Update Creating an array for distance between two 2-D arrays

Question

So I have two arrays that have x, y, z coordinates. I'm just trying to apply the 3D distance formula. Problem is, that I can't find a post that constitutes arrays with multiple values in each column and spits out an array.

print MW_FirstsubPos1 

[[  51618.7265625   106197.7578125    69647.6484375 ]
 [  33864.1953125    11757.29882812   11849.90332031]
 [  12750.09863281   58954.91015625   38067.0859375 ]
 ..., 
 [  99002.6640625    96021.0546875    18798.44726562]
 [  27180.83984375   74350.421875     78075.78125   ]
 [  19297.88476562   82161.140625      1204.53503418]]

print MW_SecondsubPos1 

[[  51850.9140625   106004.0078125    69536.5234375 ]
 [  33989.9375       11847.11425781   12255.80859375]
 [  12526.203125     58372.3046875    37641.34765625]
 ..., 
 [  98823.2734375    95837.1796875    18758.7734375 ]
 [  27047.19140625   74242.859375     78166.703125  ]
 [  19353.97851562   82375.8515625     1147.07556152]]

Yes, they are the same shape.

My attempt,

import numpy as np

xs1,ys1,zs1 = zip(*MW_FirstsubPos1)
xs11,ys11,zs11 = zip(*MW_SecondsubPos1)

squared_dist1 = (xs11 - xs1)**2 + (ys11 - ys1)**2 + (zs11 - zs1)**2
dist1 = np.sqrt(squared_dist1)

print dist1

This returns:

TypeError: unsupported operand type(s) for -: 'tuple' and 'tuple'

I'm just wanting to return a 1-D array of the same shape.

* --------------------- Update --------------------- *

Using what Sнаđошƒаӽ said,

Distance1 = []
for Fir1, Sec1 in zip(MW_FirstsubVel1, MW_SecondsubPos1):
    dist1 = 0
    for i in range(3):
        dist1 += (Fir1[i]-Sec1[i])**2
    Distance1.append(dist1**0.5)

But when comparing the distance formula for one element in my original post such as,

squared_dist1 = (xs11[0] - xs1[0])**2 + (ys11[0] - ys1[0])**2 + (zs11[0] - zs1[0])**2
dist1 = np.sqrt(squared_dist1)

print dist1

returns 322.178309762

while

result = []
for a, b in zip(MW_FirstsubVel1, MW_SecondsubPos1):
    dist = 0
    for i in range(3):
        dist += (a[i]-b[i])**2
    result.append(dist**0.5)

print result[0]

returns 137163.203004

What's wrong here?

Hei, my solution gives that expected solution too. Take a look at my updated answer that shows the outputs. — Sнаđошƒаӽ
– Sнаđошƒаӽ, Commented Jul 20, 2016 at 4:33

Neo X · Accepted Answer · 2016-07-19 02:53:51Z

2

Your solutions look good to me. A better idea is to use the linear algebra module in scipy package, as it scales with multiple dimensional data. Here are my codes.

import scipy.linalg as LA

dist1 = LA.norm(MW_FirstsubPos1 - MW_SecondsubPos1, axis=1)

answered Jul 19, 2016 at 2:53

Neo X

9577 silver badges9 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

iron2man Over a year ago

My solution does not work unfortunately. I've added more to the issue.

Sнаđошƒаӽ · Accepted Answer · 2016-07-20 04:35:46Z

2

See if this works, assuming that aaa and bbb are normal python list of lists having the x, y and z coordinates (or that you can convert to such, using tolist or something like that perhaps). result will have the 1-D array you are looking for.

Edit: aaa and bbb are python lists of lists. Only code for printing the output have been added.

aaa = [[51618.7265625, 106197.7578125, 69647.6484375],
[33864.1953125, 11757.29882812, 11849.90332031],
[12750.09863281, 58954.91015625, 38067.0859375],
[99002.6640625, 96021.0546875, 18798.44726562],
[27180.83984375, 74350.421875, 78075.78125],
[19297.88476562, 82161.140625, 1204.53503418]]

bbb = [[51850.9140625, 106004.0078125, 69536.5234375],
[33989.9375, 11847.11425781, 12255.80859375],
[12526.203125, 58372.3046875, 37641.34765625],
[98823.2734375, 95837.1796875, 18758.7734375],
[27047.19140625, 74242.859375, 78166.703125],
[19353.97851562, 82375.8515625, 1147.07556152]]

result = []
for a, b in zip(aaa, bbb):
    dist = 0
    for i in range(3):
        dist += (a[i]-b[i])**2
    result.append(dist**0.5)

for elem in result:
    print(elem)

Output:

322.178309762234
434.32361222259755
755.5206249710258
259.9327309143388
194.16071591842936
229.23543894772612

edited Jul 20, 2016 at 4:35

answered Jul 19, 2016 at 4:03

Sнаđошƒаӽ

17.8k13 gold badges78 silver badges95 bronze badges

2 Comments

iron2man Over a year ago

I appreciate the response. Trying what you posted left me at a hurdle. I've posted it on the update potion.

Sнаđошƒаӽ Over a year ago

@DarthLazar The way I understood the question is that the two arrays hold pairs of points that we need to find the distance between in the same indices. It seems now, after the update and my answer being unaccepted, that so is not the case. :-)

Divakar · Accepted Answer · 2016-07-19 18:43:02Z

1

Here's a vectorized approach using np.einsum -

diffs = MW_FirstsubPos1 - MW_SecondsubPos1
dists = np.sqrt(np.einsum('ij,ij->i',diffs,diffs))

Sample run -

In [233]: MW_FirstsubPos1
Out[233]: 
array([[2, 0, 0],
       [8, 6, 1],
       [0, 2, 8],
       [7, 6, 3],
       [3, 1, 7]])

In [234]: MW_SecondsubPos1
Out[234]: 
array([[3, 4, 7],
       [0, 8, 4],
       [4, 7, 4],
       [2, 5, 6],
       [5, 0, 6]])

In [235]: diffs = MW_FirstsubPos1 - MW_SecondsubPos1

In [236]: np.sqrt(np.einsum('ij,ij->i',diffs,diffs))
Out[236]: array([ 8.1240384 ,  8.77496439,  7.54983444,  5.91607978,  2.44948974])

edited Jul 19, 2016 at 18:43

answered Jul 19, 2016 at 7:54

Divakar

222k19 gold badges273 silver badges374 bronze badges

Collectives™ on Stack Overflow

Update Creating an array for distance between two 2-D arrays

3 Answers 3

1 Comment

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related