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?
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)
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
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])
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