numpy-equivalent of list.pop?

Question

Is there a numpy method which is equivalent to the builtin pop for python lists?

Popping obviously doesn't work on numpy arrays, and I want to avoid a list conversion.

Solution 1

There is no pop method for NumPy arrays, but you could just use basic slicing (which would be efficient since it returns a view, not a copy):

In [104]: y = np.arange(5); y
Out[105]: array([0, 1, 2, 3, 4])

In [106]: last, y = y[-1], y[:-1]

In [107]: last, y
Out[107]: (4, array([0, 1, 2, 3]))

If there were a pop method it would return the last value in y and modify y.

Above,

last, y = y[-1], y[:-1]

assigns the last value to the variable last and modifies y.

Solution 2

Here is one example using numpy.delete():

import numpy as np
arr = np.array([[1,2,3,4], [5,6,7,8], [9,10,11,12]])
print(arr)
#  array([[ 1,  2,  3,  4],
#         [ 5,  6,  7,  8],
#         [ 9, 10, 11, 12]])
arr = np.delete(arr, 1, 0)
print(arr)
# array([[ 1,  2,  3,  4],
#        [ 9, 10, 11, 12]])

Solution 3

There isn't any pop() method for numpy arrays unlike List, Here're some alternatives you can try out-

• Using Basic Slicing

>>> x = np.array([1,2,3,4,5])
>>> x = x[:-1]; x
>>> [1,2,3,4]

• Or, By Using delete()
  

Syntax - np.delete(arr, obj, axis=None)

arr: Input array
obj: Row or column number to delete
axis: Axis to delete

>>> x = np.array([1,2,3,4,5])
>>> x = x = np.delete(x, len(x)-1, 0)
>>> [1,2,3,4]

Solution 4

Pop doesn't exist for NumPy arrays, but you can use NumPy indexing in combination with array restructuring, for example hstack/vstack or numpy.delete(), to emulate popping.

Here are some example functions I can think of (which apparently don't work when the index is -1, but you can fix this with a simple conditional):

def poprow(my_array,pr):
    """ row popping in numpy arrays
    Input: my_array - NumPy array, pr: row index to pop out
    Output: [new_array,popped_row] """
    i = pr
    pop = my_array[i]
    new_array = np.vstack((my_array[:i],my_array[i+1:]))
    return [new_array,pop]

def popcol(my_array,pc):
    """ column popping in numpy arrays
    Input: my_array: NumPy array, pc: column index to pop out
    Output: [new_array,popped_col] """
    i = pc
    pop = my_array[:,i]
    new_array = np.hstack((my_array[:,:i],my_array[:,i+1:]))
    return [new_array,pop]

This returns the array without the popped row/column, as well as the popped row/column separately:

>>> A = np.array([[1,2,3],[4,5,6]])
>>> [A,poparow] = poprow(A,0)
>>> poparow
array([1, 2, 3])

>>> A = np.array([[1,2,3],[4,5,6]])
>>> [A,popacol] = popcol(A,2)
>>> popacol
array([3, 6])

Solution 5

The most 'elegant' solution for retrieving and removing a random item in Numpy is this:

import numpy as np
import random

arr = np.array([1, 3, 5, 2, 8, 7])
element = random.choice(arr)
elementIndex = np.where(arr == element)[0][0]
arr = np.delete(arr, elementIndex)

For curious coders:

The np.where() method returns two lists. The first returns the row indexes of the matching elements and the second the column indexes. This is useful when searching for elements in a 2d array. In our case, the first element of the first returned list is interesting.

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Comments

[Zeugma]

pop doesn't exist in numpy and by design it is not recommended to emulate it. You would better approach the algorithm you need to write without using a pop pattern

[parvij]

pop returns the value and the list become shorter

[Brambor]

But list.pop can take an index as a parameter. This won't do.

[sameer_nubia]

Can you explain this l1=[10,11,12,13,14,16,17,18] [l1.pop(l1.index(i)) for i in l1 if i%2==0] print("l1:",l1) output - l1: [11, 13, 16, 17]