Calculate euclidean norm of 100 points of n-dimensions?

Question

I have a list of 100 values in python where each value in the list corresponds to an n-dimensional list.

For e.g

x=[[1 2],[2 3]] is a 2d list

I want to compute euclidean norm over all such points. Is there a standard method to do this?

huh.. maybe it should be y=[[1,2,3],[1,2,3]].. assuming the first[1,2,3] is a point and the second [1,2,3] is another point in 3d.. — zzk, Mar 04 '13 at 03:28
By *distance over all such points*, do you mean distance from the origin (i.e.the magnitude of each n-dimensional vector), or do you mean distances of all the points from each other? — Warren Weckesser, Mar 04 '13 at 03:29
Possible duplicate: http://stackoverflow.com/questions/7741878/how-to-apply-numpy-linalg-norm-to-each-row-of-a-matrix — John Vinyard, Mar 04 '13 at 04:44

score 1 · Answer 1 · answered Mar 04 '13 at 03:36

1

I found this on scipy and this works. scipy

answered Mar 04 '13 at 03:36

gizgok

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I second this approach when the calculation takes too much time due to the size of the list. – eldad-a Jun 23 '13 at 20:01

Sam · Answer 2 · 2015-02-06T11:54:07.107

If I have interpreted the question correctly, then you have a list of 100 n-dimensional vectors, and you would like a list of their (Euclidean) norms.

I think using numpy is easiest (and quickest!) here,

import numpy as np
a = np.array(x)
np.sqrt((a*a).sum(axis=1))

If the vectors do not have equal dimension, or if you want to avoid numpy, then perhaps,

[sum([i*i for i in vec])**0.5 for vec in x]

or,

import math
[math.sqrt(sum([i*i for i in vec])) for vec in x]

Edit: Not entirely sure what you were asking for. So, alternatively: it looks like you have a list, each element of which is an n-dimensional vector, and you want the Euclidean distance between each consecutive pair. With numpy (assuming n is fixed),

x = [ [1,2,3], [4,5,6], [8,9,10], [13,14,15] ] # 3D example.
import numpy as np
a = np.array(x)
sqrDiff = (a[:-1] - a[1:])**2
np.sqrt(sqrDiff.sum(axis=1))

where the last line returns,

array([ 5.19615242, 6.92820323, 8.66025404])

score 0 · Answer 3 · answered Mar 04 '13 at 03:43

0

Try this code:

from math import sqrt
valueList = [[[1,2], [2,3]], [[2,2], [3,3]]]

def distance(valueList):
  resultList = []
  for (point1, point2) in valueList:
    resultList.append(sqrt(sum(map(lambda (x1, x2): (x1 - x2) * (x1 - x2), zip(point1, point2)))))

  return resultList

print distance(valueList)

output is [1.4142135623730951, 1.4142135623730951]

Here is valuelist contains 2 values, but no problem with 100 values..

answered Mar 04 '13 at 03:43

zzk

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How will you write it for 100 values? – gizgok Mar 04 '13 at 03:57
It should have already worked.. just call distance(your-100-value-list). assuming each of your value is 2 points in n dimension. – zzk Mar 04 '13 at 03:59

John Vinyard · Answer 4 · 2013-03-04T04:43:41.280

You can do this to compute the euclidean norm of each row:

>>> a = np.arange(200.).reshape((100,2))
>>> a
array([[   0.,    1.],
       [   2.,    3.],
       [   4.,    5.],
       [   6.,    7.],
       [   8.,    9.],
       [  10.,   11.],
       ...
>>> np.sum(a**2,axis=-1) ** .5
array([   1.        ,    3.60555128,    6.40312424,    9.21954446,
         12.04159458,   14.86606875,   17.69180601,   20.51828453,
         23.34523506,   26.17250466,   29.        ,   31.82766093,
         34.6554469 ,   37.48332963,   40.31128874,   43.13930922,
         45.96737974,   48.7954916 ,   51.623638  ,   54.45181356,
         ...

Calculate euclidean norm of 100 points of n-dimensions?

4 Answers4