Possible to import a module whose name is contained in a variable?

[Steven Reddie]

Hi,

I want to do something like the following, which doesn't work:

modulename = 'module'
import modulename

The error is that there is no module named 'modulename'. Is there a
way to get that variable expanded?

Regards,

Steven

 

[Hye-Shik Chang]

Hi,

I want to do something like the following, which doesn't work:

modulename = 'module'
import modulename

The error is that there is no module named 'modulename'. Is there a
way to get that variable expanded?

modulename = 'module'
module = __import__(modulename)


Hye-Shik

 

[Steven Bethard]

modulename = 'module'
cmd = 'import '+modulename
exec(cmd)

Check also the thread:
How do I import everything in a subdir?
in THIS newsgroup.

And note that it tells you to use __import__, not exec. =)

STeVe

 

[Claudio Grondi]

Hi,

I want to do something like the following, which doesn't work:

modulename = 'module'
import modulename

The error is that there is no module named 'modulename'. Is there a
way to get that variable expanded?

Regards,

Steven

modulename = 'module'
cmd = 'import '+modulename
exec(cmd)

Check also the thread:
How do I import everything in a subdir?
in THIS newsgroup.

Claudio
P.S.

MODULES = [ 'module1', 'module2' ]

def libinfo():
for m in MODULES:
__import__('libinfo.'+m)
m.libinfo()
CFLAGS+=m.CFLAGS

indentation error?

 

[Heiko Wundram]

I try to avoid using any of the
__xxxx__() functions if possible
(considering this a good
programming style).

This is never good style, at least in the case of exec. exec is evil.

What works (beware that the below code is nevertheless untested and might
contain little warts) and is the "usual" and clean way to go:

### libinfo/__init__.py

# Empty.

### libinfo/Module1.py

def libinfo():
return "I am module1."

CFLAGS = ["-DMODULE1"]

### libinfo/Module2.py

def libinfo():
return "I am module2."

CFLAGS = ["-DMODULE2"]

### Importer.py

modules = {}
CFLAGS = []

def load_modules(to_load=["module1","module2"]):
global modules, CFLAGS

for mod in to_load:
try:
modules[mod] = getattr(__import__("libinfo.%s" % mod),mod)
except ImportError:
print "Could not load %s." % mod
continue
print "Module: %s (%r)." % (mod,modules[mod])
print "Modules libinfo: %r." % modules[mod].libinfo()
CFLAGS += modules[mod].CFLAGS

print "Total CFLAGS: %s." % CFLAGS

if __name__ == "__main__":
load_modules()
print "Module container: %s." % modules

### End Importer.py

HTH!

--
--- Heiko.

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[Michael Hoffman]

STeVe,

may I ask you for more details on this?
Any disadvantages while using exec()
in this context?

I try to avoid using any of the
__xxxx__() functions if possible
(considering this a good
programming style).

Avoiding exec (which is a statement, not a function) is much more
important. Since it executes arbitrary code, you can get unpredictable
results from it.
w00t

Consider the case where z = "shutil; shutil.rmtree('/')"

 

[Claudio Grondi]

STeVe,

may I ask you for more details on this?
Any disadvantages while using exec()
in this context?

I try to avoid using any of the
__xxxx__() functions if possible
(considering this a good
programming style).

Claudio

 

[Ed Leafe]

Avoiding exec (which is a statement, not a function) is much more
important. Since it executes arbitrary code, you can get unpredictable
results from it.

Is there any way to use __import__ to replace the following:

exec("from %s import *" % modulename)

___/
/
__/
/
____/
Ed Leafe
http://leafe.com/
http://dabodev.com/
Come to PyCon!!!! http://www.python.org/pycon/2005/

 

[Gerrit Holl]

Is there any way to use __import__ to replace the following:

exec("from %s import *" % modulename)

No.

Gerrit.

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[Peter Hansen]

> No.

Between the following two recipes, it seems unlikely that a simple
"No" is really correct:

http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/223972

http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/307772

A variation on some of the code therein should certainly be
able to do the equivalent of "from xx import *". (The addition
of support for __all__ is at least required for full emulation.)

-Peter

 

[Michael Hoffman]

Is there any way to use __import__ to replace the following:

exec("from %s import *" % modulename)

I shouldn't do this but:

module = __import__(modulename)
try:
for key in module.__all__:
globals()[key] = module[key]
except AttributeError:
globals().update(module.__dict__)

But really, you shouldn't be using from x import * as it
has unpredictable results as well.

http://www.python.org/doc/faq/progr...e-best-practices-for-using-import-in-a-module