Hello all
I have a list of xyz coordinates of different points from a PDB file assigned to variable x. Here is a snippet of what it looks like
[ 8.721 15.393 22.939]
[11.2 13.355 25.025]
[11.045 15.057 28.419]
[13.356 13.814 31.169]
[12.54 13.525 34.854]
[14.038 15.691 37.608]
[16.184 12.782 38.807]
[17.496 12.053 35.319]
[18.375 15.721 34.871]
[20.066 15.836 38.288]
[22.355 12.978 37.249]
[22.959 14.307 33.724]
[24.016 17.834 34.691]
[26.63 16.577 37.161]
[29.536 18.241 35.342]
[27.953 21.667 35.829]
I would like to use these points to compute a distance matrix. I have tried to use the SciPy distance_matrix function, however it does not appear to support xyz coordinates, only x and y coordinates. Is there a quick way to compute this distance matrix manually?
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westin.kosater •
Going off of my earlier comment, I think this is one way to do it, though I havent checked to ensure the maths is correct. The numbers look OK though I think.
I had to slightly reformat your input data to be a little easier to use though:
8.721 15.393 22.939
11.2 13.355 25.025
11.045 15.057 28.419
13.356 13.814 31.169
12.54 13.525 34.854
14.038 15.691 37.608
16.184 12.782 38.807
17.496 12.053 35.319
18.375 15.721 34.871
20.066 15.836 38.288
22.355 12.978 37.249
22.959 14.307 33.724
24.016 17.834 34.691
26.63 16.577 37.161
29.536 18.241 35.342
27.953 21.667 35.829
import numpy
import sys
import itertools
with open(sys.argv[1], 'r') as ifh:
coords = [numpy.array(line.rstrip('\n').split(' ')) for line in ifh]
fl_coords = [arr.astype(float) for arr in coords]
for each in itertools.combinations(fl_coords,2):
print('{} - {} = {}'.format(each[0],each[1], numpy.linalg.norm(each[0]-each[1])))
Some example output:
[ 8.721 15.393 22.939] - [ 11.2 13.355 25.025] = 3.8275685493534914
[ 8.721 15.393 22.939] - [ 11.045 15.057 28.419] = 5.961901710025082
[ 8.721 15.393 22.939] - [ 13.356 13.814 31.169] = 9.576500717903174
[ 8.721 15.393 22.939] - [ 12.54 13.525 34.854] = 12.650747408750203
[ 8.721 15.393 22.939] - [ 14.038 15.691 37.608] = 15.60573144713185
[ 8.721 15.393 22.939] - [ 16.184 12.782 38.807] = 17.728708751626556
[ 8.721 15.393 22.939] - [ 17.496 12.053 35.319] = 15.537716209276061
[ 8.721 15.393 22.939] - [ 18.375 15.721 34.871] = 15.351870374648167
[ 8.721 15.393 22.939] - [ 20.066 15.836 38.288] = 19.091806488648473
This will give you everything in 'long' format for all the pairwise point distances, I'm not sure if you wanted all-vs-all or just all-vs-one, but it shouldnt take much to manipulate this in to a matrix now.
Thank you for the explanation. For the record, I was looking for a method to do all vs. all
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You could do this in R with package rpdb much more easily. Also, I would recommend to keep the atom names attached.
I agree with Michael that its probably best to keep it ‘intact’ with atom names etc, but what you’re looking for is the 3D Euclidean distance.
Take a look at: https://stackoverflow.com/questions/1401712/how-can-the-euclidean-distance-be-calculated-with-numpy