I am using multidimensional list with numpy
I have a list.
l = [[0 2 8] [0 2 7] [0 2 5] [2 4 5] [ 8 4 7]] I need to find square root of sum of square of columns.
0 2 8 0 2 7 0 2 5 2 4 5 8 4 7 output as,
l = [sqrt((square(0) + square(0) + square(0) + square(2) + square(8)) sqrt((square(2) + square(2) + square(2) + square(4) + square(4)) sqrt((square(8) + square(7) + square(5)) + square(5) + square(7))] 
>>> import numpy as np >>> a = np.array([[0, 2, 8], [0, 2, 7], [0, 2, 5], [2, 4, 5], [ 8, 4, 7]]) >>> np.sqrt(np.sum(np.square(a), axis=0)) array([ 8.24621125, 6.63324958, 14.56021978]) 
Use the standard function numpy.linalg.norm for this...
import numpy as np a = np.array([[0, 2, 8], [0, 2, 7], [0, 2, 5], [2, 4, 5], [ 8, 4, 7]]) np.linalg.norm(a,axis=0) gives:
array([ 8.24621125, 6.63324958, 14.56021978])
>>> import numpy as np >>> np.sum(np.array(l)**2,axis=0)**.5 array([ 10.67707825, 3.46410162, 11.74734012]) What you want to do is use map/reduce
In theory in can be done using nested for loops but could be done in a more functional way...
for l in matrix: sum all elements**2 in return the squar root of the sum A one liner:
map(lambda x: sqrt(lambda r, z: r + z**2, x), matrix) But to make it more clear, you could rewrite it as such:
def SumOfSquare(lst): return reduce(lambda r, x: r + x**2, lst) def ListOfRoot(lst): return map(lambda x: SumOfSquare(x), lst) s = ListOfRoot(matrix) Misread the question, it's without numpy.
