I get the following error when our program is run on Windows XP using ActivePerl 2.6 (although it runs fine on linux). Any ideas of what is going on? Is this a bug in the module or a problem with our code?
Error log:
File "C:\Python26\lib\site-packages\Scientific\IO\PDB.py", line 1163, in __ini
t__
self.parseFile(PDBFile(file_or_filename))
File "C:\Python26\lib\site-packages\Scientific\IO\PDB.py", line 1371, in parse
File
type, data = file.readLine()
File "C:\Python26\lib\site-packages\Scientific\IO\PDB.py", line 200, in readLi
ne
line = self.file.readline()
AttributeError: 'unicode' object has no attribute 'readline'
Exception AttributeError: "'unicode' object has no attribute 'close'" in <bound
method PDBFile.__del__ of <Scientific.IO.PDB.PDBFile instance at 0x00FD0440>> ig
nored
UPDATE: Following Joe's suggestion below, I get the following error instead:
Traceback (most recent call last):
File "NewParseV1_00.py", line 47, in <module>
PromptUser()
File "NewParseV1_00.py", line 45, in PromptUser
parseGroupFile(grpfilepath, outputPPYfilepath, sorcePDBfilepath)
File "NewParseV1_00.py", line 39, in parseGroupFile
return GeneratePPY(LL,GRPNAME,SRCPDB,OUTPPY)
File "NewParseV1_00.py", line 10, in __init__
self.PDBStruct = Scientific.IO.PDB.Structure(SRCPDB)
File "C:\Python26\lib\site-packages\Scientific\IO\PDB.py", line 1163, in __ini
t__
self.parseFile(PDBFile(file_or_filename))
File "C:\Python26\lib\site-packages\Scientific\IO\PDB.py", line 161, in __init
__
if isinstance(file_or_filename, basestr):
NameError: global name 'basestr' is not defined
Exception AttributeError: "PDBFile instance has no attribute 'open'" in <bound m
ethod PDBFile.__del__ of <Scientific.IO.PDB.PDBFile instance at 0x00FDB418>> ign
ored
I'm guessing, but it looks like a bug in Scientific.IO.PDB where they do something like:
if type(file_or_filename) == str:
file = open(file_or_filename)
else:
file = file_or_filename
rather than doing:
if isinstance(file_or_filename, basestr):
...
For whatever reason, you appear to be passing in a unicode filename on the windows side, and a raw string on the linux side.
Again, though, I'm just taking a guess. I'm assuming Scientific here is this module? http://dirac.cnrs-orleans.fr/plone/software/scientificpython/ (I'm not familiar with it...)
Edit: Yep!
Here's the relevant code from Scientific.IO.PDB:
class PDBFile:
"""
X{PDB} file with access at the record level
The low-level file access is handled by the module
L{Scientific.IO.TextFile}, therefore compressed files and URLs
(for reading) can be used as well.
"""
def __init__(self, file_or_filename, mode = 'r', subformat = None):
"""
#param file_or_filename: the name of the PDB file, or a file object
#type file_or_filename: C{str} or C{file}
#param mode: the file access mode, 'r' (read) or 'w' (write)
#type mode: C{str}
#param subformat: indicates a specific dialect of the PDB format.
Subformats are defined in
L{Scientific.IO.PDBExportFilters}; they are used
only when writing.
#type subformat: C{str} or C{NoneType}
"""
if isinstance(file_or_filename, str):
self.file = TextFile(file_or_filename, mode)
else:
self.file = file_or_filename
<snip>
It should be as simple as changing isinstance(file_or_filename, str) to isinstance(file_or_filename, basestr). You might want to file a bug report...
Changing the code in Scientific.IO.PDB as explained by Joe needs one final modification to work:
On line 161, change if isinstance(file_or_filename, str): to if isinstance(file_or_filename, basestring):
(Note that Joe wrote that the second argument should be basestr but that gives the error in the update in the original question.)
Related
I was going through this question : How do I return the definition of a class in python?
But I am unable to display the class definition. I am getting the below error:
>>> class A:
... pass
...
>>> import inspect
>>> source_text = inspect.getsource(A)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "C:\Users\**\Python\Python36\lib\inspect.py", line 968, in getsource
lines, lnum = getsourcelines(object)
File "C:\Users\**\Python\Python36\lib\inspect.py", line 955, in getsourcelines
lines, lnum = findsource(object)
File "C:\Users\**\Python\Python36\lib\inspect.py", line 768, in findsource
file = getsourcefile(object)
File "C:\Users\**\Python\Python36\lib\inspect.py", line 684, in getsourcefile
filename = getfile(object)
File "C:\Users\**\Python\Python36\lib\inspect.py", line 654, in getfile
raise TypeError('{!r} is a built-in class'.format(object))
TypeError: <module '__main__' (<_frozen_importlib_external.SourceFileLoader object at 0x0000026A79293F60>)> is a built-in class
>>>
Can someone please advise what am I doing wrong here? Thanks.
The inspect.getsource() function only works if there is a text file available to load the source code.
You typed the definition of the class into the interactive interpreter, which doesn't keep the original source around when compiling that source into in-memory class and code objects.
Put your class definition into a module, import the module, and then use inspect.getsource().
inspect.getsource() works by first finding the module for a given object (for classes, by looking at the ClassObj.__module__ attribute for the module name, then getting the module via sys.modules[modulename]) then seeing if the module has a __file__ attribute from which a readable source file can be determined. If there is such a filename and it can be read, then the inspect module reads that file to then search for the class ClassName: line and giving you all lines from that point on with same or deeper indentation. The interactive interpreter executes everything in the __main__ module and there is no __file__ attribute for the interpreter, so any attempts at loading source code for objects defined there will simply fail.
If you just wanted to know what members the class defines, use dir() or help() on the object instead. You don't need to see the full source code for that information.
How to get module source code by a python 'object' of that module?
class TestClass(object):
def __init__(self):
pass
def testMethod(self):
print 'abc'
return 'abc'
It's well-known that
print inspect.getsource(TestClass)
can be used to get source code of 'TestClass'.
However, the result of
ob = TestClass()
print inspect.getsource(ob)
as below, is not as expected.
Traceback (most recent call last):
File "D:\Workspaces\WS1\SomeProject\src\python\utils\ModuleUtils.py", line 154, in <module>
print inspect.getsource(ob)
File "C:\SciSoft\WinPython-64bit-2.7.10.3\python-2.7.10.amd64\lib\inspect.py", line 701, in getsource
lines, lnum = getsourcelines(object)
File "C:\SciSoft\WinPython-64bit-2.7.10.3\python-2.7.10.amd64\lib\inspect.py", line 690, in getsourcelines
lines, lnum = findsource(object)
File "C:\SciSoft\WinPython-64bit-2.7.10.3\python-2.7.10.amd64\lib\inspect.py", line 526, in findsource
file = getfile(object)
File "C:\SciSoft\WinPython-64bit-2.7.10.3\python-2.7.10.amd64\lib\inspect.py", line 420, in getfile
'function, traceback, frame, or code object'.format(object))
TypeError: <utils.TestClass.TestClass object at 0x0000000003C337B8> is not a module, class, method, function, traceback, frame, or code object
The question is:
If there is an object like 'ob' above, how to check the module source code of ob, or 'TestClass', via a method that takes 'ob' itself as a parameter?
In short, implement the following module
def getSource(obj):
###returns the result which is exactly identical to inspect.getsource(TestClass)
ob = TestClass()
###prints the result which is exactly identical to inspect.getsource(TestClass)
print getSource(ob)
(The scenario of a method like this is much more common than inspect.getsource(); for example, check the source code of an unknown, unpickled object.)
Instances don't have source code.
Use:
print inspect.getsource(type(ob))
or:
print inspect.getsource(ob.__class__)
I'm testing ebaysdk Python library that lets you connect to ebay. Now I'm trying examples from: https://github.com/timotheus/ebaysdk-python/
So far I got stuck at this example:
from ebaysdk.shopping import Connection as Shopping
shopping = Shopping(domain="svcs.sandbox.ebay.com", config_file="ebay.yaml")
response = shopping.execute('FindPopularItems',
{'QueryKeywords': 'Python'})
print response.disct()
When I run it. It gives me this error:
Traceback (most recent call last):
File "ebay-test.py", line 13, in <module>
{'QueryKeywords': 'Python'})
File "/usr/local/lib/python2.7/dist-packages/ebaysdk-2.1.0-py2.7.egg/ebaysdk/connection.py", line 123, in execute
self.error_check()
File "/usr/local/lib/python2.7/dist-packages/ebaysdk-2.1.0-py2.7.egg/ebaysdk/connection.py", line 193, in error_check
estr = self.error()
File "/usr/local/lib/python2.7/dist-packages/ebaysdk-2.1.0-py2.7.egg/ebaysdk/connection.py", line 305, in error
error_array.extend(self._get_resp_body_errors())
File "/usr/local/lib/python2.7/dist-packages/ebaysdk-2.1.0-py2.7.egg/ebaysdk/shopping/__init__.py", line 188, in _get_resp_body_errors
dom = self.response.dom()
File "/usr/local/lib/python2.7/dist-packages/ebaysdk-2.1.0-py2.7.egg/ebaysdk/response.py", line 229, in dom
return self._dom
File "/usr/local/lib/python2.7/dist-packages/ebaysdk-2.1.0-py2.7.egg/ebaysdk/response.py", line 216, in __getattr__
return getattr(self._obj, name)
AttributeError: 'Response' object has no attribute '_dom'
Am I missing something here or it could be some kind of bug in library?
Do you have a config file? I had a lot of problems getting started with this SDK. To get the yaml config file to work, I had to specify the directory that it was in. So in your example, it would be:
shopping = Shopping(domain="svcs.sandbox.ebay.com", config_file=os.path.join(os.path.dirname(os.path.realpath(__file__)), 'ebay.yaml'))
You should also be able to specify debug=true in your Shopping() declaration as in Shopping(debug=True).
Make sure if you have not, to specify your APP_ID and other necessary values in the config file.
You have the wrong domain, it should be open.api.sandbox.ebay.com. See this page on the ebaysdk github.
I am trying to read an XML file from the Yahoo finance API. So far, I've tried the following:
from xml.dom.minidom import parse
#Start Get Employees
xml = urllib.request.urlopen('https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.stocks%20where%20symbol%3D%22wfc%22&diagnostics=true&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys')
dom = parse(xml.read())
numemployees = dom.getElementsByTagName('FullTimeEmployees')
numemployees = name[0].firstChild.nodeValue
#End Get Employees
However, this raises an exception:
AttributeError: 'bytes' object has no attribute 'read'
I think this is because if it doesn't recognize the string, it assumes I'm passing a byte pattern. However, I am passing a string so I don't know what the problem is here.
Full Stack Trace:
Traceback (most recent call last):
File "C:\Python34\lib\tkinter\__init__.py", line 1487, in __call__
return self.func(*args)
File "C:\Users\kylec\Desktop\dm\Mail Server Finder\mailserverfinder.py", line 25, in getServers
dom = parse(xml.read())
File "C:\Python34\lib\xml\dom\minidom.py", line 1960, in parse
return expatbuilder.parse(file)
File "C:\Python34\lib\xml\dom\expatbuilder.py", line 913, in parse
result = builder.parseFile(file)
File "C:\Python34\lib\xml\dom\expatbuilder.py", line 204, in parseFile
buffer = file.read(16*1024)
AttributeError: 'bytes' object has no attribute 'read'
xml.dom.minidom.parse is excepting a file-like object, not a bytes or str, as stated in its documentation:
xml.dom.minidom.parse(filename_or_file[, parser[, bufsize]])
Return a Document from the given input. filename_or_file may be either
a file name, or a file-like object.
So you just need to do this:
dom = parse(xml)
Because the http.client.HTTPResponse object returned by urlopen is file-like.
Kyle, sorry but your example isn't clear enough. I think this is what you expected to do.
from xml.dom.minidom import parseString
employees = urllib.urlopen('https://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.stocks%20where%20symbol%3D%22wfc%22&diagnostics=true&env=store%3A%2F%2Fdatatables.org%2Falltableswithkeys').read()
dom = parseString(employees)
numemployees = dom.getElementsByTagName('FullTimeEmployees')
numemployees = numeemployees[0].firstChild.nodeValue
I'm working on a Python-programm that use the modul "pptx" (edit and manipulate powerpoint). The programm works without problem in the interpretor, but once I start it as a .exe file (built with cx_freeze or py2exe) I have this error-message when I click on the main-button :
Exception in Tkinter callback
Traceback (most recent call last):
File "Tkinter.pyc", line 1470, in __call__
File "pptx_02.py", line 187, in ButtonErstellen
File "pptx\api.pyc", line 29, in __init__
File "pptx\presentation.pyc", line 87, in __init__
File "pptx\presentation.pyc", line 170, in __open
File "pptx\packaging.pyc", line 88, in open
File "pptx\packaging.pyc", line 671, in __new__
PackageNotFoundError: Package not found at 'C:\Users\Moi\Programmation\Python\build\exe.win32-2.7\library.zip\pptx\templates\default.pptx'
The problem is that the file "default.pptx" actually exists at this adress.
While looking in the functions of "pptx", I may had an idea about the reason of this problem (but no idea of the solution).
For the last error : File "pptx\packaging.pyc", line 671, in new
is the following code:
class FileSystem(object):
"""
Factory for filesystem interface instances.
A FileSystem object provides access to on-disk package items via their URI
(e.g. ``/_rels/.rels`` or ``/ppt/presentation.xml``). This allows parts to
be accessed directly by part name, which for a part is identical to its
item URI. The complexities of translating URIs into file paths or zip item
names, and file and zip file access specifics are all hidden by the
filesystem class. |FileSystem| acts as the Factory, returning the
appropriate concrete filesystem class depending on what it finds at *path*.
"""
def __new__(cls, file):
# if *file* is a string, treat it as a path
if isinstance(file, basestring):
path = file
if is_zipfile(path):
fs = ZipFileSystem(path)
elif os.path.isdir(path):
fs = DirectoryFileSystem(path)
else:
raise PackageNotFoundError("Package not found at '%s'" % path)
else:
fs = ZipFileSystem(file)
return fs
The problem may be that the final file (default.pptx) isnot a zip but is in a zip. But I have no idea how I could fix it or if it realy is the problem.
If somebody as an idea ...
Thank you !