Question

我想用三个或更多变量求解线性方程。在python中有一个很好的库吗？

Answer 1

是的，非常受欢迎的NumPy包有a function to do this。他们的例子：

求解方程组3 * x0 + x1 = 9和x0 + 2 * x1 = 8：
>>> import numpy as np
>>> a = np.array([[3,1], [1,2]])
>>> b = np.array([9,8])
>>> x = np.linalg.solve(a, b)
>>> x
array([ 2.,  3.]) 
https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.linalg.solve.html

Answer 2

请参阅http://sympy.org/和http://numpy.scipy.org/。

具体而言，http://docs.scipy.org/doc/numpy/reference/routines.linalg.html

http://docs.sympy.org/0.7.0/tutorial.html#algebra，http://docs.sympy.org/dev/modules/solvers/solvers.html

编辑：从评论添加了求解器链接。

Answer 3

你可以在python中使用最小二乘法来求解方程组例如，求解方程 3x + 4y = 7 和 5x + 6y = 8

>>> import numpy
>>> a=[[3,4],[5,6]]
>>> b=[7,8]
>>> numpy.linalg.lstsq(a,b)
(array([-5. ,  5.5]), array([], dtype=float64), 2, array([ 9.27110906,  0.21572392]))

Answer 4

使用@Jeremy的示例：

from sympy import *
x0, x1 = symbols(['x0', 'x1'])
sol = solve([3 * x0 + x1 - 9, x0 + 2 * x1 - 8], [x0, x1])
print(sol)

输出：

{x0：2，x1：3}

使用@ 004示例，符号略有不同：

from sympy import *
x, y = symbols(['x', 'y'])
system = [
    Eq(3*x + 4*y, 7),
    Eq(5*x + 6*y, 8)
]
soln = solve(system, [x, y])
print(soln)

{x：-5，y：11/2}

注意：有时人们可能会看到以下符号符号：x, y = symbols('x, y')，它似乎不像Python。

Answer 5

您还可以使用lsq_linear在x上添加约束（上限和下限）： scipy.optimize.lsq_linear

Answer 6

您可以编写一个简单的函数来求解线性方程组。

def solve(equations):
     #the constants of a system of linear equations are stored in a list for each equation in the system
     """
     for example the system below:
          2x+9y-3z+7w+8=0
          7x-2y+6z-1w-10=0
          -8x-3y+2z+5w+4=0
          0x+2y+z+w+0=0
     is expressed as the list:
          [[2,9,-3,7,8],[7,-2,6,-1,-10],[-8,-3,2,5,4],[0,2,1,1,0]]
     """
     lists=[] # I failed to name it meaningfully
     for eq in range(len(equations)):
          #print "equations 1", equations
          #find an equation whose first element is not zero and call it index
          index=0
          for i in range(len(equations)):
               if equations[i][0]<>0:
                    index=i;
                    break;
          #print "index "+str(eq)+": ",index
          #for the equation[index] calc the lists next itam  as follows
          lists.append([-1.0*i/equations[index][0] for i in equations[index][1:]])
          #print "list"+str(eq)+": ", lists[-1]
          #remve equation[index] and modify the others
          equations.pop(index)
          for i in equations:
               for j in range(len(lists[-1])):
                    i[j+1]+=i[0]*lists[-1][j]
               i.pop(0)

     lists.reverse()

     answers=[lists[0][0]]
     for i in range(1,len(lists)):
          tmpans=lists[i][-1]
          for j in range(len(lists[i])-1):
               tmpans+=lists[i][j]*answers[-1-j]
          answers.append(tmpans)
     answers.reverse()
     return answers

有没有python模块来解决线性方程？

6 个答案: