How do I get all the values from a NumPy array excluding a certain index?
Question:
I have a NumPy array, and I want to retrieve all the elements except a certain index. For example, consider the following array
a = [0,1,2,3,4,5,5,6,7,8,9]
If I specify index 3, then the resultant should be
a = [0,1,2,4,5,5,6,7,8,9]
Answers:
This should do it:
In [9]: np.hstack((a[:3], a[4:]))
Out[9]: array([0, 1, 2, 4, 5, 5, 6, 7, 8, 9])
If performance is an issue, the following will do it in place:
In [22]: a[3:-1] = a[4:]; a = a[:-1]
a_new = np.delete(a, 3, 0)
3 here is the index you wish to remove, and 0 is the axis (zero in this case if using 1D array). See np.delete
Like resizing, removing elements from an NumPy array is a slow operation (especially for large arrays since it requires allocating space and copying all the data from the original array to the new array).
It should be avoided if possible.
Often you can avoid it by working with a masked array instead. For example, consider the array a:
import numpy as np
a = np.array([0,1,2,3,4,5,5,6,7,8,9])
print(a)
print(a.sum())
# [0 1 2 3 4 5 5 6 7 8 9]
# 50
We can mask its value at index 3 and can perform a summation which ignores masked elements:
a = np.ma.array(a, mask=False)
a.mask[3] = True
print(a)
print(a.sum())
# [0 1 2 -- 4 5 5 6 7 8 9]
# 47
Masked arrays also support many operations besides sum.
If you really need to, it is also possible to remove masked elements using the compressed method:
print(a.compressed())
# [0 1 2 4 5 5 6 7 8 9]
But as mentioned above, avoid it if possible.
Here’s a one-liner if a is a NumPy array:
>>> a[np.arange(len(a))!=3]
array([0, 1, 2, 4, 5, 5, 6, 7, 8, 9])
Another solution is to use the concatenate function of NumPy:
>>> x = np.arange(0,10)
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> i = 3
>>> np.concatenate((x[:i],x[(i+1):]))
array([0, 1, 2, 4, 5, 6, 7, 8, 9])
I have a NumPy array, and I want to retrieve all the elements except a certain index. For example, consider the following array
a = [0,1,2,3,4,5,5,6,7,8,9]
If I specify index 3, then the resultant should be
a = [0,1,2,4,5,5,6,7,8,9]
This should do it:
In [9]: np.hstack((a[:3], a[4:]))
Out[9]: array([0, 1, 2, 4, 5, 5, 6, 7, 8, 9])
If performance is an issue, the following will do it in place:
In [22]: a[3:-1] = a[4:]; a = a[:-1]
a_new = np.delete(a, 3, 0)
3 here is the index you wish to remove, and 0 is the axis (zero in this case if using 1D array). See np.delete
Like resizing, removing elements from an NumPy array is a slow operation (especially for large arrays since it requires allocating space and copying all the data from the original array to the new array).
It should be avoided if possible.
Often you can avoid it by working with a masked array instead. For example, consider the array a:
import numpy as np
a = np.array([0,1,2,3,4,5,5,6,7,8,9])
print(a)
print(a.sum())
# [0 1 2 3 4 5 5 6 7 8 9]
# 50
We can mask its value at index 3 and can perform a summation which ignores masked elements:
a = np.ma.array(a, mask=False)
a.mask[3] = True
print(a)
print(a.sum())
# [0 1 2 -- 4 5 5 6 7 8 9]
# 47
Masked arrays also support many operations besides sum.
If you really need to, it is also possible to remove masked elements using the compressed method:
print(a.compressed())
# [0 1 2 4 5 5 6 7 8 9]
But as mentioned above, avoid it if possible.
Here’s a one-liner if a is a NumPy array:
>>> a[np.arange(len(a))!=3]
array([0, 1, 2, 4, 5, 5, 6, 7, 8, 9])
Another solution is to use the concatenate function of NumPy:
>>> x = np.arange(0,10)
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> i = 3
>>> np.concatenate((x[:i],x[(i+1):]))
array([0, 1, 2, 4, 5, 6, 7, 8, 9])