从问题导入 [英] import from question

问题描述

大家好


我有三个档案：


档a1.py:

========

the_number =无


文件a2.py:

====== ==

导入a1

def init（）：

a1.the_number = 100


文件a3.py：

========

来自a1 import the_number

import a2


a2.init（）

打印the_number，输入（the_number）


Runninr a3.py我得到：

无< type''NoneType''>


将a3.py更改为：

import a1

导入a2


a2.init（）

打印a1.the_number，输入（a1.the_number）


给出：

100< type''int''>


为什么它不适用于a3的第一个版本。 py？

谢谢，

iu2

解决方案

1月14日，下午4:22，iu2< isra ... @ elbit.co.ilwrote：

大家好


我有三个文件：


file a1.py：

========

the_number =无


档a2.py:

========

导入a1

def init（）：

a1.the_number = 100


文件a3.py：

========

来自a1 import the_number

导入a2


a2.init（）

打印数字，输入（the_number）


Runninr a3.py我得到：

无< type''NoneType''>


将a3.py更改为：

导入a1

导入a2

a2.init（）

打印a1.the_number，输入（a1.the_number）


给出：

100< type''int''>


为什么不呢在a3.py的第一个版本中工作？


谢谢，

iu2

试试猜猜以下代码片段打印，运行它和s如果你猜对了
：


s = {''''：无}

x = s [''a '']

s [''a''] = 1

打印x


同样的机制适用于来自... import确实如此。


HTH，

George

iu2< is ***** @ elbit.co.ilwrote：

file a3.py:
========
来自a1 import的
the_number

import a2

....

>

为什么在a3.py的第一个版本中它不起作用？

想想''import a2''是同样如下：


a2 = __import __（''a2''）


和''从a1导入the_number''大致相同as：


the_number = __import __（''a1''）。the_number


换句话说，将它们视为作业，它应该全部有意义。

Duncan Booth写道：

iu2< is **** *@elbit.co.ilwrote：

>文件a3.py:
========
从a1导入the_number
导入a2

...

>为什么它在a3.py的第一个版本中不起作用？

想想''import a2''与以下相同：


a2 = __import __（''a2''）


和''来自a1导入the_number''大致相同同样如下：


the_number = __import __（''a1''）。the_number


换句话说将它们视为作业和它应该都有意义。

这有点令人惊讶。所以来自mod import *确实将所有

标量复制到本地命名空间中的新变量中。对于我想的

对象指针也是如此，但这是透明的，因为所有副本

访问同一个对象。


我总是认为两种形式的导入是等价的，但是一种形式消除了需要明确命名空间：mod.thing

显然这个远非如此。

-

通过 http://www.teranews.com

Hi all

I''ve got three files:

file a1.py:
========
the_number = None

file a2.py:
========
import a1

def init():
a1.the_number = 100

file a3.py:
========
from a1 import the_number
import a2

a2.init()
print the_number, type(the_number)

Runninr a3.py I get:
None <type ''NoneType''>

Changing a3.py to:
import a1
import a2

a2.init()
print a1.the_number, type(a1.the_number)

gives:
100 <type ''int''>

Why doesn''t it work in the first version of a3.py?

Thanks,
iu2

解决方案

On Jan 14, 4:22 pm, iu2 <[e-mail]:

Hi all

I''ve got three files:

file a1.py:
========
the_number = None

file a2.py:
========
import a1

def init():
a1.the_number = 100

file a3.py:
========
from a1 import the_number
import a2

a2.init()
print the_number, type(the_number)

Runninr a3.py I get:
None <type ''NoneType''>

Changing a3.py to:
import a1
import a2

a2.init()
print a1.the_number, type(a1.the_number)

gives:
100 <type ''int''>

Why doesn''t it work in the first version of a3.py?

Thanks,
iu2

Try to guess what the following snippet prints, run it, and see if you
guessed correctly:

s = {''a'':None}
x = s[''a'']
s[''a''] = 1
print x

The same mechanism applies to what "from ... import" does.

HTH,
George

iu2 <is*****@elbit.co.ilwrote:

file a3.py:
========
from a1 import the_number
import a2

....

>
Why doesn''t it work in the first version of a3.py?

Think of ''import a2'' as being the same as:

a2 = __import__(''a2'')

and ''from a1 import the_number'' as roughly the same as:

the_number = __import__(''a1'').the_number

In other words think of them as assignments and it should all make sense.

Duncan Booth wrote:

iu2 <is*****@elbit.co.ilwrote:

>file a3.py:
========
from a1 import the_number
import a2

...

>Why doesn''t it work in the first version of a3.py?

Think of ''import a2'' as being the same as:

a2 = __import__(''a2'')

and ''from a1 import the_number'' as roughly the same as:

the_number = __import__(''a1'').the_number

In other words think of them as assignments and it should all make sense.

This is a little surprising. So "from mod import *" really copies all of the
scalars into new variables in the local namespace. The same is true with
object pointers I suppose, but this is transparent as all the copies
access the same object.

I always ASSumed that the two forms of import were equivalent, but that
one form did away with the need to be explicit about the namespace: mod.thing
Obviously this is far from the case.
--
Posted via a free Usenet account from http://www.teranews.com

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