I'm very new to Python & Numpy and am trying to accomplish the following:
Given, 3D Array:
arr_3d = [[[1,2,3],[4,5,6],[0,0,0],[0,0,0]],
[[3,2,1],[0,0,0],[0,0,0],[0,0,0]]
[[1,2,3],[4,5,6],[7,8,9],[0,0,0]]]
arr_3d = np.array(arr_3d)
Get the indices where [0,0,0] appears in the given 3D array.
Slice the given 3D array from where [0,0,0] appears first.
In other words, I'm trying to remove the padding (In this case: [0,0,0]) from the given 3D array.
Here is what I have tried,
arr_zero = np.zeros(3)
for index in range(0, len(arr_3d)):
rows, cols = np.where(arr_3d[index] == arr_zero)
arr_3d[index] = np.array(arr_3d[0][:rows[0]])
But doing this, I keep getting the following error:
Could not broadcast input array from shape ... into shape ...
I'm expecting something like this:
[[[1,2,3],[4,5,6]],
[[3,2,1]]
[[1,2,3],[4,5,6],[7,8,9]]]
Any help would be appreciated.
Get the first occurance of those indices with all() reduction alongwith argmax() and then slice each 2D slice off the 3D array -
In [106]: idx = (arr_3d == [0,0,0]).all(-1).argmax(-1)
# Output as list of arrays
In [107]: [a[:i] for a,i in zip(arr_3d,idx)]
Out[107]:
[array([[1, 2, 3],
[4, 5, 6]]), array([[3, 2, 1]]), array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])]
# Output as list of lists
In [108]: [a[:i].tolist() for a,i in zip(arr_3d,idx)]
Out[108]: [[[1, 2, 3], [4, 5, 6]], [[3, 2, 1]], [[1, 2, 3], [4, 5, 6], [7, 8, 9]]]
Related
There is a 2-d array like this:
img = [
[[1, 2, 3], [4, 5, 6], [7, 8, 9]],
[[2, 2, 2], [3, 2, 3], [6, 7, 6]],
[[9, 8, 1], [9, 8, 3], [9, 8, 5]]
]
And i just want to get the sum of certain indices which are like this:
indices = [[0, 0], [0, 1]] # which means img[0][0] and img[0][1]
# means here is represents
There was a similar ask about 1-d array in stackoverflow in this link, but it got a error when I tried to use print(img[indices]). Because I want to make it clear that the element of img are those which indicates by indices, and then get the mean sum of it.
Expected output
[5, 7, 9]
Use NumPy:
import numpy as np
img = np.array(img)
img[tuple(indices)].sum(axis = 0)
#array([5, 7, 9])
If the result would be [5, 7, 9] which is sum over the column of the list. Then easy:
img = np.asarray(img)
indices = [[0, 0], [0, 1]]
img[(indices)].sum(axis = 0)
Result:
array([5, 7, 9])
When you supply a fancy index, each element of the index tuple represents a different axis. The shape of the index arrays broadcasts to the shape of the output you get.
In your case, the rows of indices.T are the indices in each axis. You can convert them into an index tuple and append slice(None), which is the programmatic equivalent of :. You can take the mean of the resulting 2D array directly:
img[tuple(indices.T) + (slice(None),)].sum(0)
Another way is to use the splat operator:
img[(*indices.T, slice(None))].sum(0)
When using a numpy array as a matrix, in which order are rows and columns?
For example:
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
Is [1, 2, 3] the first row or the first column?
I cannot find this information in the documentation, perhaps because the answer is too obvious.
[1, 2, 3] is the first row.
The examples in numpy ndarray documentation actually gives you some hints:
>>> x = np.array([[1, 2, 3], [4, 5, 6]], np.int32)
>>> # The element of x in the *second* row, *third* column, namely, 6.
>>> x[1, 2] ```
I tried to create a 2D numpy ndarray using the following code:
temp = np.array([[np.mean(w2v[word]) for word in docs if word in w2v] for docs in X[:5]])
temp has a shape of (5,) instead of expected (5,x).
Also temps's data structure is like: array([list([.....],...)])
It seems that the inner list is not converted to ndarray.
Your missing np.array in there, it should be:
temp = np.array([np.array([np.mean(w2v[word]) for word in docs if word in w2v] for docs in X[:5])])
Running example:
bob
Out[70]: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
tmp = np.array([np.array([x for x in Y]) for Y in bob])
tmp
Out[72]:
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
I have a numpy 2D array [[1,2,3]].
I need to append a numpy 1D array,( say [4,5,6]) to it, so that it becomes [[1,2,3], [4,5,6]]
This is easily possible using lists, where you just call append on the 2D list.
But how do you do it in Numpy arrays?
np.concatenate and np.append dont work. they convert the array to 1D for some reason.
Thanks!
You want vstack:
In [45]: a = np.array([[1,2,3]])
In [46]: l = [4,5,6]
In [47]: np.vstack([a,l])
Out[47]:
array([[1, 2, 3],
[4, 5, 6]])
You can stack multiple rows on the condition that The arrays must have the same shape along all but the first axis.
In [53]: np.vstack([a,[[4,5,6], [7,8,9]]])
Out[53]:
array([[1, 2, 3],
[4, 5, 6],
[4, 5, 6],
[7, 8, 9]])
Try this:
np.concatenate(([a],[b]),axis=0)
when
a = np.array([1,2,3])
b = np.array([4,5,6])
then result should be:
array([[1, 2, 3],
[4, 5, 6]])
I have a numpy array that looks like this
[
[[1,2,3], [4,5,6]],
[[3,8,9], [2,9,4]],
[[7,1,3], [1,3,6]]
]
I want it like this after deleting first column
[
[[2,3], [5,6]],
[[8,9], [9,4]],
[[1,3], [3,6]]
]
so currently the dimension is 3*3*3, after removing the first column it should be 3*3*2
You can slice it as so, where 1: signifies that you only want the second and all remaining columns from the inner most array (i.e. you 'delete' its first column).
>>> a[:, :, 1:]
array([[[2, 3],
[5, 6]],
[[8, 9],
[9, 4]],
[[1, 3],
[3, 6]]])
Since you are using numpy I'll mention numpy way of doing this. First of all, the dimension you have specified for the question seems wrong. See below
x = np.array([
[[1,2,3], [4,5,6]],
[[3,8,9], [2,9,4]],
[[7,1,3], [1,3,6]]
])
The shape of x is
x.shape
(3, 2, 3)
You can use numpy.delete to remove a column as shown below
a = np.delete(x, 0, 2)
a
array([[[2, 3],
[5, 6]],
[[8, 9],
[9, 4]],
[[1, 3],
[3, 6]]])
To find the shape of a
a.shape
(3, 2, 2)