I have this numpy array
X = [[ -9.03525007 7.45325017 33.34074879][ -6.63700008 5.13299996 31.66075039][ -5.12724996 8.25149989 30.92599964][ -5.12724996 8.25149989 30.92599964]]
I want to get the norm of this array using numpy. How can I do that?
for every array inside, I need sqrt(x2+y2+z2), so my output wull be array of 4 values (since there are 4 inside arrays)
To get what you ask for (the 2-norm of each row in your array), you can use the axis argument to numpy.linalg.norm:
import numpy
x = numpy.array([[ -9.03525007, 7.45325017, 33.34074879],
[ -6.63700008, 5.13299996, 31.66075039],
[ -5.12724996, 8.25149989, 30.92599964],
[ -5.12724996, 8.25149989, 30.92599964]])
print numpy.linalg.norm(x, axis=1)
=>
array([ 35.33825423, 32.75363451, 32.41594355, 32.41594355])
Why don't use numpy.linalg.norm
import numpy
x = [[ -9.03525007, 7.45325017 , 33.34074879], [ -6.63700008 , 5.13299996 , 31.66075039], [ -5.12724996 , 8.25149989 , 30.92599964], [ -5.12724996 , 8.25149989 , 30.92599964]]
print numpy.linalg.norm(x)
Output:
66.5069889437
Did you mean matrix norm(s)? If so:
import numpy as np
>>> xs = [[ -9.03525007, 7.45325017, 33.34074879], [-6.63700008, 5.13299996, 31.66075039], [-5.12724996, 8.25149989, 30.92599964], [-5.12724996, 8.25149989, 30.92599964]]
>>> np.linalg.norm(xs)
66.506988943656381
See: http://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.norm.html
Other people have already given you the norm() function. You are probably looking to map() the norm() function within the array.
Just do:
from numpy.linalg import norm
norms = map(norm, x)
Related
I am stacking point clouds into one array using np.stack but when i see the result the values are shown in e notation. if i only stack x and y then the values are not shown in e notation only if i add the z column its shown in e notation
for stacking two values
import laspy
import numpy as np
las = laspy.read("f:\\lidar\\pointcloud.laz")
arr = np.stack([las.x, las.y], axis=0).transpose((1, 0))
arr
array([[ 368230.825, 5807507.866],
[ 368231.821, 5807508.151],
[ 368232.935, 5807508.114],
...,
[ 368496.193, 5807742.345],
[ 368495.747, 5807741.916],
[ 368495.412, 5807742.06 ]])
values when stacking with z values.
import laspy
import numpy as np
las = laspy.read("f:\\lidar\\pointcloud.laz")
arr = np.stack([las.x, las.y,las.z], axis=0).transpose((1, 0))
arr
array([[3.68230825e+05, 5.80750787e+06, 3.10790000e+01],
[3.68231821e+05, 5.80750815e+06, 3.11140000e+01],
[3.68232935e+05, 5.80750811e+06, 3.11320000e+01],
...,
[3.68496193e+05, 5.80774234e+06, 3.33270000e+01],
[3.68495747e+05, 5.80774192e+06, 3.29020000e+01],
[3.68495412e+05, 5.80774206e+06, 3.13750000e+01]])
I'm trying to scale the following NumPy array based on its minimum and maximum values.
array = [[17405.051 17442.4 17199.6 17245.65 ]
[17094.949 17291.75 17091.15 17222.75 ]
[17289. 17294.9 17076.551 17153. ]
[17181.85 17235.1 17003.9 17222. ]]
Formula used is:
m=(x-xmin)/(xmax-xmin)
wherein m is an individually scaled item, x is an individual item, xmax is the highest value and xmin is the smallest value of the array.
My question is how do I print the scaled array?
P.S. - I can't use MinMaxScaler as I need to scale a given number (outside the array) by plugging it in the mentioned formula with xmin & xmax of the given array.
I tried scaling the individual items by iterating over the array but I'm unable to put together the scaled array.
I'm new to NumPy, any suggestions would be welcome.
Thank you.
Use method ndarray.min(), ndarray.max() or ndarray.ptp()(gets the range of the values in the array):
>>> ar = np.array([[17405.051, 17442.4, 17199.6, 17245.65 ],
... [17094.949, 17291.75, 17091.15, 17222.75 ],
... [17289., 17294.9, 17076.551, 17153. ],
... [17181.85, 17235.1, 17003.9, 17222. ]])
>>> min_val = ar.min()
>>> range_val = ar.ptp()
>>> (ar - min_val) / range_val
array([[0.91482554, 1. , 0.44629418, 0.55131129],
[0.2076374 , 0.65644242, 0.19897377, 0.4990878 ],
[0.65017104, 0.663626 , 0.16568073, 0.34002281],
[0.40581528, 0.527252 , 0. , 0.49737742]])
I think you should learn more about the basic operation of numpy.
import numpy as np
array_list = [[17405.051, 17442.4, 17199.6, 17245.65 ],
[17094.949, 17291.75, 17091.15, 17222.75 ],
[17289., 17294.9, 17076.551, 17153., ],
[17181.85, 17235.1, 17003.9, 17222. ]]
# Convert list into numpy array
array = np.array(array_list)
# Create empty list
scaled_array_list=[]
for x in array:
m = (x - np.min(array))/(np.max(array)-np.min(array))
scaled_array_list.append(m)
# Convert list into numpy array
scaled_array = np.array(scaled_array_list)
scaled_array
My version is by iterating over the array as you said.
You can also put everything in a function and use it in future:
def scaler(array_to_scale):
# Create empty list
scaled_array_list=[]
for x in array:
m = (x - np.min(array))/(np.max(array)-np.min(array))
scaled_array_list.append(m)
# Convert list into numpy array
scaled_array = np.array(scaled_array_list)
return scaled_array
# Here it is our input
array_list = [[17405.051, 17442.4, 17199.6, 17245.65 ],
[17094.949, 17291.75, 17091.15, 17222.75 ],
[17289., 17294.9, 17076.551, 17153., ],
[17181.85, 17235.1, 17003.9, 17222. ]]
# Convert list into numpy array
array = np.array(array_list)
scaler(array)
Output:
Out:
array([[0.91482554, 1. , 0.44629418, 0.55131129],
[0.2076374 , 0.65644242, 0.19897377, 0.4990878 ],
[0.65017104, 0.663626 , 0.16568073, 0.34002281],
[0.40581528, 0.527252 , 0. , 0.49737742]])
I'm currently writing a code, and I have to extract from a numpy array.
For example: [[1,1] , [0.6,0.6], [0,0]]), given the condition for the extracted points [x,y] must satisfy x>=0.5 and y >= 0.5
I've tried to use numpy extract, with the condition arr[0]>=0.5 & arr[1]>=0.5 however that does not seem to work
It applied the condition on all the elements, and I just want it to apply to the points inside my array.
Thanks in advance!
You can use multiple conditions to slice an array as follows:
import numpy as np
a = np.array([[1, 1] , [0.6, 0.6], [0, 0]])
new = a[(a[:, 0] >= 0.5) & (a[:, 1] >= 0.5)]
Results:
array([[1. , 1. ],
[0.6, 0.6]])
The first condition filters on column 0 and the second condition filters on column 1. Only rows where both conditions are met will be in the results.
I would do it following way: firstly look for rows full-filling condition:
import numpy as np
a = np.array([[1,1] , [0.6,0.6], [0,0]])
rows = np.apply_along_axis(lambda x:x[0]>=0.5 and x[1]>=0.5,1,a)
then use it for indexing:
out = a[rows]
print(out)
output:
[[1. 1. ]
[0.6 0.6]]
It can be solved using python generators.
import numpy as np
p = [[1,1] , [0.6,0.6], [0,0]]
result = np.array([x for x in p if x[0]>0.5 and x[1]>0.5 ])
You can read more about generators from here.
Also you can try this:-
p = np.array(p)
result= p[np.all(p>0.5, axis=1)]
This question already has answers here:
Subtract all pairs of values from two arrays
(2 answers)
Closed 4 years ago.
I have two numpy arrrays:
import numpy as np
points_1 = np.array([1.5,2.5,1,3])
points_2 = np.array([3,4])
I would like to take evey point from points_1 array and deduce whole points_2 array from it in order to get a matrix
I would like to get
[[-1.5,-2.5]
[-0.5,-1.5]
[-2 , -3]
[0 , -1]]
I know there is a way with iteration
points = [x - points_2 for x in points_1]
points = np.array(points)
However this option is not fast enough. In reality I am using much bigger arrays.
Is there some fastser way?
Thanks!
You just have to chose points_2 "better" (better means here an other dimension of you matrix), then it works as you expect it:
so do not use points_2 = np.array([3, 4]) but points_2 = np.array([[3],[4]]):
import numpy as np
points_1 = np.array([1.5,2.5,1,3])
points_2 = np.array([[3],[4]])
points = (points_1 - points_2).transpose()
print(points)
results in:
[[-1.5 -2.5]
[-0.5 -1.5]
[-2. -3. ]
[ 0. -1. ]]
If you don't the whole array at once. You can use generators and benefit from lazy evaluation:
import numpy as np
points_1 = np.array([1.5,2.5,1,3])
points_2 = np.array([3,4])
def get_points():
def get_points_internal():
for p1 in points_1:
for p2 in points_2:
yield [p1 - p2]
x = len(points_1) * len(points_2)
points_1d = get_points_internal()
for i in range(0, int(x/2)):
yield [next(points_1d), next(points_1d)]
points = get_points()
Make use of numpy's broadcasting feature. This will provide the following:
import numpy as np
points_1 = np.array([1.5,2.5,1,3])
points_2 = np.array([3,4])
points = points_1[:, None] - points_2
print(points)
Output:
[[-1.5 -2.5]
[-0.5 -1.5]
[-2. -3. ]
[ 0. -1. ]]
It works by repeating the operation over the 1 dimension injected by the None index. For more info see the link.
You can do it in one line :
np.subtract.outer(points_1,points_2)
This is vectored so very fast.
You need to use tranposed matrix.
points_1-np.transpose([points_2])
and for your result
np.tanspose(points_1-np.transpose([points_2]))
Using some experimental data, I cannot for the life of me work out how to use splrep to create a B-spline. The data are here: http://ubuntuone.com/4ZFyFCEgyGsAjWNkxMBKWD
Here is an excerpt:
#Depth Temperature
1 14.7036
-0.02 14.6842
-1.01 14.7317
-2.01 14.3844
-3 14.847
-4.05 14.9585
-5.03 15.9707
-5.99 16.0166
-7.05 16.0147
and here's a plot of it with depth on y and temperature on x:
Here is my code:
import numpy as np
from scipy.interpolate import splrep, splev
tdata = np.genfromtxt('t-data.txt',
skip_header=1, delimiter='\t')
depth = tdata[:, 0]
temp = tdata[:, 1]
# Find the B-spline representation of 1-D curve:
tck = splrep(depth, temp)
### fails here with "Error on input data" returned. ###
I know I am doing something bleedingly stupid, but I just can't see it.
You just need to have your values from smallest to largest :). It shouldn't be a problem for you #a different ben, but beware readers from the future, depth[indices] will throw a TypeError if depth is a list instead of a numpy array!
>>> indices = np.argsort(depth)
>>> depth = depth[indices]
>>> temp = temp[indices]
>>> splrep(depth, temp)
(array([-7.05, -7.05, -7.05, -7.05, -5.03, -4.05, -3. , -2.01, -1.01,
1. , 1. , 1. , 1. ]), array([ 16.0147 , 15.54473241, 16.90606794, 14.55343229,
15.12525673, 14.0717599 , 15.19657895, 14.40437622,
14.7036 , 0. , 0. , 0. , 0. ]), 3)
Hat tip to #FerdinandBeyer for the suggestion of argsort instead of my ugly "zip the values, sort the zip, re-assign the values" method.