issue on internal import in a package
Here is a simple example: [app] [module] __init__.py -- empty a.py -- import b b.py -- defined a function foo() test.py In the test.py, contains the below statement: from module import a Execute the test.py will get error: Traceback (most recent call last): File stdin, line 1, in module File module\a.py, line 1, in module import b ImportError: No module named b Why the b.py can not be found by a.py? -- http://mail.python.org/mailman/listinfo/python-list
Re: issue on internal import in a package
On Feb 27, 8:11 pm, 人言落日是天涯，望极天涯不见家 kelvin@gmail.com wrote: Here is a simple example: [app] [module] __init__.py -- empty a.py -- import b b.py -- defined a function foo() test.py In the test.py, contains the below statement: from module import a Execute the test.py will get error: Traceback (most recent call last): File stdin, line 1, in module File module\a.py, line 1, in module import b ImportError: No module named b Why the b.py can not be found by a.py? PS. This issue occurred on Python3.2 It's okay in the Python2.6 Python2.5 -- http://mail.python.org/mailman/listinfo/python-list
Re: issue on internal import in a package
On Feb 27, 8:40 pm, Ben Finney ben+pyt...@benfinney.id.au wrote: 人言落日是天涯，望极天涯不见家 kelvin@gmail.com writes: Here is a simple example: [app] [module] __init__.py -- empty a.py -- import b b.py -- defined a function foo() test.py In the test.py, contains the below statement: from module import a Execute the test.py will get error: This works fine for me:: $ mkdir --parents app/module/ $ touch app/module/__init__.py $ printf import b\n app/module/a.py $ printf def foo(): pass\n app/module/b.py $ printf from module import a\n app/test.py $ find . . ./app ./app/module ./app/module/__init__.py ./app/module/a.py ./app/module/b.py ./app/test.py $ python app/test.py Traceback (most recent call last): File stdin, line 1, in module File module\a.py, line 1, in module import b ImportError: No module named b Why the b.py can not be found by a.py? I get no errors; the code appears to run fine. Perhaps the scenario is not exactly as you describe? -- \ “If we listen only to those who are like us, we will squander | `\ the great opportunity before us: To live together peacefully in | _o__) a world of unresolved differences.” —David Weinberger | Ben Finney Thanks for your reply. What's the version of your Python? My version is 3.2. Python 2.5/2.6 doesn't have this issue. -- http://mail.python.org/mailman/listinfo/python-list
Re: issue on internal import in a package
On Feb 27, 9:22 pm, Frank Millman fr...@chagford.com wrote: Ben Finney ben+pyt...@benfinney.id.au wrote in message news:87ei6t646h@benfinney.id.au... 人言落日是天涯，望极天涯不见家 kelvin@gmail.com writes: Here is a simple example: [app] [module] __init__.py -- empty a.py -- import b b.py -- defined a function foo() test.py In the test.py, contains the below statement: from module import a Execute the test.py will get error: This works fine for me:: $ mkdir --parents app/module/ $ touch app/module/__init__.py $ printf import b\n app/module/a.py $ printf def foo(): pass\n app/module/b.py $ printf from module import a\n app/test.py $ find . . ./app ./app/module ./app/module/__init__.py ./app/module/a.py ./app/module/b.py ./app/test.py $ python app/test.py Traceback (most recent call last): File stdin, line 1, in module File module\a.py, line 1, in module import b ImportError: No module named b Why the b.py can not be found by a.py? I get no errors; the code appears to run fine. Perhaps the scenario is not exactly as you describe? I get exactly the same result as the OP, using python 3.2 on both windows and linux. It works using python 2.6. I can fix it by changing a.py from 'import b' to 'from . import b'. As I understand it, the reason is that python 3.x will no longer look for an absolute import in the current package - it will only look in sys.path. Frank Millman This behavior is by design or just a bug for Python3.x ? -- http://mail.python.org/mailman/listinfo/python-list
Re: issue on internal import in a package
On Feb 27, 9:38 pm, Frank Millman fr...@chagford.com wrote: 人言落日是天涯，望极天涯不见家 kelvin@gmail.com wrote in message news:fa94323b-d859-4599-b236-c78a22b3d...@t19g2000prd.googlegroups.com... On Feb 27, 9:22 pm, Frank Millman fr...@chagford.com wrote: This behavior is by design or just a bug for Python3.x ? Definitely by design. Have a look at PEP 328 -http://www.python.org/dev/peps/pep-0328/ In Python 2.4 and earlier, if you're reading a module located inside a package, it is not clear whether import foo refers to a top-level module or to another module inside the package. As Python's library expands, more and more existing package internal modules suddenly shadow standard library modules by accident. It's a particularly difficult problem inside packages because there's no way to specify which module is meant. To resolve the ambiguity, it is proposed that foo will always be a module or package reachable from sys.path. This is called an absolute import. HTH Frank Yes, it's okay with the change in a.py with below line: from . import b But another issue occurred if I want to run the a.py separately. $ cd module $ python a.py Traceback (most recent call last): File a.py, line 1, in module from . import b ValueError: Attempted relative import in non-package Does that mean the relative import only allowed in the package. And it cannot be run as __main__ program unless I change the relative import back to absolute import? I think this behavior is strange and difficult to use. Doesn't -- http://mail.python.org/mailman/listinfo/python-list
Re: default value for __init__ doesn't work
On Sep 11, 1:14 pm, Benjamin Kaplan benjamin.kap...@case.edu wrote:
On Sat, Sep 11, 2010 at 12:38 AM, 人言落日是天涯，望极天涯不见家 kelvin@gmail.com
wrote:
Please look at below code snippet:
class test():
def __init__(self, a, dic={}):
self.a = a
self.dic = dic
print('__init__ params:',a, dic)
This is a pretty popular mistake to make. Default arguments aren't
evaluated when you call the method. They're created when the method is
created (meaning when you first run the file and the class itself is
defined), and that's it. Because you do self.dic = dic, this means
that every instance of the object will receive the same dict object.
Change it for one object, and the change will show up in all of them.
The solution to this is to use a sentinel value, like None
def __init__(self, a, dic=None) :
if dic is None :
self.dic = {}
else :
self.dic = dic
If None is a valid value for the parameter, make a sentinel object and use
that
sentinel = object()
def __init__(self, a, dic=sentinel) :
if dic is sentinel : #you want to use is here, not ==
...
Got it. Thanks for point out my mistake. You are very nice.
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Re: default value for __init__ doesn't work
On Sep 11, 1:55 pm, 人言落日是天涯，望极天涯不见家 kelvin@gmail.com wrote:
On Sep 11, 1:14 pm, Benjamin Kaplan benjamin.kap...@case.edu wrote:
On Sat, Sep 11, 2010 at 12:38 AM, 人言落日是天涯，望极天涯不见家 kelvin@gmail.com
wrote:
Please look at below code snippet:
class test():
def __init__(self, a, dic={}):
self.a = a
self.dic = dic
print('__init__ params:',a, dic)
This is a pretty popular mistake to make. Default arguments aren't
evaluated when you call the method. They're created when the method is
created (meaning when you first run the file and the class itself is
defined), and that's it. Because you do self.dic = dic, this means
that every instance of the object will receive the same dict object.
Change it for one object, and the change will show up in all of them.
The solution to this is to use a sentinel value, like None
def __init__(self, a, dic=None) :
if dic is None :
self.dic = {}
else :
self.dic = dic
If None is a valid value for the parameter, make a sentinel object and use
that
sentinel = object()
def __init__(self, a, dic=sentinel) :
if dic is sentinel : #you want to use is here, not ==
...
Got it. Thanks for point out my mistake. You are very nice.
I remember the same issue was occurred in my C++ program. There I have
a function with a parameter referenced a default object . May be C++
also constructs the the default arguments before the function is
called.
Thank you again to help me so much!
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default value for __init__ doesn't work
Please look at below code snippet:
class test():
def __init__(self, a, dic={}):
self.a = a
self.dic = dic
print('__init__ params:',a, dic)
def get(self):
self.dic[1] = 2
self.dic[4] = 5
def foo():
print('in foo function')
bar = test(1)
bar.get()
if __name__ == '__main__':
foo()
foo()
---
Result:
in foo function
__init__ params: 1 {}
in foo function
__init__ params: 1 {1: 2, 4: 5}
But my expect result is :
in foo function
__init__ params: 1 {}
in foo function
__init__ params: 1 {}
it seems that the default value for dic doesn't work on the second
call for the class test.
It's wired. Who can give a explaination for this scenario?
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Floating Number format problem
How could I format the float number like this: (keep 2 digit precision) 1.002 = 1 1.12 = 1.12 1.00 = 1 1.567 = 1.57 2324.012 = 2324.01 I can not find any Formatting Operations is able to meet my requirement. Any suggestion will be appreciated. -- http://mail.python.org/mailman/listinfo/python-list
Why can not catch the inner exception
Please see the follow code, I can not catch the exception IOError raised from shutil.copyfile() , why? try: if (DEST_TYPE TYPE_FTP): fn = oname ftpc.UploadFile(f, fn) else: fn = os.path.join(dst, oname) shutil.copyfile(f, fn) other code except [IOError, FtpcException],why: num = 0 print sys.stderr, can not copy '%s' to '%s': %s%(f, fn, why) ERR_NUM += 1 I must do like this: try: if (DEST_TYPE TYPE_FTP): fn = oname ftpc.UploadFile(f, fn) else: fn = os.path.join(dst, oname) try: shutil.copyfile(f, fn) except IOError: other code except [IOError, FtpcException],why: num = 0 print sys.stderr, can not copy '%s' to '%s': %s%(f, fn, why) ERR_NUM += 1 Thanks! -- http://mail.python.org/mailman/listinfo/python-list
Does unicode() equal to unicode(sys.getfilesystemencoding()) ?
The follow statement comes from the Python 2.5 documentation
--
encode( [encoding[,errors]])
Return an encoded version of the string. Default encoding is the
current default string encoding. errors may be given to set a
different error handling scheme.
---
what's the Default encoding mean ? Does it equal to the
sys.getfilesystemencoding()?
If yes, but :
unicode('中国', sys.getfilesystemencoding())
u'\u4e2d\u56fd'
unicode('中国')
Traceback (most recent call last):
File input, line 1, in module
UnicodeDecodeError: 'ascii' codec can't decode byte 0xd6 in position
0: ordinal not in range(128)
It seems the Default encoding is not equal to the
sys.getfilesystemencoding(). And then, what is it ?
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Re: How to print this character u'\u20ac' to DOS terminal
On 5月30日, 下午1时23分, Martin v. Lowis [EMAIL PROTECTED] wrote: 人言落日是天涯，望极天涯不见家 schrieb: Who could explain the follow issue ? print u'\u0394' Δ print u'\u20ac' Traceback (most recent call last): File stdin, line 1, in module UnicodeEncodeError: 'gbk' codec can't encode character u'\u20ac' in position 0: illegal multibyte sequence My terminal is cmd.exe under windows XP. what's the different between the two character ? what can I do if I want to print the u'\u20ac'? The problem is that your terminal uses (some form of) the GBK encoding; seehttp://zh.wikipedia.org/wiki/GBKfor details on GBK. It seems that GBK (or, rather, code page 936) supports the delta character, but not the euro sign. To change that, you can use chcp in your terminal window. For example, if you do chcp 850, you should be able to display the euro sign (but will simultaneously use the ability to display the letter delta, and the chinese letters). I don't know whether the terminal supports an UTF-8 code page; you can try setting the terminal's code page to 65001 (which should be UTF-8). Regards, Martin Thanks, but it seems not work yet. C:\WINDOWSchcp 850 Active code page: 850 C:\WINDOWSpython Python 2.5.1 (r251:54863, Apr 18 2007, 08:51:08) [MSC v.1310 32 bit (Intel)] on win32 Type help, copyright, credits or license for more information. print u'\u20ac' Traceback (most recent call last): File stdin, line 1, in module File C:\Python25\lib\encodings\cp850.py, line 12, in encode return codecs.charmap_encode(input,errors,encoding_map) UnicodeEncodeError: 'charmap' codec can't encode character u'\u20ac' in position 0: character maps to undefined C:\WINDOWSchcp 65001 Active code page: 65001 C:\WINDOWSpython Python 2.5.1 (r251:54863, Apr 18 2007, 08:51:08) [MSC v.1310 32 bit (Intel)] on win32 Type help, copyright, credits or license for more information. print u'\u20ac' Traceback (most recent call last): File stdin, line 1, in module LookupError: unknown encoding: cp65001 --- I find that the u'\u20ac' related 'mbcs' encode is 0x80, I could print it directly print '\x80' � But the string contained the u'\u20ac' is get from remote host. Is there any method to decode it to the local 'mbcs'? -- http://mail.python.org/mailman/listinfo/python-list
Re: How to print this character u'\u20ac' to DOS terminal
On 5月30日, 下午9时03分, Tijs [EMAIL PROTECTED] wrote:
??? wrote:
But the string contained the u'\u20ac' is get from remote host. Is
there any method to decode it to the local 'mbcs'?
remote_string = u'\u20ac'
try:
local_string = remote_string.encode('mbcs')
except:
# no mbcs equivalent available
print encoding error
else:
# local_string is now an 8-bit string
print result:, local_string
# if console is not mbcs, you should see incorrect result
assert result == '\x80'
Mbcs is windows-only so I couldn't test this.
If your application handles text, it may be easier to just leave everything
in Unicode and encode to utf-8 for storage?
Regards,
Tijs
Yes, it works, thank you.
But I doubt this way may not work on linux. Maybe I should write some
additional code for supporting both windows and linux OS.
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Re: how to print the GREEK CAPITAL LETTER delta under utf-8 encoding
On 5月29日, 下午1时34分, Martin v. Lowis [EMAIL PROTECTED] wrote:
人言落日是天涯，望极天涯不见家 schrieb:
I lookup the utf-8 form of delta from the link.
http://www.fileformat.info/info/unicode/char/0394/index.htm
and then I want to print it in the python ( I work under windows)
#!/usr/bin/python
#coding=utf-8
print \xce\x94
but the result is not the 'delta' but an unknown character.
I assume you print to the terminal (cmd.exe). This cannot work;
the terminal (usually) does not interpret the characters in UTF-8.
Instead, you should print a Unicode string, e.g.
print u\N{GREEK CAPITAL LETTER DELTA}
or
print u'\u0394'
This should work as long as your terminal supports printing
the letter at all.
Regards,
Martin
yes, it could print to the terminal(cmd.exe), but when I write these
string to file. I got the follow error:
File E:\Tools\filegen\filegen.py, line 212, in write
self.file.write(data)
UnicodeEncodeError: 'ascii' codec can't encode character u'\u0394' in
position 0
: ordinal not in range(128)
but other text, in which include chinese characters got from
os.listdir(...), are written to the file OK. why?
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Re: how to print the GREEK CAPITAL LETTER delta under utf-8 encoding
On 5月29日, 下午3时05分, Martin v. Lowis [EMAIL PROTECTED] wrote: yes, it could print to the terminal(cmd.exe), but when I write these string to file. I got the follow error: File E:\Tools\filegen\filegen.py, line 212, in write self.file.write(data) UnicodeEncodeError: 'ascii' codec can't encode character u'\u0394' in position 0 : ordinal not in range(128) Yes, when writing to a file, you need to define an encoding, e.g. self.file.write(data.encode(utf-8)) You can use codecs.open() instead of open(), so that you can just use self.file.write(data) Alternatively, you can find out what sys.stdout.encoding is, and use that when encoding data for the terminal (falling back to utf-8 when .encoding is not available on the file). but other text, in which include chinese characters got from os.listdir(...), are written to the file OK. why? Your version of Windows uses a code page that supports Chinese characters in the byte-oriented character set. These are normally encoded using the mbcs encoding (except that the terminal likely uses a different encoding). So if you use mbcs instead of utf-8, you might be able to read the text as well. Regards, Martin Thanks a lot! I want to just use the utf-8. how could I convert my 'mbcs' encoding to the utf-8 and write it to the file? I have replaced the open() to codecs.open() but it still can not decode the 'mbcs', the error is as follow: File E:\Tools\filegen\filegen.py, line 213, in write self.file.write(data) File C:\Python25\lib\codecs.py, line 638, in write return self.writer.write(data) File C:\Python25\lib\codecs.py, line 303, in write data, consumed = self.encode(object, self.errors) UnicodeDecodeError: 'ascii' codec can't decode byte 0xcc in position 32: ordinal not in range(128) -- http://mail.python.org/mailman/listinfo/python-list
How to print this character u'\u20ac' to DOS terminal
Who could explain the follow issue ? print u'Δ' Δ print u'�' Traceback (most recent call last): File stdin, line 1, in module UnicodeEncodeError: 'gbk' codec can't encode character u'\x80' in position 0: il legal multibyte sequence or I just put the unicode number print u'\u0394' Δ print u'\u20ac' Traceback (most recent call last): File stdin, line 1, in module UnicodeEncodeError: 'gbk' codec can't encode character u'\u20ac' in position 0: illegal multibyte sequence My terminal is cmd.exe under windows XP. what's the different between the two character ? what can I do if I want to print the u'\u20ac'? -- http://mail.python.org/mailman/listinfo/python-list
Issue of redirecting the stdout to both file and screen
I wanna print the log to both the screen and file, so I simulatered a
'tee'
class Tee(file):
def __init__(self, name, mode):
file.__init__(self, name, mode)
self.stdout = sys.stdout
sys.stdout = self
def __del__(self):
sys.stdout = self.stdout
self.close()
def write(self, data):
file.write(self, data)
self.stdout.write(data)
Tee('logfile', 'w')
print sys.stdout, 'abcdefg'
I found that it only output to the file, nothing to screen. Why?
It seems the 'write' function was not called when I *print* something.
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Re: Issue of redirecting the stdout to both file and screen
I see. Many thanks to you! -- http://mail.python.org/mailman/listinfo/python-list
how to print the GREEK CAPITAL LETTER delta under utf-8 encoding
I lookup the utf-8 form of delta from the link. http://www.fileformat.info/info/unicode/char/0394/index.htm and then I want to print it in the python ( I work under windows) #!/usr/bin/python #coding=utf-8 print \xce\x94 but the result is not the 'delta' but an unknown character. -- http://mail.python.org/mailman/listinfo/python-list
Re: how to refer to partial list, slice is too slow?
I make a sample here for the more clearly explanation s = . - this is a large string data - ... def parser1(data) # do some parser ... # pass the remainder to next parser parser2(data[100:]) def parser2(data) # do some parser ... # pass the remainder to next parser parser3(data[100:]) def parser3(data) # do some parser ... # pass the remainder to next parser parser4(data[100:]) ... -- http://mail.python.org/mailman/listinfo/python-list
how to refer to partial list, slice is too slow?
I'm a python newbie. It seems the slice operation will do copy. for example: a = [1,2,3,4,5,6,7,8,9,0] b = a[7:] b [8, 9, 0] a.remove(9) a [1, 2, 3, 4, 5, 6, 7, 8, 0] b [8, 9, 0] if the list have large members, the slice operations will consume many times. for instance, I have a long string named it as S, the size is more than 100K I want to parser it one part-to-part. first, I process the first 100 byte, and pass the remainder to the next parser function. I pass the S[100:] as an argument of the next parser function. but this operation will cause a large bytes copy. Is there any way to just make a reference to the remainder string not copy? -- http://mail.python.org/mailman/listinfo/python-list
What's the life time of the variable defined in a class function?
Please see the followed example: class A: def __init__(self): pass class X: def __init__(self): n = 200 if True: j = 200 m = j k = A() print m, j a = X() # ?? what about the m, n and j? is it still alive? del a -- In C/C++, the life time of m,n and j was the nearest block. but obviously, python doen't have this syntax, but I would like to know that whether the life time of m, n, j is base on function range or the object range. We can not access the m, n, and j from the outside of class X. Now I'm writing a program base on the wxpython. In the __init__ function of wx.Panel, I use normal varable(just like the m,n and j) created some widgets. It could be show in the window. Does it indicated the life time of varable m,n,j is base on the object range? Sorry for my poor english! It seems -- http://mail.python.org/mailman/listinfo/python-list
Re: What's the life time of the variable defined in a class function?
On Apr 30, 5:20 pm, Diez B. Roggisch [EMAIL PROTECTED] wrote: 人言落日是天涯，望极天涯不见家 wrote: Please see the followed example: class A: def __init__(self): pass class X: def __init__(self): n = 200 if True: j = 200 m = j k = A() print m, j a = X() # ?? what about the m, n and j? is it still alive? del a -- In C/C++, the life time of m,n and j was the nearest block. but obviously, python doen't have this syntax, but I would like to know that whether the life time of m, n, j is base on function range or the object range. We can not access the m, n, and j from the outside of class X. Now I'm writing a program base on the wxpython. In the __init__ function of wx.Panel, I use normal varable(just like the m,n and j) created some widgets. It could be show in the window. Does it indicated the life time of varable m,n,j is base on the object range? Python has no variables. It has objects, which can be bound to names. Each binding to a name will increase a reference counter. Each unbinding will decrease it. so a = SomeObject() b = a del a will result in the SomeObject-instance still be alive. But when you add del b it will be garbage collected. Now in your example A() bound to k will not survive the exit of the method, as that means that k goes out of scope, and the object is bound to - the A-instance - gets its reference-counter decreased, resulting in it being freed. The wxwidgets example though is a different thing. If the panel stores a reference to the object, e.g. via a list (being part of a list or dict also increases the reference count), it will be kept around. Diez- Hide quoted text - - Show quoted text - Yes, I see. Many thanks for you ! -- http://mail.python.org/mailman/listinfo/python-list
Re: How do I parse a string to a tuple??
On Apr 30, 5:47 pm, Soren [EMAIL PROTECTED] wrote:
Hi!
I have a string that contains some text and newline characters. I want
to parse the string so that the string just before a newline character
goes in as an element in the tuple.
ex:
text1 \n text2 \n text3 \n text4 -- (text1, text2, text3, text4)
Is there an easy way to do this?
Thanks!,
Soren
tuple(text1 \n text2 \n text3 \n text4.split('\n'))
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Could zipfile module process the zip data in memory?
I made a C/S network program, the client receive the zip file from the server, and read the data into a variable. how could I process the zipfile directly without saving it into file. In the document of the zipfile module, I note that it mentions the file-like object? what does it mean? class ZipFile( file[, mode[, compression[, allowZip64]]]) Open a ZIP file, where file can be either a path to a file (a string) or a file-like object. Thanks! -- http://mail.python.org/mailman/listinfo/python-list
Re: Could zipfile module process the zip data in memory?
On Apr 29, 7:37 pm, Daniel Nogradi [EMAIL PROTECTED] wrote: I made a C/S network program, the client receive the zip file from the server, and read the data into a variable. how could I process the zipfile directly without saving it into file. In the document of the zipfile module, I note that it mentions the file-like object? what does it mean? class ZipFile( file[, mode[, compression[, allowZip64]]]) Open a ZIP file, where file can be either a path to a file (a string) or a file-like object. Yes it is possible to process the content of the zipfile without saving every file: [untested] from zipfile import ZipFile from StringIO import StringIO zipp = ZipFile( this_is_the_zip_file_from_your_server, 'r' ) for name in zipp.namelist( ): content = zipp.read( name ) s = StringIO( ) s.write( content ) # now the file 'name' is in 's' (in memory) # you can process it further # s.close( ) zipp.close( ) HTH, Daniel Thanks! Maybe my poor english makes you confusion:-). The client receive the zip file data from the server, and keep these data as a variable, not as a file in harddisk. such as zipFileData, but the first argument of the ZipFile is filename. I would like to let the ZipFile() open the file from zipFileData directly but not file in harddisk zipp = ZipFile( this_is_the_zip_file_from_your_server, 'r' ) ^ I don't have this file, all its data is in a variable. -- http://mail.python.org/mailman/listinfo/python-list
Re: Could zipfile module process the zip data in memory?
On Apr 29, 7:37 pm, Daniel Nogradi [EMAIL PROTECTED] wrote: I made a C/S network program, the client receive the zip file from the server, and read the data into a variable. how could I process the zipfile directly without saving it into file. In the document of the zipfile module, I note that it mentions the file-like object? what does it mean? class ZipFile( file[, mode[, compression[, allowZip64]]]) Open a ZIP file, where file can be either a path to a file (a string) or a file-like object. Yes it is possible to process the content of the zipfile without saving every file: [untested] from zipfile import ZipFile from StringIO import StringIO zipp = ZipFile( this_is_the_zip_file_from_your_server, 'r' ) for name in zipp.namelist( ): content = zipp.read( name ) s = StringIO( ) s.write( content ) # now the file 'name' is in 's' (in memory) # you can process it further # s.close( ) zipp.close( ) HTH, Daniel OK, I see, thank you! -- http://mail.python.org/mailman/listinfo/python-list
Re: Could zipfile module process the zip data in memory?
On Apr 29, 8:14 pm, Daniel Nogradi [EMAIL PROTECTED] wrote: I made a C/S network program, the client receive the zip file from the server, and read the data into a variable. how could I process the zipfile directly without saving it into file. In the document of the zipfile module, I note that it mentions the file-like object? what does it mean? class ZipFile( file[, mode[, compression[, allowZip64]]]) Open a ZIP file, where file can be either a path to a file (a string) or a file-like object. Yes it is possible to process the content of the zipfile without saving every file: [untested] from zipfile import ZipFile from StringIO import StringIO zipp = ZipFile( this_is_the_zip_file_from_your_server, 'r' ) for name in zipp.namelist( ): content = zipp.read( name ) s = StringIO( ) s.write( content ) # now the file 'name' is in 's' (in memory) # you can process it further # s.close( ) zipp.close( ) HTH, Daniel Thanks! Maybe my poor english makes you confusion:-). The client receive the zip file data from the server, and keep these data as a variable, not as a file in harddisk. such as zipFileData, but the first argument of the ZipFile is filename. I would like to let the ZipFile() open the file from zipFileData directly but not file in harddisk zipp = ZipFile( this_is_the_zip_file_from_your_server, 'r' ) ^ I don't have this file, all its data is in a variable. Well, as you correctly pointed out in your original post ZipFile can receive a filename or a file-like object. If the zip archive data is in zipFileData then you might do: from StringIO import StringIO from zipfile import ZipFile data = StringIO( ) data.write( zipFileData ) data.close( ) zipp = ZipFile( data ) . and continue in the same way as before. Daniel- Hide quoted text - - Show quoted text - Thanks all of your kindly help! -- http://mail.python.org/mailman/listinfo/python-list
ctypes: how to make a structure pointer to point to a buffer
first, I'm try the POINTER to convesion the pointer type. but failed.
class STUDENT(Structure):
_fields_ = [('name', c_int),
('id', c_int),
('addition',c_ubyte)]
buffer = c_byte * 1024
student_p = cast(buffer, POINTER(STUDENT))
The parameter of the POINTER must be ctypes type.
How could I attach the buffer pointer to the structure STUDENT ?
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Re: ctypes: how to make a structure pointer to point to a buffer
On Apr 23, 5:42 pm, Diez B. Roggisch [EMAIL PROTECTED] wrote:
人言落日是天涯，望极天涯不见家 wrote:
first, I'm try the POINTER to convesion the pointer type. but failed.
class STUDENT(Structure):
_fields_ = [('name', c_int),
('id', c_int),
('addition', c_ubyte)]
buffer = c_byte * 1024
student_p = cast(buffer, POINTER(STUDENT))
The parameter of the POINTER must be ctypes type.
How could I attach the buffer pointer to the structure STUDENT ?
I think it should work like this:
from ctypes import *
class STUDENT(Structure):
_fields_ = [('name', c_int),
('id', c_int),
('addition', c_ubyte)]
buffer = (c_byte * 1024)()
buffer_p = pointer(buffer)
student_p = cast(buffer_p, POINTER(STUDENT))
print student_p
Diez
yes, it should add the bracket
buffer = (c_byte * 1024)()
Thank you !
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How to generate a continuous string
How to generate a continuous string, like this aaa the number of characters is dynamic. Is there a module or function implement this string ? such as: duplicate_string(char, num) -- http://mail.python.org/mailman/listinfo/python-list
Re: Shutting down windows using win32api
On Apr 11, 7:49 pm, Tim Golden [EMAIL PROTECTED] wrote: On a whim, given the terseness of your post, I cut-and-pasted your subject line into Google, added python for good measure, and looked at the results. I suggest you might do the same. Granted, maybe this will raise more questions, but at least it shows willing :) TJG use ctypes module to execute the windows api. Shutdown windows can use the follow api BOOL ExitWindowsEx( UINT uFlags, DWORD dwReason ); about the usage, u can refer to the microsoft msdn -- http://mail.python.org/mailman/listinfo/python-list
Does python have the static function member like C++
I define the class like this: class AAA: counter = 0 def __init__(self): pass def counter_increase(): AAA.counter += 1 print couter now :, AAA.counter But how could I call the function counter_incrrease ? Thanks ! -- http://mail.python.org/mailman/listinfo/python-list
Re: Does python have the static function member like C++
On Apr 11, 11:19 am, 7stud [EMAIL PROTECTED] wrote: On Apr 10, 9:08 pm, 人言落日是天涯，望极天涯不见家 [EMAIL PROTECTED] wrote: I define the class like this: class AAA: counter = 0 def __init__(self): pass def counter_increase(): AAA.counter += 1 print couter now :, AAA.counter But how could I call the function counter_incrrease ? Thanks ! 1) class AAA: counter = 0 def __init__(self): pass @staticmethod def counter_increase(): AAA.counter += 1 print couter now :, AAA.counter AAA.counter_increase() 2) class AAA: counter = 0 def __init__(self): pass def counter_increase(): AAA.counter += 1 print couter now :, AAA.counter counter_increase = staticmethod(counter_increase) AAA.counter_increase() 3) class AAA: counter = 0 def __init__(self): pass def counter_increase(self): AAA.counter += 1 print couter now :, AAA.counter aaa = AAA() AAA.counter_increase(aaa) Many thanks for you! I've never heard of the staticmethod , that's great! -- http://mail.python.org/mailman/listinfo/python-list
Is there a simple function to generate a list like ['a', 'b', 'c', ... 'z']?
Is there a simple function to generate a list like ['a', 'b', 'c', ... 'z']? The range() just can generate the numeric list. -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a simple function to generate a list like ['a', 'b', 'c', ... 'z']?
On Apr 9, 4:35 pm, Michael Bentley [EMAIL PROTECTED] wrote: On Apr 9, 2007, at 3:29 AM, 人言落日是天涯，望极天涯不 见家 wrote: Is there a simple function to generate a list like ['a', 'b', 'c', ... 'z']? The range() just can generate the numeric list. import string list(string.lowercase) Thanks a lot! -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a simple function to generate a list like ['a', 'b', 'c', ... 'z']?
On Apr 9, 4:39 pm, Thomas Krüger [EMAIL PROTECTED] wrote: 人言落日是天涯，望极天涯不见家 schrieb: Is there a simple function to generate a list like ['a', 'b', 'c', ... 'z']? The range() just can generate the numeric list. There is: [ chr(i) for i in range(97, 123) ] Thomas Thanks you too! I'm a beginner of python. -- http://mail.python.org/mailman/listinfo/python-list
how to use the string '\\.\'
print r'\\.\' This line will cause error. I just want to print the \\.\ why the prefix character r isn't effective. Thanks! -- http://mail.python.org/mailman/listinfo/python-list
