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Message: 2
Date: Mon, 03 Apr 2017 11:36:10 +0200
From: Peter Otten <__pete...@web.de>
To: tutor@python.org
Subject: Re: [Tutor] Euclidean Distances between Atoms in a Molecule.
Message-ID: <obt525$ths$1...@blaine.gmane.org>
Content-Type: text/plain; charset="ISO-8859-1"
Stephen P. Molnar wrote:
> I am trying to port a program that I wrote in FORTRAN twenty years ago
> into Python 3 and am having a hard time trying to calculate the
> Euclidean distance between each atom in the molecule and every other
> atom in the molecule.
>
> Here is a typical table of coordinates:
>
>
> MASS X Y Z
> 0 12.011 -3.265636 0.198894 0.090858
> 1 12.011 -1.307161 1.522212 1.003463
> 2 12.011 1.213336 0.948208 -0.033373
> 3 14.007 3.238650 1.041523 1.301322
> 4 12.011 -5.954489 0.650878 0.803379
> 5 12.011 5.654476 0.480066 0.013757
> 6 12.011 6.372043 2.731713 -1.662411
> 7 12.011 7.655753 0.168393 2.096802
> 8 12.011 5.563051 -1.990203 -1.511875
> 9 1.008 -2.939469 -1.327967 -1.247635
> 10 1.008 -1.460475 2.993912 2.415410
> 11 1.008 1.218042 0.451815 -2.057439
> 12 1.008 -6.255901 2.575035 1.496984
> 13 1.008 -6.560562 -0.695722 2.248982
> 14 1.008 -7.152500 0.390758 -0.864115
> 15 1.008 4.959548 3.061356 -3.139100
> 16 1.008 8.197613 2.429073 -2.588339
> 17 1.008 6.503322 4.471092 -0.543939
> 18 1.008 7.845274 1.892126 3.227577
> 19 1.008 9.512371 -0.273198 1.291080
> 20 1.008 7.147039 -1.365346 3.393778
> 21 1.008 4.191488 -1.928466 -3.057804
> 22 1.008 5.061650 -3.595015 -0.302810
> 23 1.008 7.402586 -2.392148 -2.374554
>
> What I need for further calculation is a matrix of the Euclidean
> distances between the atoms.
>
> So far in searching the Python literature I have only managed to confuse
> myself and would greatly appreciate any pointers towards a solution.
>
> Thanks in advance.
>
Stitched together with heavy use of a search engine:
$ cat data.txt
MASS X Y Z
0 12.011 -3.265636 0.198894 0.090858
1 12.011 -1.307161 1.522212 1.003463
2 12.011 1.213336 0.948208 -0.033373
3 14.007 3.238650 1.041523 1.301322
4 12.011 -5.954489 0.650878 0.803379
5 12.011 5.654476 0.480066 0.013757
6 12.011 6.372043 2.731713 -1.662411
7 12.011 7.655753 0.168393 2.096802
8 12.011 5.563051 -1.990203 -1.511875
9 1.008 -2.939469 -1.327967 -1.247635
10 1.008 -1.460475 2.993912 2.415410
11 1.008 1.218042 0.451815 -2.057439
12 1.008 -6.255901 2.575035 1.496984
13 1.008 -6.560562 -0.695722 2.248982
14 1.008 -7.152500 0.390758 -0.864115
15 1.008 4.959548 3.061356 -3.139100
16 1.008 8.197613 2.429073 -2.588339
17 1.008 6.503322 4.471092 -0.543939
18 1.008 7.845274 1.892126 3.227577
19 1.008 9.512371 -0.273198 1.291080
20 1.008 7.147039 -1.365346 3.393778
21 1.008 4.191488 -1.928466 -3.057804
22 1.008 5.061650 -3.595015 -0.302810
23 1.008 7.402586 -2.392148 -2.374554
$ python3
Python 3.4.3 (default, Nov 17 2016, 01:08:31)
[GCC 4.8.4] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import numpy, pandas, scipy.spatial.distance as dist
>>> df = pandas.read_table("data.txt", sep=" ", skipinitialspace=True)
>>> a = numpy.array(df[["X", "Y", "Z"]])
>>> dist.squareform(dist.pdist(a, "euclidean"))
<snip big matrix>
Here's an example with just the first 4 atoms:
>>> dist.squareform(dist.pdist(a[:4], "euclidean"))
array([[ 0. , 2.53370139, 4.54291701, 6.6694065 ],
[ 2.53370139, 0. , 2.78521357, 4.58084922],
[ 4.54291701, 2.78521357, 0. , 2.42734737],
[ 6.6694065 , 4.58084922, 2.42734737, 0. ]])
See
https://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.pdist.html[https://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.pdist.html]
There may be a way to do this with pandas.pivot_table(), but I didn't manage
to find that.
As Alan says, this is not the appropriate forum for the topic you are
belabouring.
Work your way through a Python tutorial to pick up the basics (we can help
you with this), then go straight to where the (numpy/scipy) experts are.
------------------------------
An alternative starting from the numpy array "a" from Peter answer:
import numpy as np
#Taking the number of rows and columns of the array
anrows, ancols = np.shape(a)
# Gather the coordinates as one dimensional arrays
a_new = a.reshape(anrows, 1, ancols)
# Takes the difference between each of the elements (one Vs all)
diff = a_new - a
# Computes the sum of the squared difference
D = (diff ** 2).sum(2)
# Takes the square root
D = np.sqrt(D)
#Check that both answers are the same:
print(D-dist.squareform(dist.pdist(a, "euclidean")))
Salut,
Sergio
Avoiding for loops section of:
https://www.packtpub.com/big-data-and-business-intelligence/numerical-and-scientific-computing-scipy-video
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