索引所有*除外* python中的一项

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.,

  • mylist[3] will return the item in position 3

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


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[]()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]

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。

  1. 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]
  2. 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]
  3. 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]
  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]
  5. 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.

  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]
    

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!