I am doing work on parallelization .and I have numpy array in between.
I want to check the index of numpy array
for example :
pos is a ndarray and its value is
pos :
ndarray[[ 0. 44. 2367.]
[ 0. 73. 2301.]
[ 0. 38. 2304.]
[ 0. 35. 2349.]
[ 0. 6. 0.]
[ 0. 43. 2319.]
[ 0. 95. 2381.]
[ 0. 189. 3.]
[ 0. 0. 2339.]
[ 0. 22. 2335.]
[ 0. 44. 2345.]
[ 0. 44. 2345.]
[ 0. 52. 2348.]
[ 0. 50. 2348.]]
dtype :float64
max : 2381.0
min :0.0
Shape = {tuple}(14, 3)
size :42
I want to consider for my calculation only 2nd position of pos ndarray for example 2367,2301,2304...etc.
so how can I indexing them and how can I sort 2nd position value only? means if 2nd position of array is 3,4,2,1 then I want 2nd position is 1,2,3,4 like that..
give me suggestion..
If you want to refer to the column with index 2 (i.e. the rightmost column) of the array, you can just do
pos[:,2]
If you want to sort (in place) the pos array by the second column, you can do:
pos[:,2].sort()
Related
I've the following array:
np.array([[0.07704314, 0.46752589, 0.39533099, 0.35752864],
[0.45813299, 0.02914078, 0.65307364, 0.58732429],
[0.32757561, 0.32946822, 0.59821108, 0.45585825],
[0.49054429, 0.68553148, 0.26657932, 0.38495586]])
I want to find the minimum value in each row of the array. How can I achieve this?
Expected answer:
[[0.07704314 0. 0. 0. ]
[0. 0.02914078 0. 0. ]
[0.32757561 0 0. 0. ]
[0. 0. 0.26657932 0. ]]
You can use np.where like so:
np.where(a.argmin(1)[:,None]==np.arange(a.shape[1]), a, 0)
Or (more lines but potentially more efficient):
out = np.zeros_like(a)
idx = a.argmin(1)[:, None]
np.put_along_axis(out, idx, np.take_along_axis(a, idx, 1), 1)
IIUC first find out out the min value of each line , then we base on the min value mask all min value in original array as True, using multiple(matrix) , get what we need as result
np.multiply(a,a==np.min(a,1)[:,None])
Out[225]:
array([[0.07704314, 0. , 0. , 0. ],
[0. , 0.02914078, 0. , 0. ],
[0.32757561, 0. , 0. , 0. ],
[0. , 0. , 0.26657932, 0. ]])
np.amin(a, axis=1) where a is your np array
Please have a look at this code:
import numpy as np
from scipy.spatial import distance
#1
X = [[0,0], [0,1], [0,2], [0,3], [0,4], [0,5]]
c = [[0,0], [0,1], [0,3]]
#2
dists = distance.cdist(X, c)
print(dists)
#3
dmini = np.argmin(dists, axis=1)
print(dmini)
#4
mindists = dists[:, dmini]
print(mindists)
(#1) So I have my data X, some other points (centroids) c, then (#2) I compute the distance from each point in X to all the centroids c, and store the result in dists.
(#3) Then I select the index of the minimum distances with argmin.
(#4) Now I only want to select the value of the minimum values, using the indexes computed in step #3.
However, I get a strange output.
# dists
[[ 0. 1. 3.]
[ 1. 0. 2.]
[ 2. 1. 1.]
[ 3. 2. 0.]
[ 4. 3. 1.]
[ 5. 4. 2.]]
#dmini
[0 1 1 2 2 2]
#mindists
[[ 0. 1. 1. 3. 3. 3.]
[ 1. 0. 0. 2. 2. 2.]
[ 2. 1. 1. 1. 1. 1.]
[ 3. 2. 2. 0. 0. 0.]
[ 4. 3. 3. 1. 1. 1.]
[ 5. 4. 4. 2. 2. 2.]]
Reading here and there, it seems possible to select specific columns by giving a list of integers (indexes). In this case I should use the dmini values for indexing columns along rows.
I was expecting mindists to be (6,) in shape. What am I doing wrong?
I have the following code where I have been trying to create a tridiagonal matrix x using if-conditions.
#!/usr/bin/env python
# import useful modules
import numpy as np
N=5
x=np.identity(N)
#x=np.zeros((N,N))
print x
# Construct NxN matrix
for i in range(N):
for j in range(N):
if i-j==1:
x[i][j]=1
elif j-1==1:
x[i][j]=-1
else:
x[i][j]=0
print "i= ",i," j= ",j
print x
I desire to get
[[ 0. -1. 0. 0. 0.]
[ 1. 0. -1. 0. 0.]
[ 0. 1. 0. -1 0.]
[ 0. 0. 1. 0. -1.]
[ 0. 0. 0. 1. 0.]]
However, I obtain
[[ 0. 0. -1. 0. 0.]
[ 1. 0. -1. 0. 0.]
[ 0. 1. -1. 0. 0.]
[ 0. 0. 1. 0. 0.]
[ 0. 0. -1. 1. 0.]]
What's going wrong?
Bonus question : Can I forcefully index from 1 to 5 instead of 0 to 4 in this example, or Python never allows that?
elif j-1==1: should be elif j-i==1:.
And no, lists/arrays etc. are always indexed from 0.
As for the bonus question, the first element of a sequence in Python has always the index 0. However, if for some particular reason (for example to prevent off-by-one errors) you wish to count the elements of a sequence from a value other than 0, you could use the built-in function enumerate() and set the value of the optional parameter start to fit your needs:
>>> seq = ['a', 'b', 'c']
>>> for count, item in enumerate(seq, start=1):
... print(count, item)
...
1 a
2 b
3 c
I wrote a simple Linear Algebra code in Python Numpy to calculate the Diagonal of EigenValues by calculating $M^{-1}.A.M$ (M is the Modal Matrix) and it's working strange.
Here's the Code :
import numpy as np
array = np.arange(16)
array = array.reshape(4, -1)
print(array)
[[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]
[12 13 14 15]]
eigenvalues, eigenvectors = np.linalg.eig(array)
print eigenvalues
[ 3.24642492e+01 -2.46424920e+00 1.92979794e-15 -4.09576009e-16]
print eigenvectors
[[-0.11417645 -0.7327781 0.54500164 0.00135151]
[-0.3300046 -0.28974835 -0.68602671 0.40644504]
[-0.54583275 0.15328139 -0.2629515 -0.8169446 ]
[-0.76166089 0.59631113 0.40397657 0.40914805]]
inverseEigenVectors = np.linalg.inv(eigenvectors) #M^(-1)
diagonal= inverseEigenVectors.dot(array).dot(eigenvectors) #M^(-1).A.M
print(diagonal)
[[ 3.24642492e+01 -1.06581410e-14 5.32907052e-15 0.00000000e+00]
[ 7.54951657e-15 -2.46424920e+00 -1.72084569e-15 -2.22044605e-16]
[ -2.80737213e-15 1.46768503e-15 2.33547852e-16 7.25592561e-16]
[ -6.22319863e-15 -9.69656080e-16 -1.38050658e-30 1.97215226e-31]]
the final 'diagonal' matrix should be a diagonal matrix with EigenValues on the main diagonal and zeros elsewhere. but it's not... the two first main diagonal values ARE eigenvalues but the two second aren't (although just like the two second eigenvalues, they are nearly zero).
and by the way a number like $-1.06581410e-14$ is literally zero so how can I make numpy show them as zero?
What am I doing wrong?
Thanks...
Just round the final result to the desired digits :
print(diagonal.round(5))
array([[ 32.46425, 0. , 0. , 0. ],
[ 0. , -2.46425, 0. , 0. ],
[ 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. ]])
Don't confuse precision of computation and printing policies.
>>> diagonal[np.abs(diagonal)<0.0000000001]=0
>>> print diagonal
[[ 32.4642492 0. 0. 0. ]
[ 0. -2.4642492 0. 0. ]
[ 0. 0. 0. 0. ]
[ 0. 0. 0. 0. ]]
>>>
This question already has answers here:
How do I print the full NumPy array, without truncation?
(22 answers)
Closed 9 years ago.
I exported a matrix of 127x127 values as a txt, but the output appears as
answer:[[ 44. 1. 0. ..., 12. 13. 2.]
[ 51. 7. 0. ..., 5. 14. 4.]
[ 0. 1. 4. ..., 0. 0. 1.]
...,
[ 22. 110. 70. ..., 5. 0. 0.]
[ 12. 36. 12. ..., 0. 0. 2.]
[ 0. 0. 0. ..., 24. 177. 53.]]
I need access to all values, as input on Support Vector Machines
Thank You
Use numpy.set_printoptions to change the number of items returned when the array is printed.
>>> import numpy as np
>>> a = np.arange(127*127).reshape(127, 127)
>>> np.set_printoptions(edgeitems=127)
>>> print a
As expected this will flood your screen.
I'm guessing you're using numpy. If that's the case, I suggest you use the savetxt() function (http://docs.scipy.org/doc/numpy/reference/generated/numpy.savetxt.html).