How to sort the rows of a numpy matrix with respect to their norm?
Question:
I have a numpy array
array([[ 4.14022471, 3.96360618],
[ 0.37601032, 1.25528411],
[ 3.49313049, 0.94909878]])
Now I wish to sort this array’s rows with respect to the norm of the rows, i.e
norm([ 4.14022471, 3.96360618]) = 5.73163455
norm([ 0.37601032, 1.25528411]) = 1.31039
norm([ 3.49313049, 0.94909878]) = 3.61977197
Hence the first row remains in first position, while 2nd and 3rd rows need to be swapped.
The solution will be
array([[ 4.14022471, 3.96360618],
[ 3.49313049, 0.94909878],
[ 0.37601032, 1.25528411]])
How can I do this?
Answers:
Python’s built-in sorted() function has a key parameter that accepts a function, simply pass in the norm function and reverse=True to sort largest to smallest
>>> arr = array([[ 4.14022471,3.96360618],[ 0.37601032,1.25528411],[3.49313049, 0.94909878]])
>>> array(sorted(arr, key=norm, reverse = True))
array([[4.14022471, 3.96360618],
[3.49313049, 0.94909878],
[0.37601032, 1.25528411]])
With numpy.argsort and numpy.take:
arr = np.array([[ 4.14022471, 3.96360618], [ 0.37601032, 1.25528411], [ 3.49313049, 0.94909878]])
norms = np.linalg.norm(arr, axis=1)
arr_sorted = np.take(arr, np.argsort(norms)[::-1], axis=0)
[[4.14022471 3.96360618]
[3.49313049 0.94909878]
[0.37601032 1.25528411]]
For your task it is enough to use the only library – numpy.
Solution:
a = np.array([[ 4.14022471, 3.96360618], [ 0.37601032, 1.25528411], [ 3.49313049, 0.94909878]])
def get_norm(array):
return np.linalg.norm(array)
sorted(a, reverse=True, key = get_norm)
It will be appropriate to use the lambda function. It’s more concise and easier to read the code.
a = np.array([[ 4.14022471, 3.96360618], [ 0.37601032, 1.25528411], [ 3.49313049, 0.94909878]])
sorted(a, reverse=True, key = lambda x:np.linalg.norm(x))
In both cases the program will output:
[array([4.14022471, 3.96360618]),
array([3.49313049, 0.94909878]),
array([0.37601032, 1.25528411])]
I have a numpy array
array([[ 4.14022471, 3.96360618],
[ 0.37601032, 1.25528411],
[ 3.49313049, 0.94909878]])
Now I wish to sort this array’s rows with respect to the norm of the rows, i.e
norm([ 4.14022471, 3.96360618]) = 5.73163455
norm([ 0.37601032, 1.25528411]) = 1.31039
norm([ 3.49313049, 0.94909878]) = 3.61977197
Hence the first row remains in first position, while 2nd and 3rd rows need to be swapped.
The solution will be
array([[ 4.14022471, 3.96360618],
[ 3.49313049, 0.94909878],
[ 0.37601032, 1.25528411]])
How can I do this?
Python’s built-in sorted() function has a key parameter that accepts a function, simply pass in the norm function and reverse=True to sort largest to smallest
>>> arr = array([[ 4.14022471,3.96360618],[ 0.37601032,1.25528411],[3.49313049, 0.94909878]])
>>> array(sorted(arr, key=norm, reverse = True))
array([[4.14022471, 3.96360618],
[3.49313049, 0.94909878],
[0.37601032, 1.25528411]])
With numpy.argsort and numpy.take:
arr = np.array([[ 4.14022471, 3.96360618], [ 0.37601032, 1.25528411], [ 3.49313049, 0.94909878]])
norms = np.linalg.norm(arr, axis=1)
arr_sorted = np.take(arr, np.argsort(norms)[::-1], axis=0)
[[4.14022471 3.96360618]
[3.49313049 0.94909878]
[0.37601032 1.25528411]]
For your task it is enough to use the only library – numpy.
Solution:
a = np.array([[ 4.14022471, 3.96360618], [ 0.37601032, 1.25528411], [ 3.49313049, 0.94909878]])
def get_norm(array):
return np.linalg.norm(array)
sorted(a, reverse=True, key = get_norm)
It will be appropriate to use the lambda function. It’s more concise and easier to read the code.
a = np.array([[ 4.14022471, 3.96360618], [ 0.37601032, 1.25528411], [ 3.49313049, 0.94909878]])
sorted(a, reverse=True, key = lambda x:np.linalg.norm(x))
In both cases the program will output:
[array([4.14022471, 3.96360618]),
array([3.49313049, 0.94909878]),
array([0.37601032, 1.25528411])]