在列表中的特定索引处插入元素并返回更新列表

Question

我有这个:

>>> a = [1, 2, 4]
>>> print a
[1, 2, 4]

>>> print a.insert(2, 3)
None

>>> print a
[1, 2, 3, 4]

>>> b = a.insert(3, 6)
>>> print b
None

>>> print a
[1, 2, 3, 6, 4]

无论如何我可以获得更新的列表作为结果,而不是更新原始列表？

Answer 1

l.insert(index, obj)实际上并没有返回任何内容,它只是更新列表.正如ATO所说,你可以做到b = a[:index] + [obj] + a[index:].但是,另一种方式是:

a = [1, 2, 4]
b = a[:]
b.insert(2, 3)

Answer 2

最节能的方法

您还可以使用列表中的切片索引插入元素.例如:

>>> a = [1, 2, 4]
>>> insert_at = 2  # index at which you want to insert item

>>> b = a[:]   # created copy of list "a" as "b"
               # skip this step if you are ok with modifying original list

>>> b[insert_at:insert_at] = [3]  # insert "3" within "b"
>>> b
[1, 2, 3, 4]

要在给定索引处将多个元素一起插入,您需要做的就是使用list要插入的多个元素.例如:

>>> a = [1, 2, 4]
>>> insert_at = 2   # index starting from which multiple elements will be inserted 

# List of elements that you want to insert together at "index_at" (above) position
>>> insert_elements = [3, 5, 6]

>>> a[insert_at:insert_at] = insert_elements
>>> a   # [3, 5, 6] are inserted together in `a` starting at index "2"
[1, 2, 3, 5, 6, 4]

使用列表理解的替代方案 (但在性能方面非常慢):

作为替代方案,它可以使用来实现清单理解与enumerate过.(但请不要这样做.这只是为了说明):

>>> a = [1, 2, 4]
>>> insert_at = 2

>>> b = [y for i, x in enumerate(a) for y in ((3, x) if i == insert_at else (x, ))]
>>> b
[1, 2, 3, 4]

所有解决方案的性能比较

以下是timeitPython 3.4.5的所有答案与1000个元素列表的比较:

• 使用切片插入我的答案 - 最快(每循环3.08 usec)

  mquadri$ python3 -m timeit -s "a = list(range(1000))" "b = a[:]; b[500:500] = [3]"
  100000 loops, best of 3: 3.08 usec per loop
  

• ATOzTOA基于切片列表合并的接受答案 - 第二个(每个循环6.71次使用)

  mquadri$ python3 -m timeit -s "a = list(range(1000))" "b = a[:500] + [3] + a[500:]"
  100000 loops, best of 3: 6.71 usec per loop
  

• Rushy Panchal使用大多数选票的回答list.insert(...) - 第三(每循环26.5次使用)

  python3 -m timeit -s "a = list(range(1000))" "b = a[:]; b.insert(500, 3)"
  10000 loops, best of 3: 26.5 usec per loop
  

• 我的回答是列表理解和enumerate- 第四(非常慢,每个循环168 usec)

  mquadri$ python3 -m timeit -s "a = list(range(1000))" "[y for i, x in enumerate(a) for y in ((3, x) if i == 500 else (x, )) ]"
  10000 loops, best of 3: 168 usec per loop
  

Answer 3

最短的我得到: b = a[:2] + [3] + a[2:]

>>> 
>>> a = [1, 2, 4]
>>> print a
[1, 2, 4]
>>> b = a[:2] + [3] + a[2:]
>>> print a
[1, 2, 4]
>>> print b
[1, 2, 3, 4]