conditional print statement ?

[Stef Mientki]

hello,


As part of a procedure I've a number sequences like this:

<Python>
if Print_Info: print Datafile.readline()
else: Datafile.readline()
</Python>

Is there a more compressed way to write such a statement,
especially I dislike the redundancy "Datafile.readline()".

thanks,
Stef Mientki

 

[=?ISO-8859-1?Q?=22Martin_v=2E_L=F6wis=22?=]

hello,


As part of a procedure I've a number sequences like this:

<Python>
if Print_Info: print Datafile.readline()
else: Datafile.readline()
</Python>

Is there a more compressed way to write such a statement,
especially I dislike the redundancy "Datafile.readline()".

d=Datafile.readline()
if Print_info: print d

It's still two lines, but only has a single call to .readline().

HTH,
Martin

 

[Terry Reedy]

| if Print_Info: print Datafile.readline()
| else: Datafile.readline()

Since both branches discard the data read, I presume Martin's fix is what
you really want.

| Is there a more compressed way to write such a statement,
| especially I dislike the redundancy "Datafile.readline()".

But for future reference, if you really do need to call a method in
multiple places (or even just multiple times in a loop) you can condense
like so:

dread = Datafile.readline # followed by
....
dread() # as needed

Terry Jan Reedy

 

[Antoon Pardon]

hello,


As part of a procedure I've a number sequences like this:

<Python>
if Print_Info: print Datafile.readline()
else: Datafile.readline()
</Python>

Is there a more compressed way to write such a statement,
especially I dislike the redundancy "Datafile.readline()".

thanks,
Stef Mientki

You could consider the following

def Print(arg):
print arg

def Noop(arg):
pass

(Print if Print_Info else Noop) (Datafile.readline())

 

[stef]

You could consider the following

def Print(arg):
print arg

def Noop(arg):
pass

(Print if Print_Info else Noop) (Datafile.readline())

thank you all for your answers,
I'll play a little with the suggested solutions.

cheers,
Stef Mientki

 

[Dustan]

You could consider the following

def Print(arg):
print arg

def Noop(arg):
pass

or (untested):

if Print_Info:
def printOrNot(arg):
print arg
else:
def printOrNot(arg):
pass

printOrNot(Datafile.readline())

 

[stef]

or (untested):

if Print_Info:
def printOrNot(arg):
print arg
else:
def printOrNot(arg):
pass

printOrNot(Datafile.readline())

thanks for the creative solution, and indeed it does work ;-)

cheers,
Stef Mientki

 

[Paul McGuire]

or (untested):

if Print_Info:
def printOrNot(arg):
print arg
else:
def printOrNot(arg):
pass

printOrNot(Datafile.readline())




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The Enable/Disable decorators on the Python wiki (http://
wiki.python.org/moin/PythonDecoratorLibrary?highlight=%28decorator
%29#head-8298dbf9ac7325d9ef15e7130e676378bbbda572) help you do
something very similar, without having to replicate the function being
enabled/disabled.

@(disabled,enabled)[Print_Info]
def printOrNot(arg):
print arg

-- Paul

 

[Duncan Booth]

The Enable/Disable decorators on the Python wiki (http://
wiki.python.org/moin/PythonDecoratorLibrary?highlight=%28decorator
%29#head-8298dbf9ac7325d9ef15e7130e676378bbbda572) help you do
something very similar, without having to replicate the function being
enabled/disabled.

@(disabled,enabled)[Print_Info]
def printOrNot(arg):
print arg

Pardon me for asking, but isn't that a syntax error? Decorator syntax is:

"@" dotted_name ["(" [argument_list [","]] ")"] NEWLINE

and you don't have a dotted_name.

 

[Paul McGuire]

The Enable/Disable decorators on the Python wiki (http://
wiki.python.org/moin/PythonDecoratorLibrary?highlight=%28decorator
%29#head-8298dbf9ac7325d9ef15e7130e676378bbbda572) help you do
something very similar, without having to replicate the function being
enabled/disabled.

@(disabled,enabled)[Print_Info]
def printOrNot(arg):
print arg

Pardon me for asking, but isn't that a syntax error? Decorator syntax is:

"@" dotted_name ["(" [argument_list [","]] ")"] NEWLINE

and you don't have a dotted_name.

My bad. The wiki example assigns the appropriate decorator to another
name, and then uses that name, like this:

debugFlag = int(False)
state = (disabled,enabled)[debugFlag] # <-- proper way to do this

@state
def debugPrint(s):
print s

print "here comes some debug output"
debugPrint("xyzzy is the secret word")
print "that was it"


I think early in the decorator syntax discussions, there were some
proposals that decorators could be expressions, but I guess I forgot
which way that was decided. The example in this post does work (and
so does the one on the wiki) .

-- Paul