如何在python中打印100个随机数的“set”并将它们添加到空集（）(how to print 100 random numbers of “set” in python and add them to empty set())

 当你定义x(t) ，由于它依赖于t它最终成为符号函数（ symfun ）而不是符号对象。 然后将这种依赖性转移到矩阵N ，使其成为依赖于t的符号函数（这解释了为什么它只依赖于t而不是L ）。 
 
>> syms t x(t) L
>> N = ...
>> whos
   Name      Size            Bytes  Class     Attributes

   L         1x1               112  sym                 
   t         1x1               112  sym                 
   x         1x1               112  symfun  
   N         1x1               112  symfun 
 
 您可以通过上面的变通方法避免自动转换为symfun ，也可以在创建矩阵N时明确定义它，如下所示： 
 
>> N = sym(char([    0,                        0, ...
                     0,                        0;
                     0,                        0, ...
                     0,                        0;
                     1 - 3*(x/L)^2 + 2*(x/L)^3, -x + 2*x^2/L - x^3/(L^2), ...
                     3*(x/L)^2 - 2*(x/L)^3,        x^2/L - x^3/(L^2)]));
 
 这里的技巧是sym()和char()函数的组合使用。 如果您只使用sym()而不将矩阵转换为字符串，则无法使用。 
 
 话虽这么说，我个人发现你的第二种方法，你手动填充元素，以便更清晰，更容易阅读。 

When you define x(t) it ends up as a symbolic function (symfun) instead of a symbolic object due to its dependency on t. This dependency is then carried over to your matrix N, making it a symbolic function dependent on t (which explains why it is only dependent on t and not L). 
 
>> syms t x(t) L
>> N = ...
>> whos
   Name      Size            Bytes  Class     Attributes

   L         1x1               112  sym                 
   t         1x1               112  sym                 
   x         1x1               112  symfun  
   N         1x1               112  symfun 
 
You can avoid the automatic conversion to symfun by the workarounds you do above, or you can define it explicitly when you create you matrix N, like this:
 
>> N = sym(char([    0,                        0, ...
                     0,                        0;
                     0,                        0, ...
                     0,                        0;
                     1 - 3*(x/L)^2 + 2*(x/L)^3, -x + 2*x^2/L - x^3/(L^2), ...
                     3*(x/L)^2 - 2*(x/L)^3,        x^2/L - x^3/(L^2)]));
 
The trick here is the combined use of the sym() and char() functions. If you only use sym() without turning the matrix into a string it won't work.
 
That being said, I personally find your second approach where you manually fill the elements to be more clear and easier to read.