Using NumPy, I want to create an n-by-2 array, by starting with an empty array, and adding on some 1-by-2 arrays.
Here's what I have tried so far:
x = np.array([1, 2])
y = np.array([3, 4])
z = np.array([])
z = np.append(z, x)
z = np.append(z, y)
However, this gives me:
z = [1, 2, 3, 4]
What I want is:
z = [[1, 2], [3, 4]]
How can I achieve this?
import numpy as np
x = np.array([1, 2])
y = np.array([3, 4])
z = np.append([x],[y], axis=0)
print(z)
>>> [[1 2]
[3 4]]
No need to create the array before appending, axis=0 will allow you to append row wise.
The previous works if z is not an array already. From then on you specify z as the original array and append the other array as such:
t = np.array([5, 6])
z = np.append(z,[t], axis=0)
print(z)
[[1 2]
[3 4]
[5 6]]
You can simply use np.array :
>>> np.array((x,y))
array([[1, 2],
[3, 4]])
Related
If I have a 2D numpy array A:
[[6 9 6]
[1 1 2]
[8 7 3]]
And I have access to array [1 1 2]. Clearly, [1 1 2] belongs to index 1 of array A. But how do I do this?
Access the second row using the following operator:
import numpy as np
a = np.array([[6, 9, 6],
[1, 1, 2],
[8, 7, 3]])
row = [1, 1, 2]
i = np.where(np.all(a==row, axis=1))
print(i[0][0])
np.where will return a tuple of indices (lists), which is why you need to use the operators [0][0] consecutively in order to obtain an int.
One option:
a = np.array([[6, 9, 6],
[1, 1, 2],
[8, 7, 3]])
b = np.array([1, 1, 2])
np.nonzero((a == b).all(1))[0]
output: [1]
arr1 = [[6,9,6],[1,1,2],[8,7,3]]
ind = arr1.index([1,1,2])
Output:
ind = 1
EDIT for 2D np.array:
arr1 = np.array([[6,9,6],[1,1,2],[8,7,3]])
ind = [l for l in range(len(arr1)) if (arr1[l,:] == np.array([1,1,2])).all()]
import numpy as np
a = np.array([[6, 9, 6],
[1, 1, 2],
[8, 7, 3]])
b = np.array([1, 1, 2])
[x for x,y in enumerate(a) if (y==b).all()] # here enumerate will keep the track of index
#output
[1]
I have the following Arrays
import numpy as np
A = np.array([[1,2], [3,4]])
B = np.array([5,6])
Now I want to add the second element of A to B using the append function:
B = np.append(B, A[1])
What I want to get is:
B = np.array([[5, 6],[3,4]])
But what I get is:
B = np.array([5, 6, 3, 4])
How can I solve this? Should I use another function instead of numpy.append?
import numpy as np
A = np.array([[1,2], [3,4]])
B = np.array([[5,6]])
B = np.append(B, [A[1]], axis=0)
print(B)
Output
array([[5, 6],
[3, 4]])
This would be one of the way using np.append() specifying the axis = 0.
You can use np.stack for this instead of np.append:
>>> np.stack([B, A[1]])
array([[5, 6],
[3, 4]])
You could also add [:, None] to your two arrays and append with axis=1:
>>> np.append(B[:, None], A[1][:, None], axis=1)
array([[5, 3],
[6, 4]])
Is there any way that I can shift specific row in python? (It would be nice if numpy is used).
I want
[[1,2],
[3,4]]
to be
[[1,2],
[4,3]].
Also for the column would be nice!
[[1,2],
[3,4]]
to be
[[1,4],
[3,2]]
.
Thank you.
np.roll is your friend.
>>> import numpy as np
>>> x = np.array([[1,2],[3,4]])
>>> x
array([[1, 2],
[3, 4]])
>>> x[1]
array([3, 4])
>>> np.roll(x[1],1)
array([4, 3])
>>> x[1] = np.roll(x[1],1)
>>> x
array([[1, 2],
[4, 3]])
>>> x[1] = np.roll(x[1],1)
>>> x[:,1]
array([2, 4])
>>> x[:,1] = np.roll(x[:,1],1)
>>> x
array([[1, 4],
[3, 2]])
>>>
For the row shifting you can do:
import numpy as np
matrix = np.random.rand(5, 5)
row_no = 2
matrix[row_no, :] = np.array([matrix[row_no, -1]] + list(matrix[row_no, :-1]))
and similarly for column shifting you can simply switch the dimension orders.
import numpy as np
matrix = np.random.rand(5, 5)
col_no = 2
matrix[:, col_no] = np.array([matrix[-1, col_no]] + list(matrix[:-1, col_no]))
I have this set of equations I want to perform:
x = np.linspace(0, 2, 3)
y = np.linspace(x, x+2, 3)
I then want to populate the 2D array with a calculation that does:
a = 2*x + y
So for example, given an array:
x = [0, 1, 2]
Then, the array y is:
y = [[0, 1, 2],
[1, 2, 3],
[2, 3, 4]]
When I perform the operation a = 2*x + y I should get the array:
a = [[0, 1, 2],
[3, 4, 5],
[6, 7, 8]]
How do I do this, keeping in mind I want to perform this operation quickly for array of size up to 10000x10000 (or larger)?
Or do your code adding two Ts:
print((2*x+y.T).T)
Output:
[[0 1 2]
[3 4 5]
[6 7 8]]
How can I split an array's columns into three arrays x, y, z without manually writing each of the [:,0],[:,1],[:,2] separately?
Example
# Create example np array
import numpy as np
data = np.array([[1,2,3],[4,5,6],[7,8,9]])
Now data is
[[1 2 3]
[4 5 6]
[7 8 9]]
What I want to do:
x, y, z = data[:,0], data[:,1], data[:,2] ## Help me here!
print(x)
Wanted output:
array([1, 4, 7])
Transpose, then unpack:
>>> x, y, z = data.T
>>> x
array([1, 4, 7])
You don't need to slice it.
>>> import numpy as np
>>> data = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> x, y, z = data.T
>>> x
array([1, 4, 7])
>>> y
array([2, 5, 8])
>>> z
array([3, 6, 9])