英文:
Replacing values in a string expression based on a matrix and iterating over columns
问题
在列X中，将有每一列j中的值的变量，这种情况下只有U1、X4和U2有值，属于列表['B', 'X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2']的其余变量都将具有值0
# 示例矩阵new_matrix = [['C', 'X', 'B', 'X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2'],[0.0, 'U1', 8, 2.0, 1.0, -1.0, 0, 0, 1.0, 0],['+M', 'X4', 2, 1.0, 1.0, 0, 1.0, 0, 0, 0],['+M', 'U2', 8, 1.0, 2.0, 0, 0, -1.0, 0, 1.0]]variables = ['X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2', 'B'] # 选择第一行（只有变量）variables_j_col_values = [['B', 'X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2'], []] # --> ['B', 'X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2']
问题是我需要创建以下变量的值矩阵（不使用库），其中我会有以下内容：
variables_j_col_values = [['B', 'X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2'],[0, 0, 0, 0, 2, 0, 8, 8], #列 new_matrix[][2][0, 0, 0, 0, 1.0, 0, 2.0, 1.0], #列 new_matrix[][3][0, 0, 0, 0, 1.0, 0, 1.0, 2.0], #列 new_matrix[][4][0, 0, 0, 0, 0, 0, -1.0, 0], #列 new_matrix[][5][0, 0, 0, 0, 1.0, 0, 0, 0], #列 new_matrix[][6][0, 0, 0, 0, 0, 0, 0, -1.0], #列 new_matrix[][7][0, 0, 0, 0, 0, 0, 1.0, 0], #列 new_matrix[][8][0, 0, 0, 0, 0, 0, 0, 1.0], ] #列 new_matrix[][9]
创建了variables_j_col_values后，继续替换funcion_obj_z字符串中的行值（除了variables_j_col_values数组的第0行，因为它是标题）。逻辑是使用循环遍历行，并执行.replace(new_matrix[][n], this_element)。
funcion_obj_z = 'Z = 3 * X1 + 2 * X2 + 0 * X3 + 0 * X4 + 0 * X5 + M * U1 + M * U2'
这样，如果在每个j迭代中打印j_func的值，则会在控制台中获得以下所需的正确输出：
# 对于循环，打印替换字符串中的 j 值j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 2 + 0 * 0 + M * 8 + M * 8' # 迭代 1j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 1.0 + 0 * 0 + M * 2.0 + M * 1.0' # 迭代 2j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 1.0 + 0 * 0 + M * 1.0 + M * 2.0' # 迭代 3j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 0 + 0 * 0 + M * -1.0 + M * 0' # 迭代 4j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 1.0 + 0 * 0 + M * 0 + M * 0' # 迭代 5j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 0 + 0 * 0 + M * 0 + M * -1.0' # 迭代 6j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 0 + 0 * 0 + M * 1.0 + M * 0' # 迭代 7j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 0 + 0 * 0 + M * 0 + M * 1.0' # 迭代 8
英文:
In column X are those variables that will have values in each column j, in this case only U1, X4 and U2 have values, the rest of the variables belonging to the list ['B', 'X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2'] will all have their values 0
#example matrixnew_matrix = [[ 'C', 'X', 'B', 'X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2'],[ 0.0, 'U1', 8, 2.0, 1.0, -1.0, 0, 0, 1.0, 0],['+M', 'X4', 2, 1.0, 1.0, 0, 1.0, 0, 0, 0],['+M', 'U2', 8, 1.0, 2.0, 0, 0, -1.0, 0, 1.0]]variables = ['X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2', 'B'] #select the first row (only variables)variables_j_col_values = [[variables.pop(variables.index('B'))] + variables, []] # --> ['B', 'X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2']
The problem with this is that I need create the following matrix of values of the variables (without using libraries) where I would have the following:
variables_j_col_values = [['B', 'X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2'],[ 0 , 0, 0, 0, 2, 0, 8, 8], #column new_matrix[][2][ 0 , 0, 0, 0, 1.0, 0, 2.0, 1.0], #column new_matrix[][3][ 0 , 0, 0, 0, 1.0, 0, 1.0, 2.0], #column new_matrix[][4][ 0 , 0, 0, 0, 0, 0, -1.0, 0], #column new_matrix[][5][ 0 , 0, 0, 0, 1.0, 0, 0, 0], #column new_matrix[][6][ 0 , 0, 0, 0, 0, 0, 0, -1.0], #column new_matrix[][7][ 0 , 0, 0, 0, 0, 0, 1.0, 0], #column new_matrix[][8][ 0 , 0, 0, 0, 0, 0, 0, 1.0], ] #column new_matrix[][9]
After create the variables_j_col_values, go replacing the values of the rows (except for row 0 of the variables_j_col_values array because it is a header) in the string inside funcion_obj_z
The logic would be to use a loop that goes through the rows, and does a
.replace(new_matrix[][n], this_element)
funcion_obj_z = 'Z = 3 * X1 + 2 * X2 + 0 * X3 + 0 * X4 + 0 * X5 + M * U1 + M * U2'
In this way, using said string as an expression, it would obtain these prints in the console if it printed the value of j_func in each j iteration. These would be the desired correct output:
#for loop, print the j string replacement the values in the stringj_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 2 + 0 * 0 + M * 8 + M * 8' #iteration 1j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 1.0 + 0 * 0 + M * 2.0 + M * 1.0' #iteration 2j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 1.0 + 0 * 0 + M * 1.0 + M * 2.0' #iteration 3j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 0 + 0 * 0 + M * -1.0 + M * 0' #iteration 4j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 1.0 + 0 * 0 + M * 0 + M * 0' #iteration 5j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 0 + 0 * 0 + M * 0 + M * -1.0' #iteration 6j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 0 + 0 * 0 + M * 1.0 + M * 0' #iteration 7j_func = 'Z = 3 * 0 + 2 * 0 + 0 * 0 + 0 * 0 + 0 * 0 + M * 0 + M * 1.0' #iteration 8
答案1
得分: 0
尽管这个解决方案不够美观，但它应该为您提供所需的转换：
new_matrix = [['C', 'X', 'B', 'X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2'],[0.0, 'U1', 8, 2.0, 1.0, -1.0, 0, 0, 1.0, 0],['+M', 'X4', 2, 1.0, 1.0, 0, 1.0, 0, 0, 0],['+M', 'U2', 8, 1.0, 2.0, 0, 0, -1.0, 0, 1.0]]variables = ['X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2', 'B']# 创建一个空矩阵variables_j_col_values = [[0 for _ in range(len(variables))] for _ in range(len(new_matrix[0])-1)]# 根据new_matrix的标头替换第一行的变量variables_j_col_values[0] = sorted(variables, key=lambda x: new_matrix[0].index(x))# 循环遍历所有值行for row in new_matrix[1:]:# 根据variables_j_col_values的第一行获取正确的列col = variables_j_col_values[0].index(row[1])# 将值和行进行配对并进行相应的更新for val, target in zip(row[2:], variables_j_col_values[1:]):target[col] = val
英文:
Albeit an ugly solution, this should give you the transformation you need:
new_matrix = [[ 'C', 'X', 'B', 'X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2'],[ 0.0, 'U1', 8, 2.0, 1.0, -1.0, 0, 0, 1.0, 0],['+M', 'X4', 2, 1.0, 1.0, 0, 1.0, 0, 0, 0],['+M', 'U2', 8, 1.0, 2.0, 0, 0, -1.0, 0, 1.0]]variables = ['X1', 'X2', 'X3', 'X4', 'X5', 'U1', 'U2', 'B']# create empty matrixvariables_j_col_values = [[0 for _ in range(len(variables))] for _ in range(len(new_matrix[0])-1)]# replace first row with sorted variables based on new_matrix headersvariables_j_col_values[0] = sorted(variables, key=lambda x: new_matrix[0].index(x))# loop over all value rowsfor row in new_matrix[1:]# get correct column in variables_j_col_values basedcol = variables_j_col_values[0].index(row[1])# zip the values and rows and update accordinglyfor val, target in zip(row[2:], variables_j_col_values[1:]):target[col] = val
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论