I have a list of 32 numpy arrays, each of which has shape (n, 108, 108, 2), where n is different in each array. I want to stack all of them to create a numpy array of shape (32, m, 108, 108, 2), where m is the maximum among the ns, and the shorter arrays are padded with zeros.
How do I do this?
I asked something similar yesterday, but the answers there seem to break when using deep arrays like in my case.
Concretely, I went with this solution in the end, which produced the cleanest code:
data = np.column_stack(zip_longest(*data, fillvalue=0))
But now it is throwing this error:
ValueError: setting an array element with a sequence.
I have found a godly answer in this webpage.
The pad_sequences function is exactly what I needed.
from tensorflow.python.keras.preprocessing.sequence import pad_sequences
result = pad_sequences(imgs, padding='post')
In my case I needed to stack images with different width and padded with zeros to the left side.
for me this works well:
np.random.seed(42)
image_batch = []
for i in np.random.randint(50,500,size=10):
image_batch.append(np.random.randn(32,i))
for im in image_batch:
print(im.shape)
output: (32, 152)
(32, 485)
(32, 398)
(32, 320)
(32, 156)
(32, 121)
(32, 238)
(32, 70)
(32, 152)
(32, 171)
def stack_images_rows_with_pad(list_of_images):
func = lambda x: np.array(list(zip_longest(*x, fillvalue=0))) # applied row wise
return np.array(list(map(func, zip(*list_of_images)))).transpose(2,0,1)
res = stack_images_rows_with_pad(image_batch)
for im in rez:
print(im.shape)
output: (32, 485)
(32, 485)
(32, 485)
(32, 485)
(32, 485)
(32, 485)
(32, 485)
(32, 485)
(32, 485)
(32, 485)
Try this:
# Create matrices with random first axis length.
depth = np.random.randint(3,20,size=32)
l = []
lmax = 0
for i in depth:
l.append(np.ones((i,10,10,2)))
lmax = i if i > lmax else lmax
# Join the matrices:
new_l = []
for m in l:
new_l.append(np.vstack([m, np.zeros((lmax-m.shape[0], 10, 10, 2))]))
master = np.stack(new_l, axis=0)
master.shape
>>> (32, 19, 10, 10, 2)
I find np.pad almost impossible to work with on higher dimensional matrix - luckily, what you asked was simple, where only one of the dimension will have to extended, such that it's easy to use np.vstack to stack a zeros array that make it conform to a new shape.
A = np.ones((4,3))
border_top_bottom = np.zeros((A.shape[1])).reshape(1,A.shape[1])
print(np.vstack([border_top_bottom,A,border_top_bottom]))
temp = np.vstack([border_top_bottom,A,border_top_bottom])
border_right_left = np.zeros((temp.shape[0])).reshape(temp.shape[0],1)
print(np.hstack([np.hstack([border_right_left,temp,border_right_left])]))
Related
I have a 2D array of shape (10, 3) and an image represented as a 3D array of shape (480, 640, 3). I'd like to perform a difference between each pixel and each element of the 2D array, to get a final result of shape (10, 480, 640, 3).
For now, my code looks like this:
arr_2d = np.random.rand(10, 3)
arr_3d = np.random.rand(480, 640, 3)
res = np.ones_like(arr_3d)
res = np.tile(res, (10, 1, 1, 1))
for i in range(10):
res[i] = arr_3d - arr_2d[i]
My question is if there's a way to do this without the for loop, only using numpy operations.
You can try broadcasting with np.array like this
arr_2d = arr_2d.reshape(-1,1,1,3)
arr_3d = arr_3d.reshape((-1,*arr_3d.shape))
res = arr_3d - arr_2d
This should give the same result as your original code
I have a numpy array np_arr with a shape as such
(3787, 256, 256)
I want to sample 20 random arrays from the existing array such that
(20, 256, 256)
I tried
import random
new_array = random.sample(np_arr,20)
but that didn't work.
How do I go about it
You can use np.random.choiceto randomly choose indices of the first dimension, then index your array with it:
import numpy as np
# Generating random data
rand_arr = np.random.rand(3787, 256, 256)
rand_idx = np.random.choice(rand_arr.shape[0], 20)
rand_arr[rand_idx]
# > shape = (20, 256, 256)
# In one line:
rand_arr[np.random.choice(rand_arr.shape[0], 20)]
I have a numpy array of dimension:
(1, 1, 188, 621, 32) corresponding to the (batch, depth, height, width, features) dimensions
And I want to convert this array to have the shape (1, 48, 188, 621, 32), i.e., have a depth of 48, where the rest of the arrays in the depth dimension will be the array that's (height, width + x), for x = 1 to x = 47, distance away from the current one?
In case of overflow, the array can be full of zeros.
If you have 48 arrays of shape (1, 1, 188, 621, 32) then just
bigarr = np.concatenate([arr1, arr2, ... , arr48], axis=1)
will produce the (1,48, 188, 621, 32). The concatenation part is trivial.
Now you just need to figure out how to generate the 47 added arrays from the first (at least that seems to be what you are trying to do).
I have a numpy array v with shape (1000, 68), v is supposed to padding to 100 dimension with 0s. As a result, the v's shape will be (1000, 100)
I tried to use the following approaches:
t = np.lib.pad(v, (16, 16), 'minimum') # numpy method
t = sequence.pad_sequences(v, maxlen = 100, padding = 'post') # Keras text processing method
Above two methods returned the t with correct shape (1000, 100), but each array t[n] (n from 0 to 99) is a zero vector [0, 0, 0, ....0]
Following numpy.pad documentation, I tried
np.pad(v, [(0,0), (16,16)], 'constant')
with the expected result: 16 columns of zeros added on the left, and 16 on the right.
How can I align the 4d z array and the 4d QCLOUD array and to then find out the z value of when QCLOUD max occurs?
print(z.shape)
print(qcloud.shape)
out: (6, 100, 128, 128)
(6, 99, 128, 128)
Ignoring the fact that (np.array(z.shape) > np.array(qcloud.shape)).any(), you want argmax:
idx = np.argmax(qcloud)
result = z[tuple(idx)]