I have no idea what the difference is between these two codes. When I rum those codes in the scoring python code, it marks mine wrong. I would appreciate it if you can tell me the different between using a variable to make a new empty list and appending values verses just making a list with values in it. The first code is mine and the second code is the code from the answer sheet. Thank you very much :)
class Server:
def __init__(self):
self.list = []
def makeOrder(self,orderNum, orderList):
existance = False
for order in self.list:
if order[0]==orderNum:
existance = True
if existance == True:
return -1
else:
self.list.append([orderNum,orderList])
return [orderNum,orderList]
class Server:
def __init__(self):
self.q = []
# 1
def makeOrder(self, orderNumber, orderList):
existAlready = False
for order in self.q:
if order[0] == orderNumber:
existAlready = True
if existAlready == True:
return -1
else:
tmp = []
tmp.append(orderNumber)
tmp.append(orderList)
self.q.append(tmp)
return tmp
CodePudding user response:
Functionally, these are both very similar. The only difference of substance I see (other than obvious variable names, etc) is in the return value.
In the first option, you can see that one list is appended to self.list
, and a new (albeit identical in value) list is returned. This means that this is a different object.
self.list.append([orderNum,orderList])
return [orderNum,orderList]
However in the second option, you can clearly see that tmp
is both pushed AND returned:
tmp = []
tmp.append(orderNumber)
tmp.append(orderList)
self.q.append(tmp)
return tmp
This is a single object that gets appended and returned.
Fundamentally, this mens that any modification to the returned list in the second option will be reflected inside self.q
, while in the first option, the returned value is wholly independent. If you wanted to more or less replicate the behavior of option 1, you can change:
return tmp
with
return list(tmp)
Although keep in mind that if orderList
is itself a mutable list, the same behavior will occur if you modify the returned value's [1]
element. Since it is a reference to a data structure, modification of that list will also affect self.q
(or self.list
in the first option)
Example:
>>> class Foo:
... def __init__(self):
... self.q = []
... def bar(self, i, l):
... tmp = [i, l]
... self.q.append(tmp)
... return list(tmp)
...
>>> f = Foo()
>>> f.q
[]
>>> x = f.bar(1, [1,2,3])
>>> x
[1, [1, 2, 3]]
>>> x[1].append(4)
>>> x
[1, [1, 2, 3, 4]]
>>> f.q # it changed here as well
[[1, [1, 2, 3, 4]]]