Why does adding the 'type' field to Argparse change it's behavior?
Question:
I have been following the example outlined in this previous question. But the behavior changes when I specify a type and I don’t understand why?
parser.add_argument('--bar', nargs='?', default=None, const=True)
args = parser.parse_args(['--bar=False'])
#Prints False as a string
print(args.bar)
parser.add_argument('--bar', nargs='?', default=None, const=True, type=bool)
args = parser.parse_args(['--bar=False'])
#Prints True as a bool
print(args.bar)
It’s not clear to me why in the first example ‘False’ overrides the const value of True but in the second example it does not?
Answers:
When you say "type=bool", the criterion becomes "is it present or not". You specified --bar, so it was present, and the result is True. If you leave it out, the result will be False.
Usually, that’s exactly what you want. You don’t want people to type --bar==True or --bar==False. You want either --bar or no bar.
The Documentation is often helpful…
By default, the parser reads command-line arguments in as simple strings. However, quite often the command-line string should instead be interpreted as another type, such as a float or int. The type keyword for add_argument() allows any necessary type-checking and type conversions to be performed.
Also, you should note that:
The bool() function is not recommended as a type converter. All it does is convert empty strings to False and non-empty strings to True. This is usually not what is desired.
I have been following the example outlined in this previous question. But the behavior changes when I specify a type and I don’t understand why?
parser.add_argument('--bar', nargs='?', default=None, const=True)
args = parser.parse_args(['--bar=False'])
#Prints False as a string
print(args.bar)
parser.add_argument('--bar', nargs='?', default=None, const=True, type=bool)
args = parser.parse_args(['--bar=False'])
#Prints True as a bool
print(args.bar)
It’s not clear to me why in the first example ‘False’ overrides the const value of True but in the second example it does not?
When you say "type=bool", the criterion becomes "is it present or not". You specified --bar, so it was present, and the result is True. If you leave it out, the result will be False.
Usually, that’s exactly what you want. You don’t want people to type --bar==True or --bar==False. You want either --bar or no bar.
The Documentation is often helpful…
By default, the parser reads command-line arguments in as simple strings. However, quite often the command-line string should instead be interpreted as another type, such as a float or int. The type keyword for add_argument() allows any necessary type-checking and type conversions to be performed.
Also, you should note that:
The bool() function is not recommended as a type converter. All it does is convert empty strings to False and non-empty strings to True. This is usually not what is desired.