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] will return the item in position 3

• milist[~3] will return the whole list except for 3

Answer 1

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 2

The simplest way I found was:

mylist[:x] + mylist[x+1:]

that will produce your mylist without the element at index x.

Example

mylist = [0, 1, 2, 3, 4, 5]
x = 3
mylist[:x] + mylist[x+1:]

Result produced

mylist = [0, 1, 2, 4, 5]

Answer 3

>>> l = range(1,10)
>>> l
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> l[:2] 
[1, 2]
>>> l[3:]
[4, 5, 6, 7, 8, 9]
>>> l[:2] + l[3:]
[1, 2, 4, 5, 6, 7, 8, 9]
>>> 

See also

Explain Python's slice notation

Answer 4

If you are using numpy, the closest, I can think of is using a mask

>>> import numpy as np
>>> arr = np.arange(1,10)
>>> mask = np.ones(arr.shape,dtype=bool)
>>> mask[5]=0
>>> arr[mask]
array([1, 2, 3, 4, 5, 7, 8, 9])

Something similar can be achieved using itertools without numpy

>>> from itertools import compress
>>> arr = range(1,10)
>>> mask = [1]*len(arr)
>>> mask[5]=0
>>> list(compress(arr,mask))
[1, 2, 3, 4, 5, 7, 8, 9]

Answer 5

Use np.delete ! It does not actually delete anything inplace

In your example, "mylist[~3]" would be written like that: mylist.delete(3)

A more complex example:

import numpy as np
a = np.array([[1,4],[5,7],[3,1]])                                       
 
# a: array([[1, 4],
#           [5, 7],
#           [3, 1]])

ind = np.array([0,1])                                                   

# ind: array([0, 1])

# a[ind]: array([[1, 4],
#                [5, 7]])

all_except_index = np.delete(a, ind, axis=0)                                              
# all_except_index: array([[3, 1]])

# a: (still the same): array([[1, 4],
#                             [5, 7],
#                             [3, 1]])

Answer 6

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 7

For 1D numpy, np.concatenate is faster than np.arange method.

Benchmark:

x = np.arange(1000) * 2
i = 3

%timeit np.concatenate((x[:i],x[i+1:]))
%timeit x[np.arange(len(x))!=i]

# 9.21 µs ± 467 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# 32.8 µs ± 7.46 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Answer 8

idxs = list(range(len(array))).remove(idx)
array[idxs]

This should work, it removes the not needed index and then slices the array

Answer 9

If you don't know the index beforehand here is a function that will work

def reverse_index(l, index):
    try:
        l.pop(index)
        return l
    except IndexError:
        return False

Answer 10

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

Answer 11

If you want to cut out the last or the first do this:

list = ["This", "is", "a", "list"]
listnolast = list[:-1]
listnofirst = list[1:]

If you change 1 to 2 the first 2 characters will be removed not the second. Hope this still helps!

Answer 12

arr=[1,3,5,7,9]
for i in range(len(arr)):
    arsum = arr[:i] + arr[i + 1:]