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Need an OOP method for share variable between function?

I have a class and some functions. In the 'check_reflexive()' function, there is a variable called reflexive_list. I want to use this variable also in the 'antisymmetric' function.
I checked some examples about class but didn't find a specific example to solve this problem.
I'll be waiting for your advice. Hope you have a nice day
class MyClass():
def __init__(self):
def checkif_pair(k):
for a in k:
if a%2 == 0:
None
else:
return False
return True
def check_reflexive(k):
j = 0
z = 0
reflexive_list = []
while j < len(k):
i = 0
while i < len(k):
if k[i] == k[j]:
tup = k[j],k[i]
reflexive_list.append(tup)
i += 1
else:
None
j = j + 1
else:
None
print(reflexive_list)
if len(reflexive_list) == len(self.list1):
return True
else:
return False
def antisymmetric(k):
antisymettric_list = []
for b in k:
swap1 = b[0]
swap2 = b[1]
newtuple = (swap2, swap1)
antisymettric_list.append(newtuple)
for ü in reflexive_list:
if ü in antisymettric_list:
antisymettric_list.remove(ü)
else:
None
print(antisymettric_list)
for q in antisymettric_list:
if q in k:
print("The system is not Anti-Symmetric.")
break
print("The system is Anti-Symmetric.")
def transitive(k):
result = {}
for first, second in k:
result.setdefault(first, []).append(second)
print(result)
for a, b in k:
for x in result[b]:
if x in result[a]:
None
else:
print("There is no {} in the {}".format(x, result[a]))
return False
return True

You can just use reflexive_list as an instance variable. Just add a constructor where the variable is defined:
class MyClass():
def __init__(self):
self.reflexive_list = []
And everytime you want to use it inside the function, you use self.reflexive_list

Is there a way to simplify this if else block?

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)

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.

Remove unused variables in Python source code

The Question
Is there a straightforward algorithm for figuring out if a variable is "used" within a given scope?
In a Python AST, I want to remove all assignments to variables that are not otherwise used anywhere, within a given scope.
Details
Motivating example
In the following code, it is obvious to me (a human), that _hy_anon_var_1 is unused, and therefore the _hy_anon_var_1 = None statements can be removed without changing the result:
# Before
def hailstone_sequence(n: int) -> Iterable[int]:
while n != 1:
if 0 == n % 2:
n //= 2
_hy_anon_var_1 = None
else:
n = 3 * n + 1
_hy_anon_var_1 = None
yield n
# After
def hailstone_sequence(n: int) -> Iterable[int]:
while n != 1:
if 0 == n % 2:
n //= 2
else:
n = 3 * n + 1
yield n
Bonus version
Extend this to []-lookups with string literals as keys.
In this example, I would expect _hyx_letXUffffX25['x'] to be eliminated as unused, because _hyx_letXUffffX25 is local to h, so _hyx_letXUffffX25['x'] is essentially the same thing as a local variable. I would then expect _hyx_letXUffffX25 itself to be eliminated once there are no more references to it.
# Before
def h():
_hyx_letXUffffX25 = {}
_hyx_letXUffffX25['x'] = 5
return 3
# After
def h():
return 3
From what I can tell, this is somewhat of an edge case, and I think the basic algorithmic problem is the same.
Definition of "used"
Assume that no dynamic name lookups are used in the code.
A name is used if any of these are true in a given scope:
It is referenced anywhere in an expression. Examples include: an expression in a return statement, an expression on the right-hand side of an assignment statement, a default argument in a function definition, being referenced inside a local function definition, etc.
It is referenced on the left-hand side of an "augmented assignment" statement, i.e. it is an augtarget therein. This might represent "useless work" in a lot of programs, but for the purpose of this task that's OK and distinct from being an entirely unused name.
It is nonlocal or global. These might be useless nonlocals or globals, but because they reach beyond the given scope, it is OK for my purposes to assume that they are "used".
Please let me know in the comments if this seems incorrect, or if you think I am missing something.
Examples of "used" and "unused"
Example 1: unused
Variable i in f is unused:
def f():
i = 0
return 5
Example 2: unused
Variable x in f is unused:
def f():
def g(x):
return x/5
x = 10
return g(100)
The name x does appear in g, but the variable x in g is local to g. It shadows the variable x created in f, but the two x names are not the same variable.
Variation
If g has no parameter x, then x is in fact used:
def f():
x = 10
def g():
return x/5
return g(100)
Example 3: used
Variable i in f is used:
def f():
i = 0
return i
Example 4: used
Variable accum in silly_map and silly_sum is used in both examples:
def silly_map(func, data):
data = iter(data)
accum = []
def _impl():
try:
value = next(data)
except StopIteration:
return accum
else:
accum.append(value)
return _impl()
return _impl()
def silly_any(func, data):
data = iter(data)
accum = False
def _impl():
nonlocal accum, data
try:
value = next(data)
except StopIteration:
return accum
else:
if value:
data = []
accum = True
else:
return _impl()
return _impl()

The solution below works in two parts. First, the syntax tree of the source is traversed and all unused target assignment statements are discovered. Second, the tree is traversed again via a custom ast.NodeTransformer class, which removes these offending assignment statements. The process is repeated until all unused assignment statements are removed. Once this is finished, the final source is written out.
The ast traverser class:
import ast, itertools, collections as cl
class AssgnCheck:
def __init__(self, scopes = None):
self.scopes = scopes or cl.defaultdict(list)
#classmethod
def eq_ast(cls, a1, a2):
#check that two `ast`s are the same
if type(a1) != type(a2):
return False
if isinstance(a1, list):
return all(cls.eq_ast(*i) for i in itertools.zip_longest(a1, a2))
if not isinstance(a1, ast.AST):
return a1 == a2
return all(cls.eq_ast(getattr(a1, i, None), getattr(a2, i, None))
for i in set(a1._fields)|set(a2._fields) if i != 'ctx')
def check_exist(self, t_ast, s_path):
#traverse the scope stack and remove scope assignments that are discovered in the `ast`
s_scopes = []
for _ast in t_ast:
for sid in s_path[::-1]:
s_scopes.extend(found:=[b for _, b in self.scopes[sid] if AssgnCheck.eq_ast(_ast, b) and \
all(not AssgnCheck.eq_ast(j, b) for j in s_scopes)])
self.scopes[sid] = [(a, b) for a, b in self.scopes[sid] if b not in found]
def traverse(self, _ast, s_path = [1]):
#walk the ast object itself
_t_ast = None
if isinstance(_ast, ast.Assign): #if assignment statement, add ast object to current scope
self.traverse(_ast.targets[0], s_path)
self.scopes[s_path[-1]].append((True, _ast.targets[0]))
_ast = _ast.value
if isinstance(_ast, (ast.ClassDef, ast.FunctionDef, ast.AsyncFunctionDef)):
s_path = [*s_path, (nid:=(1 if not self.scopes else max(self.scopes)+1))]
if isinstance(_ast, (ast.FunctionDef, ast.AsyncFunctionDef)):
self.scopes[nid].extend([(False, ast.Name(i.arg)) for i in _ast.args.args])
_t_ast = [*_ast.args.defaults, *_ast.body]
self.check_exist(_t_ast if _t_ast is not None else [_ast], s_path) #determine if any assignment statement targets have previously defined names
if _t_ast is None:
for _b in _ast._fields:
if isinstance((b:=getattr(_ast, _b)), list):
for i in b:
self.traverse(i, s_path)
elif isinstance(b, ast.AST):
self.traverse(b, s_path)
else:
for _ast in _t_ast:
self.traverse(_ast, s_path)
Putting it all together:
class Visit(ast.NodeTransformer):
def __init__(self, asgn):
super().__init__()
self.asgn = asgn
def visit_Assign(self, node):
#remove assignment nodes marked as unused
if any(node.targets[0] == i for i in self.asgn):
return None
return node
def remove_assgn(f_name):
tree = ast.parse(open(f_name).read())
while True:
r = AssgnCheck()
r.traverse(tree)
if not (k:=[j for b in r.scopes.values() for k, j in b if k]):
break
v = Visit(k)
tree = v.visit(tree)
return ast.unparse(tree)
print(remove_assgn('test_name_assign.py'))
Output Samples
Contents of test_name_assign.py:
def hailstone_sequence(n: int) -> Iterable[int]:
while n != 1:
if 0 == n % 2:
n //= 2
_hy_anon_var_1 = None
else:
n = 3 * n + 1
_hy_anon_var_1 = None
yield n
Output:
def hailstone_sequence(n: int) -> Iterable[int]:
while n != 1:
if 0 == n % 2:
n //= 2
else:
n = 3 * n + 1
yield n
Contents of test_name_assign.py:
def h():
_hyx_letXUffffX25 = {}
_hyx_letXUffffX25['x'] = 5
return 3
Output:
def h():
return 3
Contents of test_name_assign.py:
def f():
i = 0
return 5
Output:
def f():
return 5
Contents of test_name_assign.py:
def f():
x = 10
def g():
return x/5
return g(100)
Ouptut:
def f():
x = 10
def g():
return x / 5
return g(100)

How to clean (reset) cache memory of my decorator?

I need to write decorator to remember function results for given parameters. I did it, but how to reset the memory? I tried to reset func.cache, but without correct result.
from random import random
def cached(func):
cache = func.cache = {}
func.call_num = 0
func.eval_num = 0
def memorised_func(*args, **kwargs):
key = str(args) + str(sorted(kwargs.items(), key=lambda kv: kv[1]))
#print(key)
if key not in cache:
cache[key] = func(*args, **kwargs)
func.eval_num += 1
func.call_num += 1
print(func.call_num, func.eval_num)
return cache[key]
memorised_func.cache_status = lambda: print(f"Function {func.__name__} called {func.call_num} times, evaluated {func.eval_num} times")
#memorised_func.cache_reset = lambda : func.cache #-TODO , help
return memorised_func
#cached
def foo(x, y=1, z=4):
return random()
#EXAMPLE THAT SHOULD WORK
print(foo(3) == foo(3))
print(foo(4) == foo(4))
print(foo(3, z=-1, y=3) == foo(3, y=3, z=-1))
print(foo(3) != foo(x=3))
a = foo(3)
foo.cache_reset() #HERE DOESN'T WORK - TODO
print(a != foo(3))
print(foo.cache_status() == 'Function foo called 10 times, evaluated 5 times')

the commented out section is simply:
memorised_func.cache_reset = lambda: func.cache.clear()
You have a strong crossover with memoize from the func library, consider looking in to that, too.

Why does passing a dictionary as a parameter take more time?

I tried to a leetcode problem. I find one of the following code throws a time limit exceeded error. I created the following testing code. I found the first one pass dictionary as a parameter takes more time the the other one. 0.94s vs 0.84s.
Can anyone explain this ?
class Solution(object):
def longestPalindromeSubseq(self, x):
"""
:type s: str
:rtype: int
"""
#dic = {}
def helper(s, dic):
if len(s) == 0:
return 0
if len(s) == 1:
return 1
if s in dic:
return dic[s]
if s[0] == s[-1]:
res = helper(s[1:-1], dic)+2
else:
l1 = helper(s[:-1], dic)
l2 = helper(s[1:], dic)
res = max(l1,l2)
dic[s] = res
#print (id(dic), dic)
return res
d = {}
ans = helper(x, d)
#print (id(d), d)
return ans
class Solution1(object):
def longestPalindromeSubseq(self, x):
"""
:type s: str
:rtype: int
"""
dic = {}
def helper(s):
if len(s) == 0:
return 0
if len(s) == 1:
return 1
if s in dic:
return dic[s]
if s[0] == s[-1]:
res = helper(s[1:-1])+2
else:
l1 = helper(s[:-1])
l2 = helper(s[1:])
res = max(l1,l2)
dic[s] = res
#print (id(dic), dic)
return res
ans = helper(x)
#print (id(dic), dic)
return ans
import time
if __name__ == "__main__":
x = "gphyvqruxjmwhonjjrgumxjhfyupajxbjgthzdvrdqmdouuukeaxhjumkmmhdglqrrohydrmbvtuwstgkobyzjjtdtjroqpyusfsbjlusekghtfbdctvgmqzeybnwzlhdnhwzptgkzmujfldoiejmvxnorvbiubfflygrkedyirienybosqzrkbpcfidvkkafftgzwrcitqizelhfsruwmtrgaocjcyxdkovtdennrkmxwpdsxpxuarhgusizmwakrmhdwcgvfljhzcskclgrvvbrkesojyhofwqiwhiupujmkcvlywjtmbncurxxmpdskupyvvweuhbsnanzfioirecfxvmgcpwrpmbhmkdtckhvbxnsbcifhqwjjczfokovpqyjmbywtpaqcfjowxnmtirdsfeujyogbzjnjcmqyzciwjqxxgrxblvqbutqittroqadqlsdzihngpfpjovbkpeveidjpfjktavvwurqrgqdomiibfgqxwybcyovysydxyyymmiuwovnevzsjisdwgkcbsookbarezbhnwyqthcvzyodbcwjptvigcphawzxouixhbpezzirbhvomqhxkfdbokblqmrhhioyqubpyqhjrnwhjxsrodtblqxkhezubprqftrqcyrzwywqrgockioqdmzuqjkpmsyohtlcnesbgzqhkalwixfcgyeqdzhnnlzawrdgskurcxfbekbspupbduxqxjeczpmdvssikbivjhinaopbabrmvscthvoqqbkgekcgyrelxkwoawpbrcbszelnxlyikbulgmlwyffurimlfxurjsbzgddxbgqpcdsuutfiivjbyqzhprdqhahpgenjkbiukurvdwapuewrbehczrtswubthodv"
print (x)
t0 = time.time()
sol = Solution()
print (sol.longestPalindromeSubseq(x))
t1 = time.time()
print(t1- t0)
sol1 = Solution1()
print (sol1.longestPalindromeSubseq(x))
t2 = time.time()
print(t2-t1)

Python uses something that is called call by sharing. The function gets only the alias on the parameter. With that in mind, it does't matter what you pass to the function.
But the script may take different time to execute. It is not constant. Using recursion makes it even harder to predict