Question

我有一个会产生大量数据的程序。我想每秒绘制一次数据，以便我可以监控它的进度。在下面的示例中，我使用＆＃39; a＆＃39;创建了10个图表（每秒一个）。如果我绘制函数而不是数据点，这可以正常工作。

在b循环中，我想创建一组新的x-y数据然后绘制。我可以创建数据，但我无法弄清楚如何将其传递给gnuplot。

#include <iostream>
#include <stdio.h>
#include <cstdlib>
#include <unistd.h>     // usleep
using namespace std;

int main(){
// Code for gnuplot pipe

FILE *pipe = popen("gnuplot -persist", "w"); // open pipe to gnuplot
fprintf(pipe, "\n");

fprintf(pipe,"plot '-' using 1:2\n");  // so I want the first column to be x values, second column to be y

int b;
for (int a=0;a<10;a++) // 10 plots
{
    for (b=0;b<10;b++);  // 10 datapoints per plot
    {
        // this is the bit I can't get right:
        fprintf(pipe,"%d %d\n",a,b);    // passing x,y data pairs one at a time to gnuplot
    }
    fprintf(pipe,"e\n");    // finally, e
    fflush(pipe);   // flush the pipe to update the plot
    usleep(1000000);// wait a second before updating again
}

// Don't forget to close the pipe
fclose(pipe);
return 0;
}

编辑：下面的代码应该有效 - 每秒绘制一次10个数据点，持续10秒：

#include <iostream>
#include <stdio.h>
#include <cstdlib>
#include <unistd.h>     // usleep
using namespace std;

int main(){
// Code for gnuplot pipe
FILE *pipe = popen("gnuplot -persist", "w");

// set axis ranges
fprintf(pipe,"set xrange [0:11]\n");
fprintf(pipe,"set yrange [0:11]\n");

int b;
for (int a=0;a<10;a++) // 10 plots
{
    fprintf(pipe,"plot '-' using 1:2 \n");  // so I want the first column to be x values, second column to be y
    for (b=0;b<10;b++)  // 10 datapoints per plot
    {
        fprintf(pipe, "%d %d \n",a,b);  // passing x,y data pairs one at a time to gnuplot
    }
    fprintf(pipe,"e \n");    // finally, e
    fflush(pipe);   // flush the pipe to update the plot
    usleep(1000000);// wait a second before updating again
}

// Don't forget to close the pipe
fclose(pipe);
return 0;
}

Answer 1

首先，您需要将plot置于循环内，如：

for (int a=0;a<10;a++) // 10 plots
{
    fprintf(pipe,"plot '-' using 1:2\n");  // so I want the first column to be x values, second column to be y
    for (b=100;b<110;b++)  // 10 datapoints per plot
    {
        // this is the bit I can't get right:
        fprintf(pipe,"%d, %d\n",(a+1)*b,(a+1)*b*b);  // passing x,y data pairs one at a time to gnuplot
//              fprintf(pipe, "%d %d\n",b,a);  // passing x,y data pairs one at a time to gnuplot
    }
    fprintf(pipe,"e\n");    // finally, e
    fflush(pipe);   // flush the pipe to update the plot
    usleep(1000000);// wait a second before updating again
}

此外，您还可能需要设置xrange和yrange，然后才能看到实际情况：

FILE *pipe = popen("gnuplot -persist", "w"); // open pipe to gnuplot
fprintf(pipe,"set xrange [0:1115]\n");
fprintf(pipe,"set yrange [0:122500]\n");

更新没有来自c ++的文件的gnuplot数据集

1 个答案: