So I have this really long if else block and I was wondering if there is a way to simplify it?
def position(self, k): if self.b == None: self.b = k if (self.b).location_of(k) == 0: if self.aquire(0) == None: self.put(0, ABK(k)) else: self.aquire(0).insert(k) elif (self.b).location_of(k) == 1: if self.aquire(1) == None: self.put(1, ABK(k)) else: self.aquire(1).insert(k) elif (self.b).location_of(k) == 2: if self.aquire(2) == None: self.put(2, ABK(k)) else: self.aquire(2).insert(k) You could store the value of self.b.location_of(k) on a variable and use it later to call your functions.
def position(self, k): if self.b is None: self.b = k loc = self.b.location_of(k) if loc < 0 or loc > 2: return # Return in case the value is out of scope if self.aquire(loc) is None: self.put(loc, ABK(k)) else: self.aquire(loc).insert(k) Edit:
Olvin Roght pointed out that you can also save the result of the self.aquire(loc) call:
def position(self, k): if self.b is None: self.b = k loc = self.b.location_of(k) if loc < 0 or loc > 2: return aqu = self.aquire(loc) if aqu is None: self.put(loc, ABK(k)) else: aqu.insert(k) 
:= to make it even shorter and more readable.
As of python 3.8, you can achieve a result similar to the other answers in a shorter manner using the walrus operator (:=), more formally known as the assignment expression operator. This sort of inline assignment is similar to the way assignment can sometimes be done in nested ifs in C:
def position(self, k): if self.b is None: self.b = k if 0 <= (loc := self.b.location_of(k)) <= 2: if (val := self.aquire(loc)) is None: self.put(loc, ABK(k)) else: val.insert(k) In addition to the more obvious changes, I recommend using x is None rather than x == None, as it is the more idiomatic comparison.
