处理:如何只绘制每 x 帧?
Processing: How can I draw only every x frames?
我正在试验以下代码:
//3D Spectrogram with Microphone Input
//Modified by kylejanzen 2011 - https://kylejanzen.wordpress.com
//Based on script written by John Locke 2011 - http://gracefulspoon.com
//Output .DXF file at any time by pressing "r" on the keyboard
import processing.dxf.*;
import ddf.minim.analysis.*;
import ddf.minim.*;
import peasy.*;
PeasyCam cam;
FFT fftLin;
FFT fftLog;
Waveform audio3D;
Minim minim;
AudioInput microphone;
boolean record;
PFont font;
float camzoom;
float maxX = 0;float maxY = 0;float maxZ = 0;
float minX = 0;float minY = 0;float minZ = 0;
void setup()
{
size(1250,750,P3D); //screen proportions
noStroke();
minim = new Minim(this);
microphone = minim.getLineIn(Minim.STEREO, 4096); //repeat the song
cam = new PeasyCam(this, 0, 0, 0, 50);
cam.setMinimumDistance(50);
cam.setMaximumDistance(500);
background(255);
fftLog = new FFT(microphone.bufferSize(),microphone.sampleRate());
fftLog.logAverages(1,2); //adjust numbers to adjust spacing;
float w = float (width/fftLog.avgSize());
print(fftLog.avgSize());
float x = w;
float y = 0;
float z = 10;
float radius = 100;
audio3D = new Waveform(x,y,z,radius);
}
void draw()
{
background(0);
// ambientLight(102,102,102);
if (frameCount>0)
{
for(int i = 0; i < fftLog.avgSize(); i++){
float zoom = 1;
float jitter = (fftLog.getAvg(i)*2);
//println(jitter);
PVector foc = new PVector(audio3D.x+jitter, audio3D.y+jitter, 0);
PVector cam = new PVector(zoom, zoom, -zoom);
// camera(foc.x+cam.x+50,foc.y+cam.y+50,foc.z+cam.z+100,foc.x+30,foc.y+30,foc.z+100,0,0,1);
}
}
//play the song
fftLog.forward(microphone.mix);
audio3D.update();
// audio3D.textdraw();
if(record)
{
beginRaw(DXF, "output.dxf");
}
audio3D.plotTrace();
if(record)
{
endRaw();
record = false;
println("It's Done Bitches! Find your DXF!");
}
}
void stop()
{
// always close Minim audio classes when you finish with them
microphone.close();
// always stop Minim before exiting
minim.stop();
super.stop();
}
class Waveform
{
float x,y,z;
float radius;
PVector[] pts = new PVector[fftLog.avgSize()];
PVector[] trace = new PVector[0];
Waveform(float incomingX, float incomingY, float incomingZ, float incomingRadius)
{
x = incomingX;
y = incomingY;
z = incomingZ;
radius = incomingRadius;
}
void update()
{
plot();
}
void plot()
{
for(int i = 1; i < fftLog.avgSize(); i++)
{
int w = int(width/fftLog.avgSize());
x = i*w;
y = frameCount*5; // CHANGE THE SPEED
z = height/4-fftLog.getAvg(i)*4; //change multiplier to reduces height default '10'
stroke(0);
point(x, y, z);
pts[i] = new PVector(x, y, z);
//increase size of array trace by length+1
trace = (PVector[]) expand(trace, trace.length+1);
//always get the next to last
trace[trace.length-1] = new PVector(pts[i].x, pts[i].y, pts[i].z);
}
}
/* void textdraw()
{
for(int i =0; i<fftLog.avgSize(); i++){
pushMatrix();
translate(pts[i].x, pts[i].y, pts[i].z);
rotateY(PI/2);
rotateZ(PI/2);
fill(255,255);
text(round(fftLog.getAvg(i)*100),0,0,0);
popMatrix();
}
} */
void plotTrace()
{
stroke(255,100);
int inc = fftLog.avgSize();
for(int i=1; i<trace.length-inc; i++)
{
if(i%inc != 0)
{
beginShape(POINTS);
strokeWeight(2);
fill(0, 0, 0, 100);
vertex(trace[i].x, trace[i].y, trace[i].z);
vertex(trace[i-1].x, trace[i-1].y, trace[i-1].z);
vertex(trace[i+inc].x, trace[i+inc].y, trace[i+inc].z);
vertex(trace[i-1+inc].x, trace[i-1+inc].y, trace[i-1+inc].z);
endShape(CLOSE);
}
}
}
}
void keyPressed()
{
if (key == 'r') record = true;
}
我目前尝试实现的是减少(移动)(y-)轴上生成点的数量。似乎每一帧,都会产生一个点。
因此,我的问题很简单:我怎样才能让它只画,例如每5帧?我只是找不到管理它的值。
非常感谢。
您有三个选择:
选项 1:调用 frameRate()
函数以减少每秒绘制的帧数。
void setup(){
size(500, 500);
frameRate(5);
}
void draw(){
background(0);
ellipse(mouseX, mouseY, 20, 20);
}
选项 2: 使用 frameCount
variable and the modulus % operator
确定 X
帧何时过去。
void setup(){
size(500, 500);
}
void draw(){
if(frameCount % 5 == 0){
background(0);
ellipse(mouseX, mouseY, 20, 20);
}
}
选项 3:您可以创建自己的变量来存储已经过去的帧数。
int framesElapsed = 0;
void setup(){
size(500, 500);
}
void draw(){
framesElapsed++;
if(framesElapsed == 5){
background(0);
ellipse(mouseX, mouseY, 20, 20);
framesElapsed = 0;
}
}
请注意,对于简单的情况,这只是在做选项 2 中的模数运算符所做的事情。在那种情况下,模数可能更好。但是,如果您希望在不同的时间发生不同的事情,这将变得很有用。在那种情况下,您将有多个变量来跟踪您想要跟踪的任何内容的 "lifetime"。
但对于您的示例,选项 2 可能是最佳选项。
我正在试验以下代码:
//3D Spectrogram with Microphone Input
//Modified by kylejanzen 2011 - https://kylejanzen.wordpress.com
//Based on script written by John Locke 2011 - http://gracefulspoon.com
//Output .DXF file at any time by pressing "r" on the keyboard
import processing.dxf.*;
import ddf.minim.analysis.*;
import ddf.minim.*;
import peasy.*;
PeasyCam cam;
FFT fftLin;
FFT fftLog;
Waveform audio3D;
Minim minim;
AudioInput microphone;
boolean record;
PFont font;
float camzoom;
float maxX = 0;float maxY = 0;float maxZ = 0;
float minX = 0;float minY = 0;float minZ = 0;
void setup()
{
size(1250,750,P3D); //screen proportions
noStroke();
minim = new Minim(this);
microphone = minim.getLineIn(Minim.STEREO, 4096); //repeat the song
cam = new PeasyCam(this, 0, 0, 0, 50);
cam.setMinimumDistance(50);
cam.setMaximumDistance(500);
background(255);
fftLog = new FFT(microphone.bufferSize(),microphone.sampleRate());
fftLog.logAverages(1,2); //adjust numbers to adjust spacing;
float w = float (width/fftLog.avgSize());
print(fftLog.avgSize());
float x = w;
float y = 0;
float z = 10;
float radius = 100;
audio3D = new Waveform(x,y,z,radius);
}
void draw()
{
background(0);
// ambientLight(102,102,102);
if (frameCount>0)
{
for(int i = 0; i < fftLog.avgSize(); i++){
float zoom = 1;
float jitter = (fftLog.getAvg(i)*2);
//println(jitter);
PVector foc = new PVector(audio3D.x+jitter, audio3D.y+jitter, 0);
PVector cam = new PVector(zoom, zoom, -zoom);
// camera(foc.x+cam.x+50,foc.y+cam.y+50,foc.z+cam.z+100,foc.x+30,foc.y+30,foc.z+100,0,0,1);
}
}
//play the song
fftLog.forward(microphone.mix);
audio3D.update();
// audio3D.textdraw();
if(record)
{
beginRaw(DXF, "output.dxf");
}
audio3D.plotTrace();
if(record)
{
endRaw();
record = false;
println("It's Done Bitches! Find your DXF!");
}
}
void stop()
{
// always close Minim audio classes when you finish with them
microphone.close();
// always stop Minim before exiting
minim.stop();
super.stop();
}
class Waveform
{
float x,y,z;
float radius;
PVector[] pts = new PVector[fftLog.avgSize()];
PVector[] trace = new PVector[0];
Waveform(float incomingX, float incomingY, float incomingZ, float incomingRadius)
{
x = incomingX;
y = incomingY;
z = incomingZ;
radius = incomingRadius;
}
void update()
{
plot();
}
void plot()
{
for(int i = 1; i < fftLog.avgSize(); i++)
{
int w = int(width/fftLog.avgSize());
x = i*w;
y = frameCount*5; // CHANGE THE SPEED
z = height/4-fftLog.getAvg(i)*4; //change multiplier to reduces height default '10'
stroke(0);
point(x, y, z);
pts[i] = new PVector(x, y, z);
//increase size of array trace by length+1
trace = (PVector[]) expand(trace, trace.length+1);
//always get the next to last
trace[trace.length-1] = new PVector(pts[i].x, pts[i].y, pts[i].z);
}
}
/* void textdraw()
{
for(int i =0; i<fftLog.avgSize(); i++){
pushMatrix();
translate(pts[i].x, pts[i].y, pts[i].z);
rotateY(PI/2);
rotateZ(PI/2);
fill(255,255);
text(round(fftLog.getAvg(i)*100),0,0,0);
popMatrix();
}
} */
void plotTrace()
{
stroke(255,100);
int inc = fftLog.avgSize();
for(int i=1; i<trace.length-inc; i++)
{
if(i%inc != 0)
{
beginShape(POINTS);
strokeWeight(2);
fill(0, 0, 0, 100);
vertex(trace[i].x, trace[i].y, trace[i].z);
vertex(trace[i-1].x, trace[i-1].y, trace[i-1].z);
vertex(trace[i+inc].x, trace[i+inc].y, trace[i+inc].z);
vertex(trace[i-1+inc].x, trace[i-1+inc].y, trace[i-1+inc].z);
endShape(CLOSE);
}
}
}
}
void keyPressed()
{
if (key == 'r') record = true;
}
我目前尝试实现的是减少(移动)(y-)轴上生成点的数量。似乎每一帧,都会产生一个点。 因此,我的问题很简单:我怎样才能让它只画,例如每5帧?我只是找不到管理它的值。
非常感谢。
您有三个选择:
选项 1:调用 frameRate()
函数以减少每秒绘制的帧数。
void setup(){
size(500, 500);
frameRate(5);
}
void draw(){
background(0);
ellipse(mouseX, mouseY, 20, 20);
}
选项 2: 使用 frameCount
variable and the modulus % operator
确定 X
帧何时过去。
void setup(){
size(500, 500);
}
void draw(){
if(frameCount % 5 == 0){
background(0);
ellipse(mouseX, mouseY, 20, 20);
}
}
选项 3:您可以创建自己的变量来存储已经过去的帧数。
int framesElapsed = 0;
void setup(){
size(500, 500);
}
void draw(){
framesElapsed++;
if(framesElapsed == 5){
background(0);
ellipse(mouseX, mouseY, 20, 20);
framesElapsed = 0;
}
}
请注意,对于简单的情况,这只是在做选项 2 中的模数运算符所做的事情。在那种情况下,模数可能更好。但是,如果您希望在不同的时间发生不同的事情,这将变得很有用。在那种情况下,您将有多个变量来跟踪您想要跟踪的任何内容的 "lifetime"。
但对于您的示例,选项 2 可能是最佳选项。