CODEWITHSUNDEEP
python
Valeriy Solovyov
Valeriy Solovyov

Reputation: 5648

How to pass named parameters to function after if Statements?

I have a library, which similar to this sample code:

class PlayBook(object):
    def __init__(self, playbook= None, host_list= None, module_path= None,):
        print "playbook %s host_list %s module_path %s" % (playbook,host_list,module_path)

I need chose named parameters, that I will pass to function: If I try to do this:

a=1
b=2
c=3

PlayBook(
    if a>1:
        module_path=a,
        playbook=b,
    elif b<1:
        playbook=b,
    else:    
        host_list=c,
        playbook="playbook",

)

I got:

    if a>1:
     ^
SyntaxError: invalid syntax

Question: how to choose named parameters and pass them to function?

PS: I know that I can do like this:

if a>1:
    PlayBook(module_path=a,
        playbook=b,)
elif b<1:
    PlayBook(  playbook=b,)
else:    
    PlayBook( host_list=c,
        playbook="playbook",)

Sorry for my English

Upvotes: 1

Views: 53

Answers (2)

vaultah
vaultah

Reputation: 46533

I'd collect the parameters into dict and unpack it afterwards:

if a > 1:
    params = dict(module_path=a, playbook=b)
elif b < 1:
    params = dict(playbook=b)
else:
    params = dict(host_list=c, playbook="playbook")

PlayBook(**params) # ** for keyword arguments

Upvotes: 2

Daniel Roseman
Daniel Roseman

Reputation: 599630

No, you can't do that. Move the instantiation into each clause.

if a>1:
  PlayBook(
    module_path=a,
    playbook=b,
  )
elif b<1:
  PlayBook(
    playbook=b,
  )
else:    
  PlayBook(
    host_list=c,
    playbook="playbook",
  )
)

Upvotes: 1

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