每次x更改我的sympy求和时，我如何生成另一个字符串？

Question

    from sympy import summation, Piecewise, symbols, Eq, Ge,solve, Le
    import sys
    import itertools
    import string
    import random
    
    sys.setrecursionlimit(300000000)
 
    
    x = symbols('x') 
    
    # Create a dictionary to map numbers to characters
    def checksum1(args2):
        def generate_random_string(min_length, max_length):
            length = random.randint(min_length, max_length)
            characters = string.printable.strip()
            return ''.join(random.choices(characters, k=length))  #change this to generate a different random one for each x
    
    client = generate_random_string(60, 200)
        
        try:
         
            account = ord(client)
            return 1
        except Exception:
            return 0
    
    
    m = Piecewise(
        (checksum1(x), Eq(checksum1(x), 1)),
        (0, True)
    )
    
    result = summation(m, (x, 1, 900))
    print(result)

在代码中，我想在函数checksum1中每次使用其他随机字符串。它仅生成一个随机字符串，然后停止。 X的每个值都应具有不同的随机字符串，直到所有随机值都用尽。有人可以告诉我该怎么办吗？

具有代码的可运行版本

In [153]: # Create a dictionary to map numbers to characters
     ...: def checksum1(args2):
     ...:         def generate_random_string(min_length, max_length):
     ...:             length = random.randint(min_length, max_length)
     ...:             characters = string.printable.strip()
     ...:             str = ''.join(random.choices(characters, k=length))  #change this to generate a different random one for each x
     ...:             #print(str)
     ...:             return str
     ...:         client = generate_random_string(60, 100)
     ...:         print(client)
     ...:         try:
     ...:             account = ord(client)
     ...:             return 1
     ...:         except Exception:
     ...:             return 0
     ...:             

In [153]: 

In [154]: checksum1(x)
5G]7C(zqa<q|l2pHzOEc<ukA<sHM9R\uc+J')P'G~Hfq2F7]+/u/yLT^{]Z9<qvYL\YP
Out[154]: 0

In [155]: checksum1(x)
*IO\'37y:M*5c%,1*RN[>3Q#.#p<BsnFh\p-Y.=<AciA3cQ$U>U&2i557HhAF{SSOU(]4#/|E/rnIX]+N)V/q
Out[155]: 0

m

Answer 1

In [156]: m = Piecewise(
     ...: 
     ...:     (checksum1(x), Eq(checksum1(x), 1)),
     ...:     (0, True)
     ...: )
/`yN6G#rF7hx:Kwc?V%c+/VFvh(Ib~8.8bL!R'pb{7^c`GvSOq25Fx|.YS,c[6E"a>f8:W
9(Cx*%sM&Am/|x4A(Su#4kK;z@YwV<^gnFH*uT)jqM_3Yk:5y>o6\s%YB@y&$%vLZYp

In [157]: m
Out[157]: 
0

In [158]: type(m)
Out[158]: sympy.core.numbers.Zero

在

summation

中不“重新恢复”，更不用说运行了

generate_random_string

函数。

summation(m, (x, 1, 3)) 在您以前的问题时，我想知道您为什么要使用

sympy

？

没有此额外的代码，一个简单的列表理解测试多个随机字符串：PieceWise