I have been trying to create a dictionary with a string for each key and a lambda function for each value. I am not sure where I am going wrong but I suspect it is either my attempt to store a lambda in a dictionary in the first place, or the fact that my lambda is using a shortcut operator.
Code:
dict = { 'Applied_poison_rating_bonus': (lambda target, magnitude: target.equipmentPoisonRatingBonus += magnitude) }
The error being raised is SyntaxError: invalid syntax
and pointing right at my +=
. Are shortcut operators not allowed in lambdas, or am I even farther off track than I thought?
For the sake of sanity, I have omitted hundreds of very similar pairs (It isn't just a tiny dictionary.)
EDIT:
It seems my issue was with trying to assign anything within a lambda expression. Howver, my issue to solve is thus how can I get a method that only knows the key to this dictionary to be able to alter that field defined in my (broken) code?
Would some manner of call to eval() help?
EDIT_FINAL:
The functools.partial() method was recommended to this extended part of the question, and I believe after researching it, I will find it sufficient to solve my problem.
A lambda function is a small anonymous function. A lambda function can take any number of arguments, but can only have one expression.
In Python, everything is an object, so the dictionary can be passed as an argument to a function like other variables are passed.
Being anonymous, lambda functions can be easily passed without being assigned to a variable. Lambda functions are inline functions and thus execute comparatively faster.
In Python, we generally use Lambda Functions as an argument to a higher-order function (a function that takes in other functions as arguments). For Example, These are used together with built-in functions like filter(), map(), and reduce(), etc, which we will discuss later in this article.
You cannot use assignments in a expression, and a lambda
only takes an expression.
You can store lambdas in dictionaries just fine otherwise:
dict = {'Applied_poison_rating_bonus' : (lambda target, magnitude: target.equipmentPoisonRatingBonus + magnitude)}
The above lambda
of course only returns the result, it won't alter target.equimentPoisonRatingBonus
in-place.
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