak value as your degree of similarity. I don't know of a normalized cross-correlation function in Python, but there is a similar fftconvolve() function and you can do the circular cross-correlation yourself:
a = asarray(Image.open('c603225337.jpg').convert('L'))
b = asarray(Image.open('9b78f22f42.jpg').convert('L'))
f1 = rfftn(a)
f2 = rfftn(b)
g = f1 * f2
c = irfftn(g)
This won't work as written since the images are different sizes, and the output isn't weighted or normalized at all.
The location of the peak value of the output indicates the offset between the two images, and the magnitude of the peak indicates the similarity. There should be a way to weight/normalize it so that you can tell the difference between a good match and a poor match.
This isn't as good of an answer as I want, since I haven't figured out how to normalize it yet, but I'll update it if I figure it out, and it will give you an idea to look into.
If your problem is about shifted pixels, maybe you should compare against a frequency transform.
The FFT should be OK (numpy has an implementation for 2D matrices), but I'm always hearing that Wavelets are better for this kind of tasks ^_^
About the performance, if all the images are of the same size, if I remember well, the FFTW package created an specialised function for each FFT input size, so you can get a nice performance boost reusing the same code... I don't know if numpy is based on FFTW, but if it's not maybe you could try to investigate a little bit there.
Here you have a prototype... you can play a little bit with it to see which threshold fits with your images.
import Image
import numpy
import sys
def main():
img1 = Image.open(sys.argv[1])
img2 = Image.open(sys.argv[2])
if img1.size != img2.size or img1.getbands() != img2.getbands():
return -1
s = 0
for band_index, band in enumerate(img1.getbands()):
m1 = numpy.fft.fft2(numpy.array([p[band_index] for p in img1.getdata()]).reshape(*img1.size))
m2 = numpy.fft.fft2(numpy.array([p[band_index] for p in img2.getdata()]).reshape(*img2.size))
s += numpy.sum(numpy.abs(m1-m2))
print s
if __name__ == "__main__":
sys.exit(main())
Another way to proceed might be blurring the images, then subtracting the pixel values from the two images. If the difference is non nil, then you can shift one of the images 1 px in each direction and compare again, if the difference is lower than in the previous step, you can repeat shifting in the direction of the gradient and subtracting until the difference is lower than a certain threshold or increases again. That should work if the radius of the blurring kernel is larger than the shift of the images.
Also, you can try with some of the tools that are commonly used in the photography workflow for blending multiple expositions or doing panoramas, like the Pano Tools.
I have done some image processing course long ago, and remember that when matching I normally started with making the image grayscale, and then sharpening the edges of the image so you only see edges. You (the software) can then shift and subtract the images until the difference is minimal.
If that difference is larger than the treshold you set, the images are not equal and you can move on to the next. Images with a smaller treshold can then be analyzed next.
I do think that at best you can radically thin out possible matches, but will need to personally compare possible matches to determine they're really equal.
I can't really show code as it was a long time ago, and I used Khoros/Cantata for that course.
First off, correlation is a very CPU intensive rather inaccurate measure for similarity. Why not just go for the sum of the squares if differences between individual pixels?
A simple solution, if the maximum shift is limited: generate all possible shifted images and find the one that is the best match. Make sure you calculate your match variable (i.e. correllation) only over the subset of pixels that can be matched in all shifted images. Also, your maximum shift should be significantly smaller than the size of your images.
If you want to use some more advances image processing techniques I suggest you look at SIFT this is a very powerfull method that (theoretically anyway) can properly match items in images independent of translation, rotation and scale.
I guess you could do something like this:
estimate vertical / horizontal displacement of reference image vs the comparison image. a
simple SAD (sum of absolute difference) with motion vectors would do to.
shift the comparison image accordingly
compute the pearson correlation you were trying to do
Shift measurement is not difficult.
Take a region (say about 32x32) in comparison image.
Shift it by x pixels in horizontal and y pixels in vertical direction.
Compute the SAD (sum of absolute difference) w.r.t. original image
Do this for several values of x and y in a small range (-10, +10)
Find the place where the difference is minimum
Pick that value as the shift motion vector
Note:
If the SAD is coming very high for all values of x and y then you can anyway assume that the images are highly dissimilar and shift measurement is not necessary.
To get the imports to work correctly on my Ubuntu 16.04 (as of April 2017), I installed python 2.7 and these:
sudo apt-get install python-dev
sudo apt-get install libtiff5-dev libjpeg8-dev zlib1g-dev libfreetype6-dev liblcms2-dev libwebp-dev tcl8.6-dev tk8.6-dev python-tk
sudo apt-get install python-scipy
sudo pip install pillow
Then I changed Snowflake's imports to these:
import scipy as sp
from scipy.ndimage import imread
from scipy.signal.signaltools import correlate2d as c2d
How awesome that Snowflake's scripted worked for me 8 years later!
I propose a solution based on the Jaccard index of similarity on the image histograms. See: https://en.wikipedia.org/wiki/Jaccard_index#Weighted_Jaccard_similarity_and_distance
You can compute the difference in the distribution of the pixel colors. This is indeed pretty invariant to translations.
from PIL.Image import Image
from typing import List
def jaccard_similarity(im1: Image, im2: Image) -> float:
"""Compute the similarity between two images.
First, for each image an histogram of the pixels distribution is extracted.
Then, the similarity between the histograms is compared using the weighted Jaccard index of similarity, defined as:
Jsimilarity = sum(min(b1_i, b2_i)) / sum(max(b1_i, b2_i)
where b1_i, and b2_i are the ith histogram bin of images 1 and 2, respectively.
The two images must have same resolution and number of channels (depth).
See: https://en.wikipedia.org/wiki/Jaccard_index
Where it is also called Ruzicka similarity."""
if im1.size != im2.size:
raise Exception("Images must have the same size. Found {} and {}".format(im1.size, im2.size))
n_channels_1 = len(im1.getbands())
n_channels_2 = len(im2.getbands())
if n_channels_1 != n_channels_2:
raise Exception("Images must have the same number of channels. Found {} and {}".format(n_channels_1, n_channels_2))
assert n_channels_1 == n_channels_2
sum_mins = 0
sum_maxs = 0
hi1 = im1.histogram() # type: List[int]
hi2 = im2.histogram() # type: List[int]
# Since the two images have the same amount of channels, they must have the same amount of bins in the histogram.
assert len(hi1) == len(hi2)
for b1, b2 in zip(hi1, hi2):
min_b = min(b1, b2)
sum_mins += min_b
max_b = max(b1, b2)
sum_maxs += max_b
jaccard_index = sum_mins / sum_maxs
return jaccard_index
With respect to mean squared error, the Jaccard index lies always in the range [0,1], thus allowing for comparisons among different image sizes.
Then, you can compare the two images, but after rescaling to the same size! Or pixel counts will have to be somehow normalized. I used this:
import sys
from skincare.common.utils import jaccard_similarity
import PIL.Image
from PIL.Image import Image
file1 = sys.argv[1]
file2 = sys.argv[2]
im1 = PIL.Image.open(file1) # type: Image
im2 = PIL.Image.open(file2) # type: Image
print("Image 1: mode={}, size={}".format(im1.mode, im1.size))
print("Image 2: mode={}, size={}".format(im2.mode, im2.size))
if im1.size != im2.size:
print("Resizing image 2 to {}".format(im1.size))
im2 = im2.resize(im1.size, resample=PIL.Image.BILINEAR)
j = jaccard_similarity(im1, im2)
print("Jaccard similarity index = {}".format(j))
Testing on your images:
$ python CompareTwoImages.py im1.jpg im2.jpg
Image 1: mode=RGB, size=(401, 105)
Image 2: mode=RGB, size=(373, 109)
Resizing image 2 to (401, 105)
Jaccard similarity index = 0.7238955686269157
$ python CompareTwoImages.py im1.jpg im3.jpg
Image 1: mode=RGB, size=(401, 105)
Image 2: mode=RGB, size=(457, 121)
Resizing image 2 to (401, 105)
Jaccard similarity index = 0.22785529941822316
$ python CompareTwoImages.py im2.jpg im3.jpg
Image 1: mode=RGB, size=(373, 109)
Image 2: mode=RGB, size=(457, 121)
Resizing image 2 to (373, 109)
Jaccard similarity index = 0.29066426814105445
You might also consider experimenting with different resampling filters (like NEAREST or LANCZOS), as they, of course, alter the color distribution when resizing.
Additionally, consider that swapping images change the results, as the second image might be downsampled instead of upsampled (After all, cropping might better suit your case rather than rescaling.)

Custom Lambert Conformal projection of 2D data gives very large file size when saved as pdf (16MB)

I'm trying to save a figure to use in my thesis, but the figure becomes 16MB (The figures made from PlateCarree projections are order of 100-200kB)
I have provided the code, any tips, tricks and otherwise would be greatly appreciated.
SUM is a netcdf-file with coords len(y) = 949 and len(x) = 889
projection = ccrs.LambertConformal(central_latitude=63,central_longitude=15,standard_parallels=[63,63],cutoff=-30)
fig = plt.figure(figsize = (12,10))
ax = plt.subplot(projection=projection)
SUM.plot.pcolormesh(ax=ax,cbar_kwargs = {"label":"Temperature - [C]"})
ax.set_title("Mean Temperature - 750m height")
ax.coastlines("50m")
fig.savefig(outsource+"MeanT750.pdf")
I have tried giving the DPI-argument to savefig, but it does not change the filesize.
A PDF file in your case, contains vector data. You can reduce the level of details to decrease the file size. For example:
ax.coastlines("110m")
will save thousands of bytes over 50m resolution, but still represents good visualization on the map.
For the pcolormesh(), the memory varies with the value (rows*columns) of your meshed data. If you reduce number of rows/columns, you reduce the memory.
However, if contourf() can be used instead of pcolormesh(), you may save much more.

Reducing time and memory used in loop calculation when locating earthquakes

I'm trying to locate tremor, which is a type of earthquake with smaller amplitude. I use grid search, which is a method that finds the coordinate where 'the difference between theoretical value and observed value of differential time in seismic wave arrival' becomes minimum.
The code I made is as follows. First I defined two functions that calculate distance between earthquake source and each point on grid, and that calculate travel time of seismic waves using obspy.
def distance(a,i):
return math.sqrt(((ste[a].stats.sac.stla-la[i])**2)+((ste[a].stats.sac.stlo-lo[i])**2))
def traveltime(a):
return model.get_travel_times(source_depth_in_km=35, distance_in_degree=a, phase_list=["S"], receiver_depth_in_km=0)[0].time
Then I conducted grid search using following codes.
di=[(la[i],lo[i],distance(a,i), distance(b,i)) for i in range(len(lo))
for a in range(len(ste))
for b in range(len(ste)) if a<b]
didf=pd.DataFrame(di)
latot=didf[0]
lotot=didf[1]
dia=didf[2]
dib=didf[3]
tt=[]
for i in range(len(di)):
try:
tt.append((latot[i],lotot[i],traveltime(dia[i])-traveltime(dib[i])))
except IndexError:
continue
ttdf=pd.DataFrame(tt)
final=[(win[j],ttdf[0][i],ttdf[1][i],(ttdf[2][i]-shift[j])**2) for i in range(len(ttdf))
for j in range(len(ccdf))]
where la and lo are the list of latitude and longitude coordinates with 0.01 degree interval, and ste is the list of the east components seismogram of each station. I have to get the list 'final' to proceed to the next step.
However, the problem is that it takes too much time to calculate three segments of codes written above. Moreover, the result I get after tens of hours of calculation is 'out of memory' error message. Is there any solution that can reduce both time and memory?
Without access to your dataset, it's a little difficult to debug, but here are a few suggestions for you.
for i in range(len(di)):
try:
tt.append((latot[i],lotot[i],traveltime(dia[i])-traveltime(dib[i])))
except IndexError:
continue
• Given the size of these lists, I think that the Garbage Collector might be slowing down this for loop; you might consider turning it off for the duration of the loop (gc.disable()).
• In theory, the Append statement shouldn't be the source of your performance problems, since it over-allocates:
/* This over-allocates proportional to the list size, making room
* for additional growth. The over-allocation is mild, but is
* enough to give linear-time amortized behavior over a long
* sequence of appends() in the presence of a poorly-performing
* system realloc().
* The growth pattern is: 0, 4, 8, 16, 25, 35, 46, 58, 72, 88, ...
*/
new_allocated = (newsize >> 3) + (newsize < 9 ? 3 : 6);
but you already know the size of the array, so you might consider using numpy.zeroes() to fill the list before the for-loop, and use the index to directly address each element. Alternatively, you could just use list comprehensions, as you did earlier, and avoid the problem altogether.
• I see that you've tagged the question with python-3.x, so range() shouldn't be an issue like it was in 2.x (otherwise you would want to consider using xrange()).
If you update your question with more details, I could probably provide a more detailed answer...hope this helps.

Why is my notebook crashing when I run this for loop and what is the fix?

I have taken code in relation to the Kalman Filter and am attempting to iterate through each column of data. What I would like to have happen is:
The column data is fed into the filter
The filtered column data (xhat) is placed into another DataFrame (filtered)
The filtered column data (xhat) is used to produce a visual.
I have created a for loop to iterate through the column data, but when I run the cell, I crash the notebook. When it doesn't crash, I get this warning:
C:\Users\perso\Anaconda3\envs\learn-env\lib\site-packages\ipykernel_launcher.py:45: RuntimeWarning: More than 20 figures have been opened. Figures created through the pyplot interface (`matplotlib.pyplot.figure`) are retained until explicitly closed and may consume too much memory. (To control this warning, see the rcParam `figure.max_open_warning`).
Thanks in advance for any help. I hope this question is detailed enough. I bombed on the last one.
'''A Python implementation of the example given in pages 11-15 of "An
Introduction to the Kalman Filter" by Greg Welch and Gary Bishop,
University of North Carolina at Chapel Hill, Department of Computer
Science, TR 95-041,
https://www.cs.unc.edu/~welch/media/pdf/kalman_intro.pdf'''
# by Andrew D. Straw
import numpy as np
import matplotlib.pyplot as plt
# dataframe created to hold filtered data
filtered = pd.DataFrame()
# intial parameters
for column in data:
n_iter = len(data.index) #number of iterations equal to sample numbers
sz = (n_iter,) # size of array
z = data[column] # observations
Q = 1e-5 # process variance
# allocate space for arrays
xhat=np.zeros(sz) # a posteri estimate of x
P=np.zeros(sz) # a posteri error estimate
xhatminus=np.zeros(sz) # a priori estimate of x
Pminus=np.zeros(sz) # a priori error estimate
K=np.zeros(sz) # gain or blending factor
R = 1.0**2 # estimate of measurement variance, change to see effect
# intial guesses
xhat[0] = z[0]
P[0] = 1.0
for k in range(1,n_iter):
# time update
xhatminus[k] = xhat[k-1]
Pminus[k] = P[k-1]+Q
# measurement update
K[k] = Pminus[k]/( Pminus[k]+R )
xhat[k] = xhatminus[k]+K[k]*(z[k]-xhatminus[k])
P[k] = (1-K[k])*Pminus[k]
# add new data to created dataframe
filtered.assign(a = [xhat])
#create visualization of noise reduction
plt.rcParams['figure.figsize'] = (10, 8)
plt.figure()
plt.plot(z,'k+',label='noisy measurements')
plt.plot(xhat,'b-',label='a posteri estimate')
plt.legend()
plt.title('Estimate vs. iteration step', fontweight='bold')
plt.xlabel('column data')
plt.ylabel('Measurement')
This seems like a pretty straightforward error. The warning indicates that you have attempted to plot more figures than the current limit before a warning is created (a parameter you can change but which by default is set to 20). This is because in each iteration of your for loop, you create a new figure. Depending on the size of n_iter, you are opening potentially hundreds or thousands of figures. Each of these figures takes resources to generate and show, so you are creating a very large resource load on your system. Either it is processing very slowly due or is crashing altogether. In any case, the solution is to plot fewer figures.
I don't know exactly what you're plotting in your loop but it seems like each iteration of your loop corresponds to one time step and at each time step you'd like to plot the estimated and actual values. In this case, you need to define a figure and figure options once, outside of the loop, rather than at each iteration. But a better way to do this is probably to generate all of the data you want to plot ahead of time and store it in an easy-to-plot datatype like lists, then plot it once at the end.

Geospatial fixed radius cluster hunting in python

I want to take an input of millions of lat long points (with a numerical attribute) and then find all fixed radius geospatial clusters where the sum of the attribute within the circle is above a defined threshold.
I started by using sklearn BallTree to sum the attribute within any defined circle, with the intention of then expanding this out to run across a grid or lattice of circles. The run time for one circle is around 0.01s, so this is fine for small lattices, but won't scale if I want to run 200m radius circles across the whole of the UK.
#example data (use 2m rows from postcode centroid file)
df = pandas.read_csv('National_Statistics_Postcode_Lookup_Latest_Centroids.csv', usecols=[0,1], nrows=2000000)
#this will be our grid of points (or lattice) use points from same file for example
df2 = pandas.read_csv('National_Statistics_Postcode_Lookup_Latest_Centroids.csv', usecols=[0,1], nrows=2000)
#reorder lat long columns for balltree input
columnTitles=["Y","X"]
df = df.reindex(columns=columnTitles)
df2 = df2.reindex(columns=columnTitles)
# assign new columns to existing dataframe. attribute will hold the data we want to sum over (set to 1 for now)
df['attribute'] = 1
df2['aggregation'] = 0
RADIANT_TO_KM_CONSTANT = 6367
class BallTreeIndex:
def __init__(self, lat_longs):
self.lat_longs = np.radians(lat_longs)
self.ball_tree_index =BallTree(self.lat_longs, metric='haversine')
def query_radius(self,query,radius):
radius_km = radius/1000
radius_radiant = radius_km / RADIANT_TO_KM_CONSTANT
query = np.radians(np.array([query]))
indices = self.ball_tree_index.query_radius(query,r=radius_radiant)
return indices[0]
#index the base data
a=BallTreeIndex(df.iloc[:,0:2])
#begin to loop over the lattice to test performance
for i in range(0,100):
b = df2.iloc[i,0:2]
output = a.query_radius(b, 200)
accumulation = sum(df.iloc[output, 2])
df2.iloc[i,2] = accumulation
It feels as if the above code is really inefficient as I don't need to run the calculation across all circles on my lattice (as most will be well below my threshold - or will have no data points in at all).
Instead of this for loop, is there a better way of scaling this algorithm to give me the most dense circles?
I'm new to python, so any help would be massively appreciated!!
First don't try to do this on a sphere! GB is small and we have a well defined geographic projection that will work. So use the oseast1m and osnorth1m columns as X and Y. They are in metres so no need to convert (roughly) to degrees and use Haversine. That should help.
Next add a spatial index to speed up lookups.
If you need more speed there are various tricks like loading a 2R strip across the country into memory and then running your circles across that strip, then moving down a grid step and updating that strip (checking Y values against a fixed value is quick, especially if you store the data sorted on Y then X value). If you need more speed then look at any of the papers the Stan Openshaw (and sometimes I) wrote about parallelising the GAM. There are examples of implementing GAM in python (e.g. this paper, this paper) that may also point to better ways.

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