conditional print statement ?

[Stef Mientki]

hello,


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

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

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

thanks,
Stef Mientki

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

Re: conditional print statement ?

Stef Mientki schrieb:

hello,
>
>
As part of a procedure I've a number sequences like this:
>
<Python>
if Print_Info: print Datafile.readli ne()
else: Datafile.readli ne()
</Python>
>
Is there a more compressed way to write such a statement,
especially I dislike the redundancy "Datafile.readl ine()".

d=Datafile.read line()
if Print_info: print d

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

HTH,
Martin

[Terry Reedy]

Re: conditional print statement ?


"Stef Mientki" <S.Mientki-nospam@mailbox. kun.nlwrote in message
news:2c923$462f b3e0$d443bb3a$1 8429@news.speed linq.nl...
| if Print_Info: print Datafile.readli ne()
| else: Datafile.readli ne()

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.readl ine()".

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.readli ne # followed by
....
dread() # as needed

Terry Jan Reedy

[Antoon Pardon]

Re: conditional print statement ?

On 2007-04-25, Stef Mientki <S.Mientki-nospam@mailbox. kun.nlwrote:

hello,
>
>
As part of a procedure I've a number sequences like this:
>
><Python>
if Print_Info: print Datafile.readli ne()
else: Datafile.readli ne()
></Python>
>
Is there a more compressed way to write such a statement,
especially I dislike the redundancy "Datafile.readl ine()".
>
thanks,
Stef Mientki

You could consider the following

def Print(arg):
print arg

def Noop(arg):
pass

(Print if Print_Info else Noop) (Datafile.readl ine())

--
Antoon Pardon

[stef]

Re: conditional print statement ?

Antoon Pardon wrote:

On 2007-04-25, Stef Mientki <S.Mientki-nospam@mailbox. kun.nlwrote:
>

>hello,
>>
>>
>As part of a procedure I've a number sequences like this:
>>
><Python>
> if Print_Info: print Datafile.readli ne()
> else: Datafile.readli ne()
></Python>
>>
>Is there a more compressed way to write such a statement,
>especially I dislike the redundancy "Datafile.readl ine()".
>>
>thanks,
>Stef Mientki
>>

>
You could consider the following
>
def Print(arg):
print arg
>
def Noop(arg):
pass
>
(Print if Print_Info else Noop) (Datafile.readl ine())
>
>

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

cheers,
Stef Mientki

[Dustan]

Re: conditional print statement ?

On Apr 26, 1:58 am, Antoon Pardon <apar...@forel. vub.ac.bewrote:

On 2007-04-25, Stef Mientki <S.Mientki-nos...@mailbox. kun.nlwrote:
>

hello,

>

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

>

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

>

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

>

thanks,
Stef Mientki

>
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(Data file.readline() )

(Print if Print_Info else Noop) (Datafile.readl ine())
>
--
Antoon Pardon

[stef]

Re: conditional print statement ?

or (untested):
>
if Print_Info:
def printOrNot(arg) :
print arg
else:
def printOrNot(arg) :
pass
>
printOrNot(Data file.readline() )
>
>

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

cheers,
Stef Mientki

[Paul McGuire]

Re: conditional print statement ?

On Apr 26, 7:31 am, Dustan <DustanGro...@g mail.comwrote:

On Apr 26, 1:58 am, Antoon Pardon <apar...@forel. vub.ac.bewrote:
>
>
>
>
>

On 2007-04-25, Stef Mientki <S.Mientki-nos...@mailbox. kun.nlwrote:

>

hello,

>

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

>

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

>

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

>

thanks,
Stef Mientki

>

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(Data file.readline() )
>
>
>

(Print if Print_Info else Noop) (Datafile.readl ine())

>

--
Antoon Pardon- Hide quoted text -

>
- Show quoted text -- Hide quoted text -
>
- Show quoted text -

The Enable/Disable decorators on the Python wiki (http://
wiki.python.org/moin/PythonDecorator Library?highlig ht=%28decorator
%29#head-8298dbf9ac7325d 9ef15e7130e6763 78bbbda572) help you do
something very similar, without having to replicate the function being
enabled/disabled.

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

-- Paul

[Duncan Booth]

Re: conditional print statement ?

Paul McGuire <ptmcg@austin.r r.comwrote:

The Enable/Disable decorators on the Python wiki (http://
wiki.python.org/moin/PythonDecorator Library?highlig ht=%28decorator
%29#head-8298dbf9ac7325d 9ef15e7130e6763 78bbbda572) help you do
something very similar, without having to replicate the function being
enabled/disabled.
>
@(disabled,enab led)[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]

Re: conditional print statement ?

On Apr 27, 9:45 am, Duncan Booth <duncan.bo...@i nvalid.invalidw rote:

Paul McGuire <p...@austin.rr .comwrote:

The Enable/Disable decorators on the Python wiki (http://
wiki.python.org/moin/PythonDecorator Library?highlig ht=%28decorator
%29#head-8298dbf9ac7325d 9ef15e7130e6763 78bbbda572) help you do
something very similar, without having to replicate the function being
enabled/disabled.

>

@(disabled,enab led)[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,enabl ed)[debugFlag] # <-- proper way to do this

@state
def debugPrint(s):
print s

print "here comes some debug output"
debugPrint("xyz zy 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