附加到范围函数 [英] Appending to range function

问题描述

编辑:感谢您对如何完成我正在尝试做的事情的友好回复，但问题是关于为什么 range().append() 返回如果您一步尝试，则没有，为什么如果您分两步操作，它会起作用.

我正在尝试创建一个数字列表，但稍有不同.我不想在列表的开头输入几个数字:

mlist = [0, 5, 6, 7, ...]

所以我想这样做:

 mlist = range(5,n+1).append(0)

但无声地失败，因为 type(mlist) 之后等于 NoneType ?！(相关:type(range(5,10) 计算结果为 list Type)

如果我尝试分两步完成，例如:

<预><代码>>>>mlist = 范围(5,10)#然后>>>mlist.append(0)>>>列表[5, 6, 7, 8, 9, 10, 0]

发生了什么?

解决方案

list.append() 改变列表就地，并返回 None.通过将返回值分配给 mlist，而不是您想要构建的列表.这是一个标准的 Python 习惯用法；改变可变对象的方法永远不会返回改变的对象，总是None.

将两个步骤分开:

mlist = range(5, n + 1)mlist.append(0)

这将 [0] 添加到 end；如果您在开始时需要 0，请使用:

mlist = [0] + range(5, n + 1)

或者你可以使用 list.insert()，再次作为一个单独的调用:

mlist = range(5, n + 1)mlist.insert(0, 0)

但后者必须将所有元素向上移动一步，通过连接创建一个新列表对于较短的列表是更快的选择，插入在较长的列表中胜出:

<预><代码>>>>从时间导入时间>>>定义连接(n):... mlist = [0] + range(5, n)...>>>定义插入(n):... mlist = range(5, n)... mlist.insert(0, 0)...>>>timeit('concat(10)', 'from __main__ import concat')1.2668070793151855>>>timeit('insert(10)', 'from __main__ import insert')1.4820878505706787>>>timeit('concat(1000)', 'from __main__ import concat')23.53221583366394>>>timeit('insert(1000)', 'from __main__ import insert')15.84601092338562

EDIT: Thank you for your kind replies on how to accomplish what I am trying to do, but the question is about WHY range().append() returns None if you try it in one step, and WHY it works if you two-step the procedure.

Im trying to create a numerical list but with a twist. I don't want a couple of numbers at the beggining of my list:

mlist = [0, 5, 6, 7, ...]

So i thought to do the following:

 mlist = range(5,n+1).append(0)

but silently fails because type(mlist) afterwards equals to NoneType ?! (related: type(range(5,10) evaluates to list Type)

If i try to do that in two steps eg:

>>> mlist = range(5,10)
#and then
>>> mlist.append(0)
>>> mlist
[5, 6, 7, 8, 9, 10, 0]

What's happening?

解决方案

list.append() alters the list in place, and returns None. By assigning you assigned that return value to mlist, not the list you wanted to build. This is a standard Python idiom; methods that alter the mutable object never return the altered object, always None.

Separate the two steps:

mlist = range(5, n + 1)
mlist.append(0)

This adds [0] to the end; if you need the 0 at the start, use:

mlist = [0] + range(5, n + 1)

or you can use list.insert(), again as a separate call:

mlist = range(5, n + 1)
mlist.insert(0, 0)

but the latter has to shift up all elements one step and creating a new list by concatenation is the faster option for shorter lists, insertion wins on longer lists:

>>> from timeit import timeit
>>> def concat(n):
...     mlist = [0] + range(5, n)
... 
>>> def insert(n):
...     mlist = range(5, n)
...     mlist.insert(0, 0)
... 
>>> timeit('concat(10)', 'from __main__ import concat')
1.2668070793151855
>>> timeit('insert(10)', 'from __main__ import insert')
1.4820878505706787
>>> timeit('concat(1000)', 'from __main__ import concat')
23.53221583366394
>>> timeit('insert(1000)', 'from __main__ import insert')
15.84601092338562

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