我原本打算在R中创建匈牙利算法的实现，但对反应性函数的行为感到困惑，后者使用打印语句进行调试似乎是在自我调用。
不起作用的反应函数是stepthree（），我代码的粗略要点是将矩阵传递给反应函数，这些函数将执行匈牙利算法的步骤并返回结果矩阵。
以下是反应函数的代码段：
stepthree <- reactive({
num_lines <- number_of_lines();
print("Step three");
matrix_clean <- stepone(); #Taking a clean matrix from step one
matrix <- steptwo(); #Taking the matrix which has lines drawn on it
print("matrix_clean - clean matrix from step 1");
print(matrix_clean);
print("Matrix - dirty matrix from step 1");
print(matrix);
line_iteral <- data.frame(matrix(ncol=3, nrow = 2*ncol(matrix))); #create a matrix to store the number of zeros in each row and column
colnames(line_iteral) <- c("Position","Num","Type");
lines_to_draw <- lines(); #Create another matrix to store the positions of the lines (1 means line, 0 means No line); row 1 is for "rows", row 2 is for "columns"
print("Lines to draw");
print(lines_to_draw);
while(num_lines < ncol(matrix_clean)){
print("No. of lines not the same!");
smallest <- as.numeric(min(unlist(matrix))); #Take the smallest UNCOVERED value from the Matrix w lines on it (Gave the covered elements a very high value)
print("Smallest:");
print(smallest);
for(i in 1:nrow(matrix_clean)){ #Subtract the smallest value from UNCOVERED entries and add the least value to COVERED entries
for(j in 1:ncol(matrix_clean)){
if(lines_to_draw[1,i] == 0 & lines_to_draw[2,j]==0){
matrix_clean[i,j] <- matrix_clean[i,j] - smallest;
print(paste("Subtracted smallest value of: ", toString(smallest), " to row ", toString(i), "and column", toString(j)));
}
else if(lines_to_draw[1,i] == 1 & lines_to_draw[2,j] == 1){
matrix_clean[i,j] <- matrix_clean[i,j] + smallest;
print(paste("Added smallest value of: ", toString(smallest), " to row ", toString(i), "and column", toString(j)));
}
}
}
print("Saving matrix_clean");
matrix3(matrix_clean); #Store the untouched matrix in a reactive global var
lines_to_draw[1,] <- 0; #Reset the matrix which stores the position of the lines
lines_to_draw[2,] <- 0;
print("matrix_clean - edited matrix");
print(matrix_clean);
for(i in 1:nrow(matrix_clean)){
line_iteral[i,1] <- i;
line_iteral[i,2] <- sum(matrix_clean[i,] == 0);
line_iteral[i,3] <- c("Row");
} #Sum up all the zeros in the rows
for(j in 1:ncol(matrix_clean)){
line_iteral[j+ncol(matrix_clean),1] <- j;
line_iteral[j+ncol(matrix_clean),2] <- sum(matrix_clean[,j] == 0);
line_iteral[j+ncol(matrix_clean),3] <- c("Col");
} #Sum up all the zeros in the columns
total_zeros <- sum(line_iteral[1:(ncol(matrix)),2]);
print("line_iteral line 468: ");
print(line_iteral);
print("Total Zeros:");
print(total_zeros);
while(total_zeros > 0){
largest <- line_iteral %>% slice(which.max(Num));
#print(largest);
if(largest[3] == "Row"){
matrix_clean[as.numeric(largest[1]),] <- 99999999;
lines_to_draw[1,as.numeric(largest[1])] <- 1;
}
else{
matrix_clean[,as.numeric(largest[1])] <- 99999999;
lines_to_draw[2,as.numeric(largest[1])] <- 1;
}
for(i in 1:nrow(matrix_clean)){
line_iteral[i,1] <- i;
line_iteral[i,2] <- sum(matrix_clean[i,] == 0);
line_iteral[i,3] <- "Row"
} #Sum up all the zeros in the rows
for(j in 1:ncol(matrix_clean)){
line_iteral[j+ncol(matrix_clean),1] <- j;
line_iteral[j+ncol(matrix_clean),2] <- sum(matrix_clean[,j] == 0);
line_iteral[j+ncol(matrix_clean),3] <- "Col";
} #Sum up all the zeros in the columns
total_zeros <- sum(line_iteral[1:ncol(matrix_clean),2]);
print("Total Zeros");
print(total_zeros);
}
#print(line_iteral)
num_lines <- sum(lines_to_draw[1,]) + sum(lines_to_draw[2,]);
print("Number of lines:");
print(num_lines);
print("Lines to draw:");
print(lines_to_draw);
matrix <- matrix_clean;
matrix4(matrix_clean);
matrix_clean <- matrix3();
print("No of columns: ");
print(ncol(matrix_clean))
if(num_lines == ncol(matrix_clean)){
break;
}
print("Rinse and Repeat: Looping");
}
return(matrix_clean);
})
对于UI，我在输出部分仅调用一次“ stepthree（）”：
gg <- reactive({
ggplot <- ggplot(data=coordinates(),mapping = aes(x="x", y="y")) + geom_point(mapping = aes(x=coordinates()[,1], y=coordinates()[,2]));
return(ggplot);
})
output$plot <- renderPlot({
gg();
})
output$table <- renderTable({
coordinates();
})
output$dist_matrix_x <- renderTable({
distance_matrix_x();
})
output$dist_matrix_y <- renderTable({
distance_matrix_y();
})
output$combined_matrix <- renderTable({
combined_matrix();
})
output$stepone <- DT::renderDataTable({
DT::datatable(stepone(),options = list(lengthMenu = c(100,1000,10000), pageLength = 100));
})
output$steptwo <- DT::renderDataTable({
DT::datatable(transposed(), options = list(lengthMenu = c(100,1000,10000), pageLength = 100)) %>% formatStyle(colnames(lines_at_step_two())[lines_at_step_two()[2,]==1], backgroundColor = "yellow") %>% formatStyle(colnames(steptwo()), valueColumns = "transposed[, 1]", target = "row", backgroundColor = styleEqual(c(1,0), c("yellow","white")));
})
output$stepthree <- DT::renderDataTable({
DT::datatable(steptwo(), options = list(lengthMenu = c(100,1000,10000), pageLength = 100));
})
output$stepfour <- DT::renderDataTable({
DT::datatable(stepthree(), options = list(lengthMenu = c(100,1000,10000), pageLength = 100))
})
output$stepfive <- DT::renderDataTable({
DT::datatable(as.data.frame(matrix4()));
})
我已上传完整代码here
该程序的总体思想是： stepone（）
到目前为止，这部分工作正常。
steptwo（）： （尝试绘制尽可能少的行数以覆盖所有零）
（重复步骤3-6，直到总零= 0） } 6.通过在“ lines_to_draw”中找到条目的总和来求和，并将其保存在全局变量“ number_of_lines（）”中 7.返回“矩阵”
stepthree（）：（编辑矩阵，并重复执行以最少的行数覆盖零的过程，直到行数=列数
WHILE （在steptwo（）中绘制的行数小于矩阵中的列数）{
我在运行代码时遇到的问题是，反应性函数stepthree（）似乎将自己称为“ number_of_lines”和“ ncol（matrix_clean）”相等，因此使用print（）进行调试表明由于某种原因，再次调用整个step3，而不是返回“ matrix_clean”作为最终结果。我不确定响应函数的行为方式是否做错了，或者该错误是否源于系统设计算法运行方式的系统性问题。
在我刚开始使用R时，我会为代码混乱表示歉意。对您的任何帮助将不胜感激！