I have the simple Python 3 expression:
if a:
b = a
else:
print('error')
However in my case, it just so happens that 'a' is equal to a very verbose but necessary regex search command (not relevant here.) Can I simplify the above code so that:
1) I don't need to assign a value to 'a'.
2) I don't need to state the value of 'a' twice in the expression.
Thanks!
In Python 3.8, named assignments will be introduced that let you do:
if a := complicated_expression:
b = a
The suggestion
b = a or b
is valid, but requires that b is assigned a value beforehand. Also it just works if the condition only checks for the truthiness of a
For a strict one-liner with an arbitrary condition, you can always use the singleton generator-trick:
b = next((a for a in (complex_expression for _ in '_') if condition(a)), None)
But that should be considered purely academical.
You can do:
b = a or False
if b:
print("okay")
else:
print("error")
where False can be any special value you can set yourself that can be used as a default value for b if a is falsy (provided that the condition of the if changes if the default value changes).
>>> import re
>>>
>>> def check(rgx, s):
>>> b = re.search(rgx, s) or False
>>> if b:
>>> print("okay")
>>> return b
>>> else:
>>> print("error")
>>>
>>> check('\W', 'Hello, world!')
okay
<re.Match object; span=(5, 6), match=','>
>>> check('\d', 'Hello, world!')
error
b = a or b raise a NameError for the b in the RHS?if with b.
b = a or b. But I would assign the result to a temporary variable.