We Keep Coding

Generating an array of arrays in Python

I want to multiply each element of B to the whole array A to obtain P. The current and desired outputs are attached. The desired output is basically an array consisting of 2 arrays since there are two elements in B.
import numpy as np
A=np.array([[1, 2, 3],
[4, 5, 6],
[7 , 8, 9]])
t = np.linspace(0,1,2)
B = 0.02109*np.exp(-t)
P=B*A
print(P)
It currently produces an error:
ValueError: operands could not be broadcast together with shapes (2,) (3,3)
The desired output is
array(([[0.02109, 0.04218, 0.06327],
[0.08436, 0.10545, 0.12654],
[0.14763, 0.16872, 0.18981]]),
([[0.00775858, 0.01551716, 0.02327574],
[0.03103432, 0.0387929 , 0.04655148],
[0.05431006, 0.06206864, 0.06982722]]))

You can do this by:
B.reshape(-1, 1, 1) * A
or
B[:, None, None] * A
where -1 or : refer to B.shape[0] which was 2 and 1, 1 or None, None add two additional dimensions to B to get the desired result shape which was (2, 3, 3).

The easiest way i can think of is using list comprehension and then casting back to numpy.ndarray
np.asarray([A*i for i in B])
Answer :
array([[[0.02109 , 0.04218 , 0.06327 ],
[0.08436 , 0.10545 , 0.12654 ],
[0.14763 , 0.16872 , 0.18981 ]],
[[0.00775858, 0.01551715, 0.02327573],
[0.03103431, 0.03879289, 0.04655146],
[0.05431004, 0.06206862, 0.0698272 ]]])

There are many possible ways for this:
Here is an overview on their runtime for the given array (bare in mind these will change for bigger arrays):
reshape: 0.000174 sec
tensordot: 0.000550 sec
einsum: 0.000196 sec
manual loop: 0.000326 sec
See the implementation for each of these:
numpy reshape
Find documentation here:
Link
Gives a new shape to an array without changing its data.
Here we reshape the array B so we can later multiply it:
import numpy as np
A=np.array([[1, 2, 3],
[4, 5, 6],
[7 , 8, 9]])
t = np.linspace(0,1,2)
B = 0.02109*np.exp(-t)
P = B.reshape(-1, 1, 1) * A
print(P)
numpy tensordot
Find documentation here:
Link
Given two tensors, a and b, and an array_like object containing two
array_like objects, (a_axes, b_axes), sum the products of a’s and b’s
elements (components) over the axes specified by a_axes and b_axes.
The third argument can be a single non-negative integer_like scalar,
N; if it is such, then the last N dimensions of a and the first N
dimensions of b are summed over.
import numpy as np
A=np.array([[1, 2, 3],
[4, 5, 6],
[7 , 8, 9]])
t = np.linspace(0,1,2)
B = 0.02109*np.exp(-t)
P = np.tensordot(B, A, 0)
print(P)
numpy einsum (Einstein summation)
Find documentation here:
Link
import numpy as np
A=np.array([[1, 2, 3],
[4, 5, 6],
[7 , 8, 9]])
t = np.linspace(0,1,2)
B = 0.02109*np.exp(-t)
P = np.einsum('ij,k', A, B)
print(P)
Note: A has two dimensions, we assign ij for their indexes. B has one dimension, we assign k to its index
manual loop
Another simple approach would be a loop (is faster than tensordot for the given input). This approach could be made "numpy free" if you dont want to use numpy for some reason. Here is the version with numpy:
import numpy as np
A=np.array([[1, 2, 3],
[4, 5, 6],
[7 , 8, 9]])
t = np.linspace(0,1,2)
B = 0.02109*np.exp(-t)
products = []
for b in B:
products.append(b*A)
P = np.array(products)
print(P)
#or the same as one-liner: np.asarray([A * elem for elem in B])

Difference between just reshaping and reshaping and getting transpose?

I'm currently studying CS231 assignments and I've realized something confusing. When calculating gradients, when I first reshape x then get transpose I got the correct result.
x_r=x.reshape(x.shape[0],-1)
dw= x_r.T.dot(dout)
However, when I reshape directly as the X.T shape it doesn't return the correct result.
dw = x.reshape(-1,x.shape[0]).dot(dout)
Can someone explain the following question?
How does the order of getting elements with np.reshape() change?
How reshaping (N,d1,d2..dn) shaped array into N,D array differs from getting a reshaped array of (D,N) with its transpose.

While both your approaches result in arrays of same shape, there will by a difference in the order of elements due to the way numpy reads / writes elements. By default, reshape uses a C-like index order, which means the elements are read / written with the last axis index changing fastest, back to the first axis index changing slowest (taken from the documentation).
Here is an example of what that means in practice. Let's assume the following array x:
x = np.asarray([[[1, 2], [3, 4], [5, 6]], [[7, 8], [9, 10], [11, 12]]])
print(x.shape) # (2, 3, 2)
print(x)
# output
[[[ 1 2]
[ 3 4]
[ 5 6]]
[[ 7 8]
[ 9 10]
[11 12]]]
Now let's reshape this array the following two ways:
opt1 = x.reshape(x.shape[0], -1)
opt2 = x.reshape(-1, x.shape[0])
print(opt1.shape) # outptu: (2, 6)
print(opt2.shape) # output: (6, 2)
print(opt1)
# output:
[[ 1 2 3 4 5 6]
[ 7 8 9 10 11 12]]
print(opt2)
# output:
[[ 1 2]
[ 3 4]
[ 5 6]
[ 7 8]
[ 9 10]
[11 12]]
reshape first inferred the shape of the new arrays and then returned a view where it read the elements in C-like index order.
To exemplify this on opt1: since the original array x has 12 elements, it inferred that the new array opt1 must have a shape of (2, 6) (because 2*6=12). Now, reshape returns a view where:
opt1[0][0] == x[0][0][0]
opt1[0][1] == x[0][0][1]
opt1[0][2] == x[0][1][0]
opt1[0][3] == x[0][1][1]
opt1[0][4] == x[0][2][0]
opt1[0][5] == x[0][2][1]
opt1[1][0] == x[1][0][0]
...
opt1[1][5] == x[1][2][1]
So as described above, the last axis index changes fastest and the first axis index slowest. In the same way, the output for opt2 will be computed.
You can now verify that transposing the first option will result in the same shape but a different order of elements:
opt1 = opt1.T
print(opt1.shape) # output: (6, 2)
print(opt1)
# output:
[[ 1 7]
[ 2 8]
[ 3 9]
[ 4 10]
[ 5 11]
[ 6 12]]
Obviously, the two approaches do not result in the same array due to element ordering, even though they will have the same shape.

Array size difference Python

My question is about Python array shape.
What is the difference between array size (2, ) and (2, 1)?
I tried to add those two arrays together. However, I got an error as follows:
Non-broadcastable output operant with shape (2, ) doesn't match the broadcast shape (2, 2)

There is no difference in the raw memory. But logically, one is a one-dimensional array of two values, the other is a 2D array (where one of the dimensions just happens to be size 1).
The logical distinction is important to numpy; when you try to add them, it wants to make a new 2x2 array where the top row is the sum of the (2, 1) array's top "row" with each value in the (2,) array. If you use += to do that though, you're indicating that you expect to be able to modify the (2,) array in place, which is not possible without resizing (which numpy won't do). If you change your code from:
arr1 += arr2
to:
arr1 = arr1 + arr2
it will happily create a new (2, 2) array. Or if the goal was that the 2x1 array should act like a flat 1D array, you can flatten it:
alreadyflatarray += twodarray.flatten()

(2,) is an unidimensional array, (2,1) is a matrix with only one column
You can easily see the difference by crating arrays full of zeros using np.zero passing the desired shape:
>>> np.zeros((2,))
array([0., 0.])
>>> np.zeros((2,1))
array([[0.],
[0.]])

#yx131, you can have a look at the below code to just have a clear picture about tuples and it's use in defining the shape of numpy arrays.
Note: Do not forget to see the code below as it has explanation of the problems related to Broadcasting in numpy.
Also check numpy's broadcasting rule at
https://docs.scipy.org/doc/numpy/user/basics.broadcasting.html.
There's a difference between (2) and (2,). The first one is a literal value 2 where as the 2nd one is a tuple.
(2,) is 1 item tuple and (2, 2) is 2 items tuple. It is clear in the code example.
Note: In case of numpy arrays, (2,) denotes shape of 1 dimensional array of 2 items and (2, 2) denotes the shape of 2 dimensional array (matrix) with 2 rows and 2 colums. If you want to add 2 arrays then their shape should be same.
v = (2) # Assignment of value 2
t = (2,) # Comma is necessary at the end of item to define 1 item tuple, it is not required in case of list
t2 = (2, 1) # 2 item tuple
t3 = (3, 4) # 2 item tuple
print(v, type(v))
print(t, type(t))
print(t2, type(t2))
print(t3, type(t3))
print(t + t2)
print(t2 + t3)
"""
2 <class 'int'>
(2,) <class 'tuple'>
(2, 1) <class 'tuple'>
(3, 4) <class 'tuple'>
(2, 2, 1)
(2, 1, 3, 4)
"""
Now, let's have a look at the below code to figure out the error related to broadcasting. It's all related to dimensions.
# Python 3.5.2
import numpy as np
arr1 = np.array([1, 4]);
arr2 = np.array([7, 6, 3, 8]);
arr3 = np.array([3, 6, 2, 1]);
print(arr1, ':', arr1.shape)
print(arr2, ":", arr2.shape)
print(arr3, ":", arr3.shape)
print ("\n")
"""
[1 4] : (2,)
[7 6 3 8] : (4,)
[3 6 2 1] : (4,)
"""
# Changing shapes (dimensions)
arr1.shape = (2, 1)
arr2.shape = (2, 2)
arr3.shape = (2, 2)
print(arr1, ':', arr1.shape)
print(arr2, ":", arr2.shape)
print(arr3, ":", arr3.shape)
print("\n")
print(arr1 + arr2)
"""
[[1]
[4]] : (2, 1)
[[7 6]
[3 8]] : (2, 2)
[[3 6]
[2 1]] : (2, 2)
[[ 8 7]
[ 7 12]]
"""
arr1.shape = (2, )
print(arr1, arr1.shape)
print(arr1 + arr2)
"""
[1 4] (2,)
[[ 8 10]
[ 4 12]]
"""
# Code with error(Broadcasting related)
arr2.shape = (4,)
print(arr1+arr2)
"""
Traceback (most recent call last):
File "source_file.py", line 53, in <module>
print(arr1+arr2)
ValueError: operands could not be broadcast together
with shapes (2,) (4,)
"""
So in your case, the problem is related to the mismatched dimensions (acc. to numpy's broadcasting ) to be added. Thanks.

Make an array that has shape (2,)
In [164]: a = np.array([3,6])
In [165]: a
Out[165]: array([3, 6])
In [166]: a.shape
Out[166]: (2,)
In [167]: a.reshape(2,1)
Out[167]:
array([[3],
[6]])
In [168]: a.reshape(1,2)
Out[168]: array([[3, 6]])
The first displays like a simple list [3,6]. The second as a list with 2 nested lists. The third as a list with one nested list of 2 items. So there is a consistent relation between shape and list nesting.
In [169]: a + a
Out[169]: array([ 6, 12]) # shape (2,)
In [170]: a + a.reshape(1,2)
Out[170]: array([[ 6, 12]]) # shape (1,2)
In [171]: a + a.reshape(2,1)
Out[171]:
array([[ 6, 9], # shape (2,2)
[ 9, 12]])
Dimensions behave as:
(2,) + (2,) => (2,)
(2,) + (1,2) => (1,2) + (1,2) => (1,2)
(2,) + (2,1) => (1,2) + (2,1) => (2,2) + (2,2) => (2,2)
That is a lower dimensional array can be expanded to the matching number of dimensions with the addition of leading size 1 dimensions.
And size 1 dimensions can be changed to match the corresponding dimension.
I suspect you got the error when doing a a += ... (If so you should have stated that clearly.)
In [172]: a += a
In [173]: a += a.reshape(1,2)
....
ValueError: non-broadcastable output operand with shape (2,)
doesn't match the broadcast shape (1,2)
In [175]: a += a.reshape(2,1)
...
ValueError: non-broadcastable output operand with shape (2,)
doesn't match the broadcast shape (2,2)
With the a+=... addition, the result shape is fixed at (2,), the shape of a. But as noted above the two additions generate (1,2) and (2,2) results, which aren't compatible with (2,).
The same reasoning can explain these additions and errors:
In [176]: a1 = a.reshape(1,2)
In [177]: a1 += a
In [178]: a1
Out[178]: array([[12, 24]])
In [179]: a2 = a.reshape(2,1)
In [180]: a2 += a
...
ValueError: non-broadcastable output operand with shape (2,1)
doesn't match the broadcast shape (2,2)
In [182]: a1 += a2
...
ValueError: non-broadcastable output operand with shape (1,2)
doesn't match the broadcast shape (2,2)

Numpy 3D array to 2D row major array

I have a numpy array with dimensions (28, 28, 60000), containing 60000 28x28 images, represented as pixel brightness. I'm trying to transform it so that I have a 60000 x 784 array, with the 784 representing the original 28x28 image in row major format. How do I do this? I'm assuming I use numpy.reshape, but I'm unsure as to how it rearranges things. Example:
[[1,2], [[1,2,3,4],
[3,4]] [5,6,7,8]]
... ->
[[5,6],
[7,8]]

This code:
import numpy
a = numpy.array([[[1,2],[3,4]],[[5,6],[7,8]]])
print(numpy.reshape(a, (2,4)))
Returns:
[[1 2 3 4]
[5 6 7 8]]

Try to experiment with something like this:
a = np.arange(6).reshape((3, 2))
b = np.reshape(a, (1, 6))
print a
print b
a =
[[0 1]
[2 3]
[4 5]]
b = [[0 1 2 3 4 5]]

Convert a numpy array to an array of numpy arrays

How can I convert numpy array a to numpy array b in a (num)pythonic way. Solution should ideally work for arbitrary dimensions and array lengths.
import numpy as np
a=np.arange(12).reshape(2,3,2)
b=np.empty((2,3),dtype=object)
b[0,0]=np.array([0,1])
b[0,1]=np.array([2,3])
b[0,2]=np.array([4,5])
b[1,0]=np.array([6,7])
b[1,1]=np.array([8,9])
b[1,2]=np.array([10,11])

For a start:
In [638]: a=np.arange(12).reshape(2,3,2)
In [639]: b=np.empty((2,3),dtype=object)
In [640]: for index in np.ndindex(b.shape):
b[index]=a[index]
.....:
In [641]: b
Out[641]:
array([[array([0, 1]), array([2, 3]), array([4, 5])],
[array([6, 7]), array([8, 9]), array([10, 11])]], dtype=object)
It's not ideal since it uses iteration. But I wonder whether it is even possible to access the elements of b in any other way. By using dtype=object you break the basic vectorization that numpy is known for. b is essentially a list with numpy multiarray shape overlay. dtype=object puts an impenetrable wall around those size 2 arrays.
For example, a[:,:,0] gives me all the even numbers, in a (2,3) array. I can't get those numbers from b with just indexing. I have to use iteration:
[b[index][0] for index in np.ndindex(b.shape)]
# [0, 2, 4, 6, 8, 10]
np.array tries to make the highest dimension array that it can, given the regularity of the data. To fool it into making an array of objects, we have to give an irregular list of lists or objects. For example we could:
mylist = list(a.reshape(-1,2)) # list of arrays
mylist.append([]) # make the list irregular
b = np.array(mylist) # array of objects
b = b[:-1].reshape(2,3) # cleanup
The last solution suggests that my first one can be cleaned up a bit:
b = np.empty((6,),dtype=object)
b[:] = list(a.reshape(-1,2))
b = b.reshape(2,3)
I suspect that under the covers, the list() call does an iteration like
[x for x in a.reshape(-1,2)]
So time wise it might not be much different from the ndindex time.
One thing that I wasn't expecting about b is that I can do math on it, with nearly the same generality as on a:
b-10
b += 10
b *= 2
An alternative to an object dtype would be a structured dtype, e.g.
In [785]: b1=np.zeros((2,3),dtype=[('f0',int,(2,))])
In [786]: b1['f0'][:]=a
In [787]: b1
Out[787]:
array([[([0, 1],), ([2, 3],), ([4, 5],)],
[([6, 7],), ([8, 9],), ([10, 11],)]],
dtype=[('f0', '<i4', (2,))])
In [788]: b1['f0']
Out[788]:
array([[[ 0, 1],
[ 2, 3],
[ 4, 5]],
[[ 6, 7],
[ 8, 9],
[10, 11]]])
In [789]: b1[1,1]['f0']
Out[789]: array([8, 9])
And b and b1 can be added: b+b1 (producing an object dtype). Curiouser and curiouser!

Based on hpaulj I provide a litte more generic solution. a is an array of dimension N which shall be converted to an array b of dimension N1 with dtype object holding arrays of dimension (N-N1).
In the example N equals 5 and N1 equals 3.
import numpy as np
N=5
N1=3
#create array a with dimension N
a=np.random.random(np.random.randint(2,20,size=N))
a_shape=a.shape
b_shape=a_shape[:N1] # shape of array b
b_arr_shape=a_shape[N1:] # shape of arrays in b
#Solution 1 with list() method (faster)
b=np.empty(np.prod(b_shape),dtype=object) #init b
b[:]=list(a.reshape((-1,)+b_arr_shape))
b=b.reshape(b_shape)
print "Dimension of b: {}".format(len(b.shape)) # dim of b
print "Dimension of array in b: {}".format(len(b[0,0,0].shape)) # dim of arrays in b
#Solution 2 with ndindex loop (slower)
b=np.empty(b_shape,dtype=object)
for index in np.ndindex(b_shape):
b[index]=a[index]
print "Dimension of b: {}".format(len(b.shape)) # dim of b
print "Dimension of array in b: {}".format(len(b[0,0,0].shape)) # dim of arrays in b