I have a numpy array consisting of
[1,3,8,6,0,2,4,5,9,7]
This array is a random array consisting of 10 numbers 0-9.
I also have a 2D numpy array, a 10X10 2D numpy array with numerical values.
I would like to use my 1D numpy array (above) to access specific instances in my 2D numpy array, by looping through the 1D array
Loop 1: takes in 1 and 3, and finds the value at [1:3] in my 2D numpy array.
Loop 2: takes in 3 and 8, and finds the value at [3:8] in my 2D numpy array.
.
Loop 10: takes in 7 and 1, and finds the value at [7:1] in my 2D numpy array.
I would like to add up these values in my 2D numpy array.
so far I have :
array=[1,3,8,6,0,2,4,5,9,7]
values =0
for i in range (0, len(array)): #this is 10
a=array2[i,array[i]+1] #array2 is the 2D numpy array with the values
values=values+a
This works to some degree but how to I get it to access the last element to the first? i.e. find [7,1]
You can use simple slicing to make this work.
arr = np.random.randint(0, 10, (10,10))
pos = np.array([1,3,8,6,0,2,4,5,9,7])
pos = np.append(pos, pos[0])
rows = pos[0:-1]
cols = pos[1:]
result = sum(arr[rows, cols])
You can do the slicing twice to make it work.
values = 0
for i in range(len(array)):
a = Matrix[array[i],array[i+1]]
values += a
Also, the array you put has 11 elements which means the 10-th loop will not be what you intended.
I'm not sure I fully understood what you were trying to achieve but...
What about something like this?
a = np.array([1,3,8,6,0,9,2,4,5,9,7])
b = np.array(range(100)).reshape(10,10)
for i in range (len(a)):
print (a[i%len(a)],a[(i+1)%len(a)])
print (b[a[i%len(a)],a[(i+1)%len(a)]])
I removed 10 from the a array to avoid an index out of range error.
I also took the value [x,y] (and not the range [x:y] from the 2D array.
Related
I have a 3d array of format given below.
The below is the one sample of the 3D array, like it , it contain more than 1000.
sample
shape of the 3D array is (1000 x 10 x 5)
The image contain one element (10 x 5)
I want to change the value to 0 after the 3rd one on the last value
check the figure below
desired
I want to change like it for all the 1000 elements in my array.
Is there a better way to do it other than using "for loop" ?
import numpy as np
# Your array here:
arr = np.arange(50000).reshape(1000, 10, 5)
# Solution:
arr[:, 3:, -1] = 0
I have a 5D numpy array containing values, and would like to obtain a subarray with one less dimension, where the values have been selected based on a 3D array that contains indices of the forth dimension of the first array. E.g., I have the following arrays:
values = np.random.randn(3,4,5,10,2)
indices = np.random.randint(0,values.shape[3],size=values.shape[:3])
I found one solution, but find it rather complicated, and would prefer a one-liner:
x = np.arange(values.shape[0])
y = np.arange(values.shape[1])
z = np.arange(values.shape[2])
result = values[x[:,None,None],y[None,:,None],z[None, None,:],indices,:]
Is there any better solution to get this array?
You can try the following:
indices = indices[..., None, None]
result = np.take_along_axis(values, indices, axis=3).squeeze(axis=3)
i have got a 3d numpy array, what is the best way to iterate over the third dimension in a for loop returning the 2d array of the current interation?
One way would be just by iterating with range from 0 to the length of the array:
# arr = 3d numpy array
for i in range(len(arr)):
print(arr[i])
just loop with its third dim:
import numpy as np
a = np.arange(24).reshape((2,3,4))
for i in range(a.shape[2]): # index 2 is for 3rd dimension
print(a[:, :, i])
# or
print(a[..., i])
then you got it.
but using loop with numpy array is costly, you should get used to broadcasting, vectorization, indexing...
I am trying to compare a 1D array element-wise to a 2D array, and returns the elements of the 2D array which fulfils the condition in a 2D array form without using a for loop. Preferably using numpy or quicker method.
a = range(1,10)
Tna = np.random.choice(a, size=[250,10,1000], replace=True)
sum_Ta = np.sum(Tna, axis = 1)
percent = np.percentile(sum_Ta, 5, axis =0)
Now I would like to get a 2D array which contains the elements of sum_Ta if the elements are smaller the percent. Such that 250 elements of sum_Ta are comparing with 1 element of percent for 1000 times. Originally I can do, ES = sum_Ta[sum_Ta < percent[:,None]], but it only gives me a 1D array, not a 2D array.
Assuming you mean that for each row, you want the element of the row to be included if it is less than the percentage associated with its column.
Try the following:
mask = sum_Ta < (percent * np.ones((250,1)))
ES = np.zeros((250, 1000))
ES[mask] = sum_Ta[mask]
In numpy I have a 3d array and I would ike to remove some of the 2d subarrays. Think about it like this:
r = range(27)
arr = np.reshape(r, (3,3,3))
del = [[0,1,2],[0,0,2]]
flatSeam = np.ravel_multi_index(del, arr.shape)
arr = np.delete(arr, flatSeam)
So at the end I would like to have an array of the shape (3,2,3) without the elements 00, 10, 22 from the original array. My problem is that I acn not use ravel_multi_index for this, because my indices are 2d and the array shape is 3d, so the wrong indices are calculated (the code above also does not execute because the indices array and the shape have to be the same size).
Do you have any ideas how I can achieve this?
Here's an approach using advanced-indexing -
# arr: Input array, rm_idx : 2-row list/array of indices to be removed
m,n,p = arr.shape
mask = np.asarray(rm_idx[1])[:,None] != np.arange(n)
out = arr[np.arange(m)[:,None],np.where(mask)[1].reshape(m,-1)]
Alternatively, with boolean-indexing -
out = arr.reshape(-1,p)[mask.ravel()].reshape(m,-1,p)
A bit less memory-intensive approach as we try to avoid creating 2D mask -
vmask = ~np.in1d(np.arange(m*n),rm_idx[1] + n*np.arange(m))
out = arr.reshape(-1,p)[vmask].reshape(m,-1,p)