CODEWITHSUNDEEP
pythonnumpy
Milad
Milad

Reputation: 125

Rearranging the values inside a NumPy array

lets assume that i have a numpy array and its like this:

[[1, 2, 3],
 [4, 5, 6],
 [7, 8, 9],
 [10, 11, 12]]

and I like to have a new numpy array, that the first half exactly the same as array above but second half it should be started from the bottom till it reaches the half of array. how do I do that?

EDIT: The method i need it has to work for an array with 60000 elements. not for this simple example!!

Output should be like this:

[[1, 2, 3],
 [4, 5, 6],
 [10, 11, 12],
 [7, 8, 9]]

Upvotes: 0

Views: 313

Answers (5)

Rahul D
Rahul D

Reputation: 36

array = numpy.array([[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]])
n = (int)(array.shape[0]//2) # change n accordingly
# n is the row number from which you want to start the reversal
temp = array[n:]
temp = temp[::-1]
array[n:] = temp

Upvotes: 1

Josef
Josef

Reputation: 3497

Takes always half the array as requested. (Based on mathfux's answer)

array = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]])
sl = range(0, len(array)//2)
s2 = reversed(range(len(array)//2, len(array)))
b = array[(*sl, *s2), :]
print(b)

prints

[[ 1  2  3]
 [ 4  5  6]
 [10 11 12]
 [ 7  8  9]]

Of course cou could just index the last half:

array = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]])
from_to = tuple(range(len(array)//2, len(array)))
to_from = tuple(reversed(from_to))
array[from_to, :] = array[to_from, :]
print(array)

Same result.

Upvotes: 1

Gerardo Zinno
Gerardo Zinno

Reputation: 1712

Another solution would be to simply swap the rows with an assignment, like:

arr[2], arr[3] = arr[3], arr[2]

After your edit, here's a solution that inverts just the bottom half of the array:

arr = np.array([[...],...])
arr[int(len(a)/2):] = arr[int(len(a)/2):][::-1]

[::-1] returns the elements of an array in reverse order. So applying it to the bottom half of the original array and assigning this new array to the bottom half of the original array will give you an array with the first n/2 rows unchanged and last n/2 rows in reverse order.

Upvotes: 1

Patrick Beaty
Patrick Beaty

Reputation: 46

This might be kinda crude but you simply need to replace the 3rd (index = 2) row in array e with the 4th (index = 3) row. 'a' is a placeholder for switching the 3rd and 4th row.

for i in range(len(e)):
    if i == 2:
        a = e[i]
        e[i] = e[3]
    if i == 3:
        e[i] = a

Upvotes: 1

mathfux
mathfux

Reputation: 5939

Use numpy indexes:

array = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]])
array[[0,1,3,2], :]
array([[ 1,  2,  3],
       [ 4,  5,  6],
       [10, 11, 12],
       [ 7,  8,  9]])

Upvotes: 2

Related Questions