I have a multidimensional numpy array that has the shape (5, 6192, 1) so essentially 5 arrays of length 6192 into one array.
How could I add the elements of all the arrays into one array of length 6192 in the following way.
For example if the 5 arrays look like
ar1 = [1,2,3...]
ar2 = [1,2,3...]
ar3 = [1,2,3...]
ar4 = [1,2,3...]
ar5 = [1,2,3...]
I want my final array to look like:
ar = [5,10,15,...]
So for each inner array, add the values of each same position into a new value for the final array that is the sum of all the values in this position.
The shape should be, I guess shape(1,6192,1).
IIUC, simply use numpy.sum:
ar1 = [1,2,3]
ar2 = [1,2,3]
ar3 = [1,2,3]
ar4 = [1,2,3]
ar5 = [1,2,3]
arrays = [ar1, ar2, ar3, ar4, ar5]
ar = np.sum(arrays, axis=0)
output: array([ 5, 10, 15])
If really the shapes you describe are correct:
arr = np.array(arrays).reshape((5, 3, 1))
print(arr.shape)
# (5, 3, 1)
ar = np.sum(arr, axis=0)[None,:]
print(ar.shape)
# (1, 3, 1)
Related
The code below allows me to add a vector to each row of a given matrix using Numpy:
import numpy as np
m = np.array([[1,2,3], [4,5,6], [7,8,9], [10, 11, 12]])
v = np.array([1, 1, 0])
print("Original vector:")
print(v)
print("Original matrix:")
print(m)
result = np.empty_like(m)
for i in range(4):
result[i, :] = m[i, :] + v
print("\nAfter adding the vector v to each row of the matrix m:")
print(result)
How do I perform a similar addition operation, but going column by column?
I have tried the following:
import numpy as np
array1 = np.array([[5,5,3],[2,2,3]])
print(array1)
addition = np.array([[1],[1]])
print(addition)
for i in range(3):
array1[:,i] = array1[:,i] + addition
print(array1)
However, I get the following broadcasting error:
ValueError: could not broadcast input array from shape (2,2) into shape (2)
Just match the number of dimensions, numpy will broadcast the arrays as needed. In the first example, it should be:
result = m + v.reshape((1, -1))
In the second example, the addition is already 2D so it will be just:
array1 + addition
You can alternatively, add a dimension via Numpy None syntax and then do the addition:
array1 += addition[:,None]
I have an empty list: x = [].
I have a numpy array, y, of shape: (180, 161). I can't necessarily define x to be an np.empty of a particular shape, because I won't know the shape of y ahead of time.
I want to append y to x so that x will have a .shape of (1, 180, 161).
Then if I append more, I want it to be (n, 180, 161)
I tried .append and .stack, but I've had a variety of errors:
TypeError: only size-1 arrays can be converted to Python scalars
ValueError: all the input arrays must have same number of dimensions, but the array at index 0 has 3 dimension(s) and the array at index 1 has 2 dimension(s)
And so on. It seems that this should be simple, but it's strangely difficult.
Assuming all items in x have the same shape, you can first construct a list and then construct the NumPy array from the list.
There, you have two options:
np.array() which is faster but not flexible
np.stack() which is slower but allows you to choose over which axis should the stack happen (it is roughly equivalent to np.array().transpose(...).copy()
The code would look like:
import numpy as np
n = 100
x = [np.random.randint(0, 10, (10, 20)) for _ in range(n)]
# same as: y = np.stack(x, 0)
y = np.array(x)
print(y.shape)
# (100, 10, 20)
Of course this line:
x = [np.random.randint(0, 10, (10, 20)) for _ in range(n)]
can be replaced with:
x = []
for _ in range(n):
x.append(np.random.randint(0, 10, (10, 20)))
You could also use np.append(), e.g.:
def stacker(arrs):
result = arrs[0][None, ...]
for arr in arrs[1:]:
result = np.append(result, arr[None, ...], 0)
return result
but with horrific performances:
n = 1000
shape = (100, 100)
x = [np.random.randint(0, n, shape) for _ in range(n)]
%timeit np.array(x)
# 10 loops, best of 3: 21.1 ms per loop
%timeit np.stack(x)
# 10 loops, best of 3: 21.6 ms per loop
%timeit stacker(x)
# 1 loop, best of 3: 11 s per loop
and, as you can see, performance-wise, the list-based method is way faster.
You can reshape y to be (1, *y.shape).
Then for appending an array you can say:
y_1 = np.vstack((y, new_arr))
where y_1.shape produces a (2, *y.shape) numpy array.
To save memory you can say y = np.vstack((y, new_arr))
You might have to reshape your array to (1, *y.shape) however.
This is a very basic example:
import numpy as np
a = np.ones((1,2,3))
b = np.ones((1,2,3))
np.vstack((a,b)).shape # (2,2,3)
Let me know if this helps!
If you keep x as a list then if you just want to maintain the shape by appending, it is possible:
>>> import numpy as np
>>> x = []
>>> y = np.arange(12).reshape(3,4)
>>> x.append(y)
>>> np.shape(x)
(1, 3, 4)
>>> x.append(y)
>>> np.shape(x)
(2, 3, 4)
>>> for i in range(10):
... x.append(y)
>>> np.shape(x)
(12, 3, 4)
But considering you are dealing with np.arrays it may not be convenient for you to keep x as list, so you may try this:
>>> x = np.array(x)
>>> x.shape
(12, 3, 4)
>>> y[None,...].shape
(1, 3, 4)
>>> np.append(x, y[None,...],axis=0).shape
(13, 3, 4)
Word of caution:
As pointed out by #hpaulj :
np.append should be avoided, as it is extremely slow, probably only faster than:
x = np.array([*x, y])
The correct usage would be:
x = np.concatenate([x, y[None,...]], axis=0)
Either way, concatenating or appending is generally a speed bump in numpy. So unless you absolutely need to create an array this way, you should work with lists. Also most functions applied to np.arrays work on lists as well. Note, functions applied to arrays, not methods of an np.array object. For example:
>>> x = list((1, 2, 3, 4))
>>> np.shape(x)
(4,)
>>> x.shape
Traceback (most recent call last):
File "<ipython-input-100-9f2b259887ef>", line 1, in <module>
x.shape
AttributeError: 'list' object has no attribute 'shape'
So I would suggest appending to list, and then after you have done appending all the arrays, convert the list to np.array if you require.
Suppose I have a 5x10x3 array, which I interpret as 5 'sub-arrays', each consisting of 10 rows and 3 columns. I also have a seperate 1D array of length 5, which I call b.
I am trying to insert a new column into each sub-array, where the column inserted into the ith (i=0,1,2,3,4) sub-array is a 10x1 vector where each element is equal to b[i].
For example:
import numpy as np
np.random.seed(777)
A = np.random.rand(5,10,3)
b = np.array([2,4,6,8,10])
A[0] should look like:
A[1] should look like:
And similarly for the other 'sub-arrays'.
(Notice b[0]=2 and b[1]=4)
What about this?
# Make an array B with the same dimensions than A
B = np.tile(b, (1, 10, 1)).transpose(2, 1, 0) # shape: (5, 10, 1)
# Concatenate both
np.concatenate([A, B], axis=-1) # shape: (5, 10, 4)
One method would be np.pad:
np.pad(A, ((0,0),(0,0),(0,1)), 'constant', constant_values=[[[],[]],[[],[]],[[],b[:, None,None]]])
# array([[[9.36513084e-01, 5.33199169e-01, 1.66763960e-02, 2.00000000e+00],
# [9.79060284e-02, 2.17614285e-02, 4.72452812e-01, 2.00000000e+00],
# etc.
Or (more typing but probably faster):
i,j,k = A.shape
res = np.empty((i,j,k+1), np.result_type(A, b))
res[...,:-1] = A
res[...,-1] = b[:, None]
Or dstack after broadcast_to:
np.dstack([A,np.broadcast_to(b[:,None],A.shape[:2])]
In Python, I have a list and a numpy array.
I would like to multiply the array by the list in such a way that I get an array where the 3rd dimension represents the input array multiplied by each element of the list. Therefore:
in_list = [2,4,6]
in_array = np.random.rand(5,5)
result = ...
np.shape(result) ---> (3,5,5)
where (0,:,:) is the input array multiplied by the first element of the list (2);
(1,:,:) is the input array multiplied by the second element of the list (4), etc.
I have a feeling this question will be answered by broadcasting, but I'm not sure how to go around doing this.
You want np.multiply.outer. The outer method is defined for any NumPy "ufunc", including multiplication. Here's a demonstration:
In [1]: import numpy as np
In [2]: in_list = [2, 4, 6]
In [3]: in_array = np.random.rand(5, 5)
In [4]: result = np.multiply.outer(in_list, in_array)
In [5]: result.shape
Out[5]: (3, 5, 5)
In [6]: (result[1, :, :] == in_list[1] * in_array).all()
Out[6]: True
As you suggest, broadcasting gives an alternative solution: if you convert in_list to a 1d NumPy array of length 3, you can then reshape to an array of shape (3, 1, 1), and then a multiplication with in_array will broadcast appropriately:
In [9]: result2 = np.array(in_list)[:, None, None] * in_array
In [10]: result2.shape
Out[10]: (3, 5, 5)
In [11]: (result2[1, :, :] == in_list[1] * in_array).all()
Out[11]: True
I have a list with numpy.ndarrays - each of shape (33,1,8,45,3)
Problem that when i concatenate the list using a = np.concatenate(list)
The output shape of a becomes
print a.shape
(726,1,8,45,3)
instead of shape (22,33,1,8,45,3).
How do I cleanly concatenate the list, without having to change the input.
You can use numpy.array() or numpy.stack():
import numpy
a = [numpy.random.rand(33,1,8,45,3) for i in range(22)]
b = numpy.array(a)
b.shape # (22, 33, 1, 8, 45, 3)
c = numpy.stack(a, axis=0)
c.shape # (22, 33, 1, 8, 45, 3)
np.concatenate:
Join a sequence of arrays along an existing axis.
np.stack:
Stack a sequence of arrays along a new axis.
a = np.ones((3, 4))
b = np.stack([a, a])
print(b.shape) # (2, 3, 4)