Question

我的问题的实质如下：我在一个名为sol_expr的函数中构建一个字符串变量。当我打印（sol_expr）时，结果看起来与我需要的完全一样，即：

'%s = %s \\text{ or } %s' %(sym.latex(expr3), sym.latex(sol[0]),sym.latex(sol[1]))

但是，当我使用以下代码

display(Math('%s' %(sol_expr) ))

我得到的结果是：'

让我感到困惑的是，下面的代码正是我想要的：

sol_expr = '%s = %s \\text{ or } %s' %(sym.latex(expr3), sym.latex(sol[0]),sym.latex(sol[1]))
display(Math('%s' %(sol_expr) ))

我不明白区别。谁能发现盲点？

为完整起见，完整的代码

import sympy as sym
import numpy as np
from  IPython.display import display, Math, Latex
sym.init_printing() # allows the Latex rendering in Markdown cells


# function to construct the expression to be used with the Display function
def construct_sol_expr(nr_of_sols):
    # construct statement part
    dyn_expr = '%s = %s'
    for n in range(1,nr_of_sols):
        dyn_expr = dyn_expr + r' \\text{ or } %s'
    dyn_expr = "'%s'" %(sym.latex(dyn_expr))
    # construct parameter part
    par_expr = ''
    for p in range(0,nr_of_sols):
        # if there is more than 1 parameter neccessary, you need a comma
        if p != nr_of_sols-1:
                comma = ','
        else:
            comma = ''
        par_expr = par_expr + 'sym.latex(sol[' + str(p) + '])' + comma
    # tie it together
    sol_expr = dyn_expr + ' %(' + 'sym.latex(expr3), ' + par_expr + ')'
    return  sol_expr, dyn_expr, par_expr

q = sym.symbols('q')

expr1 = 3*q + 4/q + 3
expr2 = 5*q + 1/q + 1
expr3 = expr1 -expr2

sol = sym.solve(expr3, q)

sol_expr, dyn_expr, par_expr = construct_sol_expr(len(sol))
print(sol_expr)
#            '%s = %s \\text{ or } %s' %(sym.latex(expr3), sym.latex(sol[0]),sym.latex(sol[1]))

# Uncomment to see my issue
# sol_expr = '%s = %s \\text{ or } %s' %(sym.latex(expr3), sym.latex(sol[0]),sym.latex(sol[1]))

display(Math('%s' %(sol_expr) ))

Answer 1

这就是我想要完成的。

import sympy as sym
import numpy as np
from  IPython.display import display, Math, Latex
sym.init_printing() # allows the Latex rendering in Markdown cells


# Function to construct the expression to be used with the Display function

def construct_sol_expr(expr, sol):
# The trick is to build the latex statement right away instead if constructing it with %s elements

    # construct the dynamic solution(s) part of the latex expression
    dyn_expr = ''
    for d in range(0,len(sol)):
        # if there is more than 1 parameter neccessary, you need a separator

        if d != len(sol)-1:
                sprtr = '\\quad \\text{ or } \\quad '
        else:
            sprtr = ''
        sol_d = sym.latex(sol[d]) # get the dth entry in the solution list
        sol_Nd = sym.latex(sym.N(sol[d])) # get the dth entry in the solution list        
        dyn_expr = dyn_expr + sol_d + ' = ' + sol_Nd + sprtr

    # tie it together
    sol_expr = str(sym.latex(expr)) + '\\quad = \\quad' + str(dyn_expr) 

    return  sol_expr

q = sym.symbols('q')

expr1 = 3*q + 4/q + 3
expr2 = 5*q + 1/q + 1
expr3 = expr1 -expr2

sol = sym.solve(expr3, q)

sol_expr = construct_sol_expr(expr3, sol)

display(Math(sol_expr))

希望它对其他人也有帮助

Python-为什么字符串变量不能在函数中“解析”正确性？

1 个答案: