I'm going to provide a functional (immutable) way of doing it.
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]
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]
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]
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]
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]
Use np.delete
! It does not actually delete anything inplace
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]])
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
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]
The simplest way I found was:
mylist[:x] + mylist[x+1:]
that will produce your mylist
without the element at index x
.
mylist = [0, 1, 2, 3, 4, 5]
x = 3
mylist[:x] + mylist[x+1:]
Result produced
mylist = [0, 1, 2, 4, 5]
>>> 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
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
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!
Source: Stackoverflow.com