Index all *except* one item in python

Question

Is there a simple way to index all elements of a list (or array, or whatever) except for a particular index? E.g.,

• mylist[3]

Answer 1

I'm going to provide a functional (immutable) way of doing it.

1. The standard and easy way of doing it is to use slicing:

   index_to_remove = 3
   data = [*range(5)]
   new_data = data[:index_to_remove] + data[index_to_remove + 1:]
   
   print(f"data: {data}, new_data: {new_data}")
   

   Output:

   data: [0, 1, 2, 3, 4], new_data: [0, 1, 2, 4]
   

2. Use list comprehension:

   data = [*range(5)]
   new_data = [v for i, v in enumerate(data) if i != index_to_remove]
   
   print(f"data: {data}, new_data: {new_data}") 
   

   Output:

   data: [0, 1, 2, 3, 4], new_data: [0, 1, 2, 4]
   

3. Use filter function:

   index_to_remove = 3
   data = [*range(5)]
   new_data = [*filter(lambda i: i != index_to_remove, data)]
   

   Output:

   data: [0, 1, 2, 3, 4], new_data: [0, 1, 2, 4]
   

4. Using masking. Masking is provided by itertools.compress function in the standard library:

   from itertools import compress
   
   index_to_remove = 3
   data = [*range(5)]
   mask = [1] * len(data)
   mask[index_to_remove] = 0
   new_data = [*compress(data, mask)]
   
   print(f"data: {data}, mask: {mask}, new_data: {new_data}")
   

   Output:

   data: [0, 1, 2, 3, 4], mask: [1, 1, 1, 0, 1], new_data: [0, 1, 2, 4]
   

5. Use itertools.filterfalse function from Python standard library

   from itertools import filterfalse
   
   index_to_remove = 3
   data = [*range(5)]
   new_data = [*filterfalse(lambda i: i == index_to_remove, data)]
   
   print(f"data: {data}, new_data: {new_data}")
   

   Output:

   data: [0, 1, 2, 3, 4], new_data: [0, 1, 2, 4]
   

Answer 2

For a list, you could use a list comp. For example, to make b a copy of a without the 3rd element:

a = range(10)[::-1]                       # [9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
b = [x for i,x in enumerate(a) if i!=3]   # [9, 8, 7, 5, 4, 3, 2, 1, 0]

This is very general, and can be used with all iterables, including numpy arrays. If you replace [] with (), b will be an iterator instead of a list.

Or you could do this in-place with pop:

a = range(10)[::-1]     # a = [9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
a.pop(3)                # a = [9, 8, 7, 5, 4, 3, 2, 1, 0]

In numpy you could do this with a boolean indexing:

a = np.arange(9, -1, -1)     # a = array([9, 8, 7, 6, 5, 4, 3, 2, 1, 0])
b = a[np.arange(len(a))!=3]  # b = array([9, 8, 7, 5, 4, 3, 2, 1, 0])

which will, in general, be much faster than the list comprehension listed above.

Answer 3

Note that if variable is list of lists, some approaches would fail. For example:

v1 = [[range(3)] for x in range(4)]
v2 = v1[:3]+v1[4:] # this fails
v2

For the general case, use

removed_index = 1
v1 = [[range(3)] for x in range(4)]
v2 = [x for i,x in enumerate(v1) if x!=removed_index]
v2