from sympy import symbols, summation, gcdex, Piecewise, Eq, Abs, Mul, Le, mod_inverse, solve, invert, Mod, Max, log, \
floor, simplify, And, Ge
import random
x = symbols('x')
#just a placeholder function that I was trying to change how next() operates so it can go up by 1 each tine i call next()
def callFunc(args):
if args:
return args
return args
class DynamicIterator:
def __init__(self, func, a, b):
self.func = func # Function to generate values
self.counter = 0 # Keeps track of the current index
self.a = a
self.b = b
self.can_advance = True # Default to allowing automatic advancement
self.advanced = False # Flag to control when to advance
def __iter__(self):
return self
def __next__(self, ):
if not self.can_advance:
raise StopIteration # Stop iteration if cannot advance
value = self.func(self.counter, self.a, self.b) # Generate the value
self.counter += 1 # Increment the counter
return value
def generate_combinations(n, a, c):
range_size = c - a + 1
l = a + (n % range_size)
return l # Only return `l`
# Create an instance of the iterator
expressions = DynamicIterator(generate_combinations, 1, 2 ** 256 + 1)
def next_expr():
iterate = next(expressions)
return iterate
def formula109(l, integer):
random_number = random.randint(integer, integer+1) + l
random_number2 = random_number - l # i removed the x so you can see how the values are the same each time without me using x.
return random_number2
logic = formula109(x, callFunc(next_expr()))
total = summation(logic, (x, 1, 900000))
print(total)
#see what numbers are for each value of x
terms = [logic.subs(x, k) for k in range(1, 10)]
print(terms)
在运行此操作时，您可以看到，值一直保持在2。我知道，如果我将X添加到它，那将会上升，但这不是我想要的。我希望每次都可以每次使用next（）来迭代x的每个值。有人可以告诉我该怎么办吗？
输出为：
1800000
[2, 2, 2, 2, 2, 2, 2, 2, 2]
如果我将您的代码复制到
ipython会话中（就像我上个月一样），然后查看
logicIn [193]: logic = formula109(x, callFunc(next_expr()))
In [194]: logic
Out[194]:
2
是一个数字，而不是一个表达式（剥离
x脱落可以做到这一点）。 再次查看它不会改变价值。
In [195]: logic
Out[195]:
2
next查看制作
logic的组件：
In [196]: next_expr()
Out[196]: 2
In [197]: next_expr()
Out[197]: 3
In [198]: next_expr()
Out[198]: 4
ok，这是一个基础。
In [199]: callFunc(next_expr())
Out[199]: 5
In [200]: callFunc(next_expr())
Out[200]: 6
In [201]: callFunc(next_expr())
Out[201]: 7
In [202]: callFunc(next_expr())
Out[202]: 8
和
callFunc包装器一样。
In [203]: formula109(x, callFunc(next_expr()))
Out[203]:
9
In [204]: formula109(x, callFunc(next_expr()))
Out[204]:
10
In [205]: formula109(x, callFunc(next_expr()))
Out[205]:
12
In [206]: formula109(x, callFunc(next_expr()))
Out[206]:
13
和
formula109包装器。
In [207]: logic = formula109(x, callFunc(next_expr()))
In [208]: logic
Out[208]:
13
In [209]: logic = formula109(x, callFunc(next_expr()))
In [210]: logic
Out[210]:
15
新的分配获得了最新的数字。，但尝试用
logic迭代
sums不会再次分配
logicIn [211]: [logic.subs(x, k) for k in range(1, 4)]
Out[211]: [15, 15, 15]
In [212]: [logic.subs(x, k) for k in range(1, 4)]
Out[212]: [15, 15, 15]
在summation
之前是
Sum().doit()。
In [213]: Sum(logic, (x,1,5))
Out[213]:
Sum(15, (x, 1, 5))
In [214]: Sum(logic, (x,1,5))
Out[214]:
Sum(15, (x, 1, 5))
llet的put the返回中，所以
x是一个。
logic
sympy.Expr是
In [215]: def formula109(l, integer):
...: random_number = random.randint(integer, integer+1) + l
...: return random_number
...:
In [216]: logic = formula109(x, callFunc(next_expr()))
In [217]: logic
Out[217]:
x + 16
logicx：
x
1+16, 2+16, ...In [218]: [logic.subs(x, k) for k in range(1, 4)]
Out[218]: [17, 18, 19]
16和
callFunc(next_expr())中：
：
Sum现在，如果我抛弃
summationIn [219]: Sum(logic, (x,1,5))
Out[219]:
Sum(x + 16, (x, 1, 5))
In [220]: summation(logic, (x,1,5))
Out[220]:
95
，然后将您的“迭代器”作为普通的python：sympy
In [221]: [formula109(x, callFunc(next_expr())) for x in range(1,5)]
Out[221]: [17, 20, 22, 24]
In [222]: [formula109(x, callFunc(next_expr())) for x in range(1,5)]
Out[222]: [22, 24, 26, 28]
In [223]: callFunc(next_expr())
Out[223]: 24
In [224]: callFunc(next_expr())
Out[224]: 25
In [225]: [formula109(x, callFunc(next_expr())) for x in range(1,5)]
Out[225]: [27, 30, 32, 33]
每次称为步骤。 但是请注意，我必须将其包括在列表中。我不带有简单的价值分配
next_expr()如果我将其包含在功能或lambda中，我确实会踏得
In [226]: foobar = callFunc(next_expr()); foobar
Out[226]: 30
In [227]: [foobar for i in range(3)]
Out[227]: [30, 30, 30]
但是，如果我尝试将其包含在In [231]: foobar = lambda k: callFunc(next_expr())
In [232]: foobar(1)
Out[232]: 31
In [233]: foobar(1)
Out[233]: 32
In [234]: [foobar(i) for i in range(4)]
Out[234]: [33, 34, 35, 36]
In [235]: [foobar(i) for i in range(4)]
Out[235]: [37, 38, 39, 40]
中，我会遇到错误。 任何功能都可能是正确的。
Sum
我得出的结论与上个月相同。
In [239]: Sum(foobar, (x,1,5))
C:\Users\14256\miniconda3\lib\site-packages\sympy\concrete\expr_with_limits.py:26: SymPyDeprecationWarning:
The string fallback in sympify() is deprecated.
To explicitly convert the string form of an object, use
sympify(str(obj)). To add define sympify behavior on custom
objects, use sympy.core.sympify.converter or define obj._sympy_
(see the sympify() docstring).
sympify() performed the string fallback resulting in the following string:
'<function <lambda> at 0x000001B818FCE290>'
See https://docs.sympy.org/latest/explanation/active-deprecations.html#deprecated-sympify-string-fallback
for details.
This has been deprecated since SymPy version 1.6. It
will be removed in a future version of SymPy.
function = sympify(function)
ValueError: Error from parse_expr with transformed code: "<Symbol ('function' )<lambda >Symbol ('at' )Integer (0x000001B818FCE290 )>"
The above exception was the direct cause of the following exception:
File <string>:1
<Symbol ('function' )<lambda >Symbol ('at' )Integer (0x000001B818FCE290 )>
^
SyntaxError: invalid syntax
During handling of the above exception, another exception occurred:
SympifyError: Sympify of expression 'could not parse '<function <lambda> at 0x000001B818FCE290>'' failed, because of exception being raised:
SyntaxError: invalid syntax (<string>, line 1)
根本不像Python列表理解或
Sum/summationmap，但是
expr.subs(x)不能包含函数。 这不是一个
expr！
lambda在Python内（或顶部）运行。 它不会重写或重新定义Python语法或解释。