Thanks for everyone's reply. I will explain here. Suppose there is a given matrix

    x                  y                B = [5,-4,5,-6]
[[0,0,0,0],        [[0,1,0,1],[0,0,0,0],         [0,0,0,0],[0,0,0,0],         [0,0,0,1],[0,0,0,0]]         [0,0,0,0]]

for example a feasible solution is [[0,4,0,1],[0,0,0,0],[0,0,0,5],[0,0,0,0]] 4+1-0 == 5 0-4 == -4 5-0 == 5 0 - 5-1 == -6

I want to update x to make sure:

 (1) if y[i][j] == 0:x[i][j] = 0(2) x[0][0]+x[0][1]+x[0][2]+x[0][3]-x[0][0]-x[1][0]-x[2][0]-x[3][0] = B[0]x[1][0]+x[1][1]+x[1][2]+x[1][3]-x[0][1]-x[1][1]-x[2][1]-x[3][1] = B[1]...

How to program to find the feasible x?

answer updated, i wrote some code to parse variables.

B = [5,-4,5,-6]
y = [[0,1,0,1],[0,0,0,0],[0,0,0,1],[0,0,0,0],]
x = []
for i, row in enumerate(y):temp = []for j, col in enumerate(row):if col != 0:temp.append(str(col) + '*x' + str(i) + str(j))else:temp.append(col)x.append(temp)#for one in x:
#    print oneequ = []
for i in xrange(4):temp1 = []temp2 = []for j in xrange(4):temp1.append(x[i][j])temp2.append(x[j][i])temp2.append(B[i])equ.append(tuple(temp1 + temp2))equtions = []
for one in equ:s = '%s + %s + %s + %s - %s - %s - %s - %s = %s' % oneequtions.append(s)for one in equtions:print oneimport re
from copy import deepcopyequ_bak = deepcopy(equtions)p_var = re.compile(r'x\d\d')
vars = set([])
for one in equ_bak:m = p_var.findall(one)vars |= set(m)
vars = sorted(list(vars))p_ef = re.compile(r'([+-]* *\d*)\*(x\d\d)')
effs = []
for one in equ_bak:m = p_ef.findall(one)#print mtemp = [0] * len(vars)for num, var in m:try:temp[vars.index(var)] = float(num.replace(' ', ''))except:passeffs.append(tuple(temp))#for one in effs:
#    print oneimport numpy as np
A = np.array(effs)
x = np.linalg.lstsq(A,B)
print vars
print x[0]