For loop 编写嵌套for循环的正确方法?
我有一个程序,可以将图像分割成50 x 50像素块进行编辑,我已经尝试了3天没有成功地编写了一个嵌套的for循环来满足我的需要,在此之前,我尝试分别获取宽度段和高度段,然后尝试获取每个段的面积,但这根本不起作用,因此,我尝试了几种使用嵌套for循环的方法,但有一些我没有做到或做得不正确,任何帮助都将不胜感激,因为我没有想法!以下是我当前的等式,首先列出并解释了变量:For loop 编写嵌套for循环的正确方法?,for-loop,nested,gml,For Loop,Nested,Gml,我有一个程序,可以将图像分割成50 x 50像素块进行编辑,我已经尝试了3天没有成功地编写了一个嵌套的for循环来满足我的需要,在此之前,我尝试分别获取宽度段和高度段,然后尝试获取每个段的面积,但这根本不起作用,因此,我尝试了几种使用嵌套for循环的方法,但有一些我没有做到或做得不正确,任何帮助都将不胜感激,因为我没有想法!以下是我当前的等式,首先列出并解释了变量: j = (picture1.bbox_right + 1) - picture1.bbox_left; //gets image
j = (picture1.bbox_right + 1) - picture1.bbox_left; //gets image width in pixels
k = (picture1.bbox_bottom + 1) - picture1.bbox_top; //gets image height in pixels
h = j*k; //total number of pixels in image
s = j mod 50; //remaining pixels in image x plane
t = k mod 50; //remaining pixels in y plane
c = ceil(j/50); //number of segments in x plane, segments can be a maximum of 50 pixels wide
d = ceil(k/50); //number of segments in y plane segments can be a maximum of 50 pixels tall
brk = c*d; //total number of segments in image
for (i=0; i<h+1; i+=1) z[i] = 0;
for (i=0; i<=brk+1; i+=1) {v[i] = 0; w[i] = 0; z[i] = 0;} //v[] is the segment width, w[] is the segment height, z[] is the area of the segment
//drwatx[] and m[] are the starting x position of each segment
//drwaty[] and n[] are the starting y position of each segment
if(h > 2500)
{
if(d > c)
{
for(i=0; i<brk; i+=1)
{
for(a=0; a<d-1; a+=1)
{
if(a < d-1)
{
for(b=0; b<c-1; b+=1)
{
if(b < c-1)
{
m[b + i] = picture1.bbox_left + b*50;
drwatx[b + i] = picture1.bbox_left + b*50;
v[b + i] = 50;
}
if (b == c-1 && s == 0) v[b + i] = 50;
if (b == c-1 && s > 0) v[b + i] = s;
}
if(a < d-1)
{
n[a + i] = picture1.bbox_top - 1 + a*50;
drwaty[a + i] = picture1.bbox_top - 1 + a*50;
w[a + i] = 50;
}
if(a == d-1 && s == 0)
{
w[a + i] = 50;
}
if(a == d-1 && s > 0)
{
v[a + i] = s;
w[a + i] = t;
}
}
}
z[i] = v[i]*w[i];
}
}
}
j=(picture1.bbox\u right+1)-picture1.bbox\u left//获取以像素为单位的图像宽度
k=(picture1.bbox_底部+1)-picture1.bbox_顶部//获取以像素为单位的图像高度
h=j*k//图像中的像素总数
s=j模50//图像x平面中的剩余像素
t=k模50//y平面中的剩余像素
c=ceil(j/50)//x平面中的分段数,分段最大宽度为50像素
d=ceil(k/50)//y平面分段中的分段数最多可以为50像素高
brk=c*d//图像中的段总数
对于(i=0;i c)
{
对于(i=0;i你的想法太复杂了。你只需要两个for循环。找到线段的x和y位置非常容易。找到区域也会使它复杂一些
segment_width = 50;
segment_height = 50;
image_width = 73;
image_height = 183;
x = 12; // x position of image
y = 148; // y position of image
// Looping through the segment rows by incrementing the current
// y-coordinate value j by segment_height
for (j = 0; j <= image_height; j += segment_height)
{
// Segment size of the current segment
current_width = 0;
current_height = 0;
if (image_height - j < segment_height)
{
// If we are on the last row, calculate the segment height
// by subtracting the image_height by the current pixel
current_height = image_height - j;
}
else
{
// Else, we know that the segment height is 50
current_height = segment_height;
}
for (i = 0; i <= image_width; i += segment_width)
{
if (image_width - i < segment_width)
{
current_width = image_width - i;
}
else
{
current_width = segment_width;
}
// Calculate the segment area
z[i*j] = current_width*current_height;
// Calculate the segment position
drawx[floor(i/segment_width)] = x + i;
drawy[floor(j/segment_height)] = y + j;
}
}
你的想法太复杂了。你只需要两个for循环。找到线段的x和y位置非常容易。找到区域也会使它复杂一些
segment_width = 50;
segment_height = 50;
image_width = 73;
image_height = 183;
x = 12; // x position of image
y = 148; // y position of image
// Looping through the segment rows by incrementing the current
// y-coordinate value j by segment_height
for (j = 0; j <= image_height; j += segment_height)
{
// Segment size of the current segment
current_width = 0;
current_height = 0;
if (image_height - j < segment_height)
{
// If we are on the last row, calculate the segment height
// by subtracting the image_height by the current pixel
current_height = image_height - j;
}
else
{
// Else, we know that the segment height is 50
current_height = segment_height;
}
for (i = 0; i <= image_width; i += segment_width)
{
if (image_width - i < segment_width)
{
current_width = image_width - i;
}
else
{
current_width = segment_width;
}
// Calculate the segment area
z[i*j] = current_width*current_height;
// Calculate the segment position
drawx[floor(i/segment_width)] = x + i;
drawy[floor(j/segment_height)] = y + j;
}
}
再次感谢@Kake_Fisk解决了我的问题,并教会了我关于FOR循环的知识,我还找到了一种使用while循环解决问题的方法,因此对于需要此解决方案的任何人,这里有一种不同的方法:
a=0;
b=0;
bwatch=0;
c=0;
d=0;
j=(picture1.bbox_right+1)-(picture1.bbox_left);
k=(picture1.bbox_bottom+1)-(picture1.bbox_top);
g=picture1.bbox_left;
l=picture1.bbox_top-1;
h=(j*k);
i=0;
ii=0;
s=j mod 50;
t=k mod 50;
c=(ceil(j/50));
d=(ceil(k/50));
brk=c*d;
for(i=0;i<h+1;i+=1)
{
colsmake[i]=0;
}
for(i=0;i<=brk+1;i+=1)
{
v[i]=0;
w[i]=0;
z[i]=0;
drwatx[i]=0;
drwaty[i]=0;
colorize[i]=0;
}
if(h<=2500)
{drwatx[bwatch]=picture1.bbox_left;
drwaty[bwatch]=picture1.bbox_top-1;
z[bwatch]=h;
}
if(h>2500)
{
if(d>c)
{
while(a<c && b<d-1)
{
if(a<c)
{
drwatx[ii]=picture1.bbox_left+(a*50);
if s=0 v[ii]=50;
if s !=0 && a<c-1 v[ii]=50;
if s !=0 && a=c-1 v[ii]=s;
drwaty[ii]=picture1.bbox_top-1+(b*50);
w[ii]=50;
z[ii]=v[ii]*w[ii];
a+=1;
ii+=1;
}
if(a=c)
{
b+=1;
a=0;
}
}
while(a<c && b=d-1)
{
drwatx[ii]=picture1.bbox_left+(a*50);
if s=0 v[ii]=50;
if s !=0 && a<c-1 v[ii]=50;
if s !=0 && a=c-1 v[ii]=s;
drwaty[ii]=picture1.bbox_top-1+(b*50);
if t=0 w[ii]=50;
if t !=0 w[ii]=t;
z[ii]=v[ii]*w[ii];
a+=1;
ii+=1;
}
}
}
a=0;
b=0;
bwatch=0;
c=0;
d=0;
j=(图1.bbox\u右+1)-(图1.bbox\u左);
k=(图1.bbox\u底部+1)-(图1.bbox\u顶部);
g=图1.bbox_左;
l=图1.bbox\u top-1;
h=(j*k);
i=0;
ii=0;
s=j模50;
t=k模50;
c=(ceil(j/50));
d=(ceil(k/50));
brk=c*d;
对于(i=0;i再次感谢@Kake_Fisk解决了我的问题,并教会了我一些关于for循环的知识,我还找到了一种使用while循环解决问题的方法,因此对于需要此解决方案的任何人,这里有一种不同的方法:
a=0;
b=0;
bwatch=0;
c=0;
d=0;
j=(picture1.bbox_right+1)-(picture1.bbox_left);
k=(picture1.bbox_bottom+1)-(picture1.bbox_top);
g=picture1.bbox_left;
l=picture1.bbox_top-1;
h=(j*k);
i=0;
ii=0;
s=j mod 50;
t=k mod 50;
c=(ceil(j/50));
d=(ceil(k/50));
brk=c*d;
for(i=0;i<h+1;i+=1)
{
colsmake[i]=0;
}
for(i=0;i<=brk+1;i+=1)
{
v[i]=0;
w[i]=0;
z[i]=0;
drwatx[i]=0;
drwaty[i]=0;
colorize[i]=0;
}
if(h<=2500)
{drwatx[bwatch]=picture1.bbox_left;
drwaty[bwatch]=picture1.bbox_top-1;
z[bwatch]=h;
}
if(h>2500)
{
if(d>c)
{
while(a<c && b<d-1)
{
if(a<c)
{
drwatx[ii]=picture1.bbox_left+(a*50);
if s=0 v[ii]=50;
if s !=0 && a<c-1 v[ii]=50;
if s !=0 && a=c-1 v[ii]=s;
drwaty[ii]=picture1.bbox_top-1+(b*50);
w[ii]=50;
z[ii]=v[ii]*w[ii];
a+=1;
ii+=1;
}
if(a=c)
{
b+=1;
a=0;
}
}
while(a<c && b=d-1)
{
drwatx[ii]=picture1.bbox_left+(a*50);
if s=0 v[ii]=50;
if s !=0 && a<c-1 v[ii]=50;
if s !=0 && a=c-1 v[ii]=s;
drwaty[ii]=picture1.bbox_top-1+(b*50);
if t=0 w[ii]=50;
if t !=0 w[ii]=t;
z[ii]=v[ii]*w[ii];
a+=1;
ii+=1;
}
}
}
a=0;
b=0;
bwatch=0;
c=0;
d=0;
j=(图1.bbox\u右+1)-(图1.bbox\u左);
k=(图1.bbox\u底部+1)-(图1.bbox\u顶部);
g=图1.bbox_左;
l=图1.bbox\u top-1;
h=(j*k);
i=0;
ii=0;
s=j模50;
t=k模50;
c=(ceil(j/50));
d=(ceil(k/50));
brk=c*d;
对于(i=0;iI建议使用更好的变量名。您的代码几乎不可读。我建议使用更好的变量名。您的代码几乎不可读。啊,非常感谢!我确实让它变得比应该的复杂,我最终创建了一个WHILE循环并使其工作,但这也相当复杂,您的方式更复杂aight Forward,再次感谢您的帮助!啊,非常感谢!我确实让它变得比它应该的更复杂,我最终创建了一个WHILE循环并使其工作,但这也相当复杂,您的方式更直接,再次感谢您的帮助!