问题：索引所有*除外* python中的一项
有没有一种简单的方法来索引列表（或数组，或其他任何东西）中除特定索引之外的所有元素？例如，
mylist[3] 将把该物品退回第3位
milist[~3] 将返回整个列表，除了3
Is there a simple way to index all elements of a list (or array, or whatever) except for a particular index? E.g.,
回答 0
对于列表，您可以使用列表组合。例如，要制作不含第3个元素b的副本a：
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]
这是非常通用的方法，可用于所有可迭代对象，包括numpy数组。如果您替换[]为()，b将是一个迭代器，而非列表。
或者，您可以通过以下方式就地完成此操作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]
在numpy中，您可以使用布尔索引来做到这一点：
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])
通常，这比上面列出的列表理解要快得多。
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.
回答 1
>>> 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]
>>>
也可以看看
解释Python的切片符号
>>> 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
回答 2
我发现的最简单的方法是：
mylist[:x]+mylist[x+1:]
这将产生mylist没有index的元素x。
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]
回答 3
如果您使用的是numpy，则我认为最接近的是使用蒙版
>>> 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])
如果itertools没有，可以达到类似的效果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]
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]
回答 4
使用np.delete！它实际上并没有删除任何内容
例：
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]])
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]])
回答 5
我将提供一种功能（不变）的方法。
做到这一点的标准和简单方法是使用切片：
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}")
输出：
data: [0, 1, 2, 3, 4], new_data: [0, 1, 2, 4]
使用清单理解：
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}")
输出：
data: [0, 1, 2, 3, 4], new_data: [0, 1, 2, 4]
使用过滤功能：
index_to_remove = 3
data = [*range(5)]
new_data = [*filter(lambda i: i != index_to_remove, data)]
输出：
data: [0, 1, 2, 3, 4], new_data: [0, 1, 2, 4]
使用遮罩。屏蔽由标准库中的itertools.compress函数提供：
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}")
输出：
data: [0, 1, 2, 3, 4], mask: [1, 1, 1, 0, 1], new_data: [0, 1, 2, 4]
使用Python标准库中的itertools.filterfalse函数
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}")
输出：
data: [0, 1, 2, 3, 4], new_data: [0, 1, 2, 4]
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]
回答 6
如果您事先不知道索引，这里的功能将起作用
def reverse_index(l, index):
try:
l.pop(index)
return l
except IndexError:
return False
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
回答 7
请注意，如果变量是列表列表，则某些方法将失败。例如：
v1 = [[range(3)] for x in range(4)]
v2 = v1[:3]+v1[4:] # this fails
v2
对于一般情况，使用
removed_index = 1
v1 = [[range(3)] for x in range(4)]
v2 = [x for i,x in enumerate(v1) if x!=removed_index]
v2
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
回答 8
如果要剪掉最后一个或第一个，请执行以下操作：
list = ["This", "is", "a", "list"]
listnolast = list[:-1]
listnofirst = list[1:]
如果将1更改为2，则将删除前2个字符，而不是第二个。希望这对您有所帮助！
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!
