numpy array of list to 2d array - numpy

>>> print(a)
array([list([52, 584, 78]), list([35, 140, 584])], dtype=object)
I would like to convert above data to numpy 2d array with int, but when I enter np.array(a), the result was the same as before. How can I convert type?


You want to use np.stack
>>> import numpy as np
>>> a = ([1,2,3,4,5])
>>> b = ([2,3,4,5,6])
>>> c = ([3,4,5,6,7])
>>> np.array([a, b, c])
array([[1, 2, 3, 4, 5],
[2, 3, 4, 5, 6],
[3, 4, 5, 6, 7]])
>>> np.stack([a, b, c], axis=0)
array([[1, 2, 3, 4, 5],
[2, 3, 4, 5, 6],
[3, 4, 5, 6, 7]])

Related

Is there a numpy (or Python) function to correlate each columns of 2D numpy array (n,m)

I have two numpy matrices (6 rows and 3 columns) :
a = np.array([[1,2,4],[3,6,2],[3,4,7],[9,7,7],[6,3,1],[3,5,9]])
b = np.array([[4,5,2],[9,2,5],[1,5,6],[4,5,6],[1,2,6],[6,4,3]])
a = array([[1, 2, 4],
[3, 6, 2],
[3, 4, 7],
[9, 7, 7],
[6, 3, 1],
[3, 5, 9]])
b = array([[4, 5, 2],
[9, 2, 5],
[1, 5, 6],
[4, 5, 6],
[1, 2, 6],
[6, 4, 3]])
I would like to calculate the pearson correlation coefficient between the first column of a and b, the second column of a and b and the third column of a and b.
The result would be a vector of 3 (3 correlation coeff).

One way using numpy.corrcoef and diagonal:
corr = np.corrcoef(a.T, b.T).diagonal(a.shape[1])
corr
Output:
array([-0.2324843 , -0.03631365, -0.18057878])

pytorch repeat 3rd dimension

I'm following this example on doc
In [42]: x = torch.tensor([1,2,3])
In [45]: x.repeat(4,2)
Out[45]: tensor([[1, 2, 3, 1, 2, 3],
[1, 2, 3, 1, 2, 3],
[1, 2, 3, 1, 2, 3],
[1, 2, 3, 1, 2, 3]])
In [46]: x.repeat(4,2).shape
Out[46]: torch.Size([4, 6])
So far, so good.
But why does repeating just 1 time on 3rd dimension expand 3rd dim to 3 (not 1)?
[On the doc]
>>> x.repeat(4, 2, 1).size()
torch.Size([4, 2, 3])
Double checking.
In [43]: x.repeat(4,2,1)
Out[43]:
tensor([[[1, 2, 3],
[1, 2, 3]],
[[1, 2, 3],
[1, 2, 3]],
[[1, 2, 3],
[1, 2, 3]],
[[1, 2, 3],
[1, 2, 3]]])
Why does it behave this way?

It expands the size([3]) tensor it only once along first dim. The (4,2,1) is the number of times you want to repeat a (3,) tensor. So, the final tensor is (4,2,3), because you repeat the (3,) once over last axis, twice over second last and 4 times over the first axis.
x = torch.tensor([1, 2, 3])
x.shape
torch.Size([3])
Then,
xx = x.repeat(4,2,1)
xx.shape
torch.Size([4, 2, 3])

Indexing numpy array using another numpy array [duplicate]

Suppose I have a matrix A with some arbitrary values:
array([[ 2, 4, 5, 3],
[ 1, 6, 8, 9],
[ 8, 7, 0, 2]])
And a matrix B which contains indices of elements in A:
array([[0, 0, 1, 2],
[0, 3, 2, 1],
[3, 2, 1, 0]])
How do I select values from A pointed by B, i.e.:
A[B] = [[2, 2, 4, 5],
[1, 9, 8, 6],
[2, 0, 7, 8]]

EDIT: np.take_along_axis is a builtin function for this use case implemented since numpy 1.15. See #hpaulj 's answer below for how to use it.
You can use NumPy's advanced indexing -
A[np.arange(A.shape[0])[:,None],B]
One can also use linear indexing -
m,n = A.shape
out = np.take(A,B + n*np.arange(m)[:,None])
Sample run -
In [40]: A
Out[40]:
array([[2, 4, 5, 3],
[1, 6, 8, 9],
[8, 7, 0, 2]])
In [41]: B
Out[41]:
array([[0, 0, 1, 2],
[0, 3, 2, 1],
[3, 2, 1, 0]])
In [42]: A[np.arange(A.shape[0])[:,None],B]
Out[42]:
array([[2, 2, 4, 5],
[1, 9, 8, 6],
[2, 0, 7, 8]])
In [43]: m,n = A.shape
In [44]: np.take(A,B + n*np.arange(m)[:,None])
Out[44]:
array([[2, 2, 4, 5],
[1, 9, 8, 6],
[2, 0, 7, 8]])

More recent versions have added a take_along_axis function that does the job:
A = np.array([[ 2, 4, 5, 3],
[ 1, 6, 8, 9],
[ 8, 7, 0, 2]])
B = np.array([[0, 0, 1, 2],
[0, 3, 2, 1],
[3, 2, 1, 0]])
np.take_along_axis(A, B, 1)
Out[]:
array([[2, 2, 4, 5],
[1, 9, 8, 6],
[2, 0, 7, 8]])
There's also a put_along_axis.

I know this is an old question, but another way of doing it using indices is:
A[np.indices(B.shape)[0], B]
output:
[[2 2 4 5]
[1 9 8 6]
[2 0 7 8]]

Following is the solution using for loop:
outlist = []
for i in range(len(B)):
lst = []
for j in range(len(B[i])):
lst.append(A[i][B[i][j]])
outlist.append(lst)
outarray = np.asarray(outlist)
print(outarray)
Above can also be written in more succinct list comprehension form:
outlist = [ [A[i][B[i][j]] for j in range(len(B[i]))]
for i in range(len(B)) ]
outarray = np.asarray(outlist)
print(outarray)
Output:
[[2 2 4 5]
[1 9 8 6]
[2 0 7 8]]

special tiling of matrix in tensorflow or numpy

Consider 3D tensor of T(w x h x d).
The goal is to create a tensor of R(w x h x K) where K = d x k by tiling along 3rd dimension in a unique way.
The tensor should repeat each slice in 3rd dimension k times, meaning :
T[:,:,0]=R[:,:,0:k] and T[:,:,1]=R[:,:,k:2*k]
There's a subtle difference with standard tiling which gives T[:,:,0]=R[:,:,::k], repeats at every kth in 3rd dimension.

Use np.repeat along that axis -
np.repeat(T,k,axis=2)
Sample run -
In [688]: # Setup
...: w,h,d = 2,3,4
...: k = 2
...: T = np.random.randint(0,9,(w,h,d))
...:
...: # Original approach
...: R = np.zeros((w,h,d*k),dtype=T.dtype)
...: for i in range(4):
...: R[:,:,i*k:(i+1)*k] = T[:,:,i][...,None]
...:
In [692]: T
Out[692]:
array([[[4, 5, 6, 4],
[5, 4, 4, 3],
[8, 0, 0, 8]],
[[7, 3, 8, 0],
[8, 7, 0, 8],
[3, 6, 8, 5]]])
In [690]: R
Out[690]:
array([[[4, 4, 5, 5, 6, 6, 4, 4],
[5, 5, 4, 4, 4, 4, 3, 3],
[8, 8, 0, 0, 0, 0, 8, 8]],
[[7, 7, 3, 3, 8, 8, 0, 0],
[8, 8, 7, 7, 0, 0, 8, 8],
[3, 3, 6, 6, 8, 8, 5, 5]]])
In [691]: np.allclose(R, np.repeat(T,k,axis=2))
Out[691]: True
Alternatively with np.tile and reshape -
np.tile(T[...,None],k).reshape(w,h,-1)

How to simplify a numpy array indexing? [duplicate]

Suppose I have a matrix A with some arbitrary values:
array([[ 2, 4, 5, 3],
[ 1, 6, 8, 9],
[ 8, 7, 0, 2]])
And a matrix B which contains indices of elements in A:
array([[0, 0, 1, 2],
[0, 3, 2, 1],
[3, 2, 1, 0]])
How do I select values from A pointed by B, i.e.:
A[B] = [[2, 2, 4, 5],
[1, 9, 8, 6],
[2, 0, 7, 8]]

EDIT: np.take_along_axis is a builtin function for this use case implemented since numpy 1.15. See #hpaulj 's answer below for how to use it.
You can use NumPy's advanced indexing -
A[np.arange(A.shape[0])[:,None],B]
One can also use linear indexing -
m,n = A.shape
out = np.take(A,B + n*np.arange(m)[:,None])
Sample run -
In [40]: A
Out[40]:
array([[2, 4, 5, 3],
[1, 6, 8, 9],
[8, 7, 0, 2]])
In [41]: B
Out[41]:
array([[0, 0, 1, 2],
[0, 3, 2, 1],
[3, 2, 1, 0]])
In [42]: A[np.arange(A.shape[0])[:,None],B]
Out[42]:
array([[2, 2, 4, 5],
[1, 9, 8, 6],
[2, 0, 7, 8]])
In [43]: m,n = A.shape
In [44]: np.take(A,B + n*np.arange(m)[:,None])
Out[44]:
array([[2, 2, 4, 5],
[1, 9, 8, 6],
[2, 0, 7, 8]])

More recent versions have added a take_along_axis function that does the job:
A = np.array([[ 2, 4, 5, 3],
[ 1, 6, 8, 9],
[ 8, 7, 0, 2]])
B = np.array([[0, 0, 1, 2],
[0, 3, 2, 1],
[3, 2, 1, 0]])
np.take_along_axis(A, B, 1)
Out[]:
array([[2, 2, 4, 5],
[1, 9, 8, 6],
[2, 0, 7, 8]])
There's also a put_along_axis.

I know this is an old question, but another way of doing it using indices is:
A[np.indices(B.shape)[0], B]
output:
[[2 2 4 5]
[1 9 8 6]
[2 0 7 8]]

Following is the solution using for loop:
outlist = []
for i in range(len(B)):
lst = []
for j in range(len(B[i])):
lst.append(A[i][B[i][j]])
outlist.append(lst)
outarray = np.asarray(outlist)
print(outarray)
Above can also be written in more succinct list comprehension form:
outlist = [ [A[i][B[i][j]] for j in range(len(B[i]))]
for i in range(len(B)) ]
outarray = np.asarray(outlist)
print(outarray)
Output:
[[2 2 4 5]
[1 9 8 6]
[2 0 7 8]]