How do I remove separate elements in a vector without using the range function?-CodePudding

I've created vector x and I need to create a vector z by removing the 3rd and 6th elements of x. I cannot just create a vector by simply typing in the elements that should be in z. I have to index them or use a separate function.

x = [5,2,0,6,-10,12]
np.array(x)
print x

z = np.delete(x,)

I am not sure if using np.delete is best or if there is a better approach. Help?

CodePudding user response：

You can index and conact pieces of the list excluding the one you want to "delete"

x = [5,2,0,6,-10,12]
print ( x[0:2] x[3:5] )

[5, 2, 6, -10]

CodePudding user response：

You can use numpy.delete like below:

del_idx = np.array([3,6])
# Because index start from zero, We need to pass 'del_idx-1' -> [2, 5]
x_new = np.delete(np.asarray(x), del_idx-1)
print(x_new)

Or You can define a list of indexes that you want to delete and use enumerate in the original_list and keep elements that the index of them doesn't exist in delete_index.

(This approach based list_comprehensions)

import numpy as np

del_idx = [3,6]

x = [5,2,0,6,-10,12]

x_new = [item for idx, item in enumerate(x) if (idx 1) not in del_idx]

print(np.array(x_new))

Output:

[  5   2   6 -10]

CodePudding user response：

if x is numpy array, first convert to list:

x = list(x)

if not array then:

z = [x.pop(2), x.pop(-1)]

This will remove 3rd and 6th element form x and place it in z. Then convert it to numpy array if needed.

CodePudding user response：

In [69]: x = np.array([5,2,0,6,-10,12])

Using delete is straight forward:

In [70]: np.delete(x,[2,5])
Out[70]: array([  5,   2,   6, -10])

delete is a general function that takes various approaches based on the delete object, but in a case like this it uses a boolean mask:

In [71]: mask = np.ones(x.shape, bool); mask[[2,5]] = False; mask
Out[71]: array([ True,  True, False,  True,  True, False])    
In [72]: x[mask]
Out[72]: array([  5,   2,   6, -10])