问题
I took a Bitmap image and pulled the R, G, B ints from it for a specified row of pixels. Converted the int to a string so that I could print my 6 specific colors. I couldn't figure out how to do it with int.
The Problem I am able to print row 0-184 (corresponding to pixels in that row) out as a sequential data 1234... or color red, red, red, black, black, gray....
However I need to count like/same colors, display the sum of like colors, and reset a counter until that color comes up again then recount. I thought an if or if else would do it but not quite. It may be my code structure that is causing an issue?
So what I desire is:
5 red,
10 black,
2 red,
1 gray,
etc......
Here is my code, I am a beginner so criticizes on my lack of knowledge so that I may learn properly.
#include <iostream>
#include <sstream>
#include <string>
#include "EasyBMP.h"
#include "EasyBMP_BMP.h"
#include "EasyBMP_DataStructures.h"
#include "EasyBMP_VariousBMPutilities.h"
//Conversion and comparison function
void calculate(int i, int x, int p);
int main(int argc, const char * argv[])
{
BMP Image;
Image.ReadFromFile( "BMP GOES HERE 24bit" );
std::cout << "Image Height and Width: " << Image.TellHeight() << " x " << Image.TellWidth() << std::endl;
std::cout << "Enter your row: ";
int pixX = 0;
std::cin >> pixX;
//Set getpixel to top of row
int pixY = 0;
for( pixY = 0; pixY < Image.TellHeight() ; pixY++ )
{
std::cout << "Pixel: " << pixY + 1;
RGBApixel Temp = Image.GetPixel(pixX,pixY);
//Array to store pixel color ints
int pixy[3];
pixy[0] = Temp.Red;
pixy[1] = Temp.Green;
pixy[2] = Temp.Blue;
calculate(pixy[0], pixy[1], pixy[2]);
}
return 0;
}
void calculate(int rnum, int gnum, int bnum)
{
//String which will contain the result
std::string result;
//Stream used for the conversion
std::ostringstream convert;
//Insert the textual representation of 'Number' in the characters in the stream
convert << rnum;
convert << gnum;
convert << bnum;
// set 'Result' to the contents of the stream
result = convert.str();
// compare result to my given value
if (result == "25500")
{
std::cout << " RED " << std::endl;
}
if (result == "255255255")
{
std::cout << " WHITE " << std::endl;
}
if (result == "000")
{
std::cout << " BLACK" << std::endl;
}
if (result == "148148148")
{
std::cout << " GRAY " << std::endl;
}
if (result == "267326")
{
std::cout << " GREEN " << std::endl;
}
if (result == "2551260")
{
std::cout << " ORANGE " << std::endl;
}
}
Following is the working code. Note that if you use it that my image only has 6 specific colors. To change the print out one would have to modify the switch statement cases as desired.
#include <iostream>
#include <vector>
#include "EasyBMP.h"
#include "EasyBMP_BMP.h"
#include "EasyBMP_DataStructures.h"
#include "EasyBMP_VariousBMPutilities.h"
long toRGB(long red, long grn, long blu);
int main(int argc, const char * argv[])
{
BMP Image;
Image.ReadFromFile( "Your BMP HERE" );
std::cout << "Image Height and Width: " << Image.TellHeight() << " x " << Image.TellWidth() << std::endl;
std::cout << "Enter your row: ";
int pixX = 0;
std::cin >> pixX;
if (pixX != 0) //Subtract one from input if not 0, image starts at 0,0
{
pixX -= 1;
}
long pop = 0;
long pop1 = 0;
RGBApixel current = Image.GetPixel(pixX,0);
long pixy1[3]; //Array to store pixel color ints
pixy1[0] = current.Red;
pixy1[1] = current.Green;
pixy1[2] = current.Blue;
pop1 = toRGB(pixy1[0], pixy1[1], pixy1[2]);
int count = 0;
for( int pixY = 0; pixY < Image.TellHeight() ; pixY++ )
{
RGBApixel Temp = Image.GetPixel(pixX,pixY);
long pixy[3]; //Array to store pixel color ints
pixy[0] = Temp.Red;
pixy[1] = Temp.Green;
pixy[2] = Temp.Blue;
pop = toRGB(pixy[0], pixy[1], pixy[2]);
if (pop == pop1)
{
count++;
}
else
{
switch (pop1) {
case 0:
std::cout << "(" << count << ")\t" << "BLACK\n" << std::endl;
break;
case 16711680:
std::cout << "(" << count << ")\t" << "RED\n" << std::endl;
break;
case 9737364:
std::cout << "(" << count << ")\t" << "GRAY\n" << std::endl;
break;
case 16777215:
std::cout << "(" << count << ")\t" << "WHITE\n" << std::endl;
break;
case 1722650:
std::cout << "(" << count << ")\t" << "GREEN\n" << std::endl;
break;
case 16743936:
std::cout << "(" << count << ")\t" << "ORANGE\n" << std::endl;
break;
default:
std::cout << " !!!NO Specified COLOR For!!! " << pop1 << std::endl;
break;
}
pop1 = pop; //Reset the count and current color
count = 1;
}
}
if (count > 0) //Returns last color and count
{
switch (pop1) {
case 0:
std::cout << "(" << count << ")\t" << "BLACK\n" << std::endl;
break;
case 16711680:
std::cout << "(" << count << ")\t" << "RED\n" << std::endl;
break;
case 9737364:
std::cout << "(" << count << ")\t" << "GRAY\n" << std::endl;
break;
case 16777215:
std::cout << "(" << count << ")\t" << "WHITE\n" << std::endl;
break;
case 1722650:
std::cout << "(" << count << ")\t" << "GREEN\n" << std::endl;
break;
case 16743936:
std::cout << "(" << count << ")\t" << "ORANGE\n" << std::endl;
break;
default:
std::cout << " !!!NO Specified COLOR For!!! " << pop1 << std::endl;
break;
}
}
return 0;
}
long toRGB(long a, long b, long c) //Function to convert R, G, B values to unique value
{
long color = 0;
color |= (a & 255) << 16;
color |= (b & 255) << 8;
color |= (c & 255);
return color;
}
回答1:
I didn't really understand your question before, so I wrote a new answer more appropriate to what you're asking.
I think you can get what you want with something like this (modifying your code):
RGBApixel current = Image.GetPixel(pixX,0);
int count = 0;
for( pixY = 0; pixY < Image.TellHeight() ; pixY++ )
{
RGBApixel Temp = Image.GetPixel(pixX,pixY);
if (Temp == current)
{
count++;
}
else
{
// same-color sequence ended
// Add code here to print out current color and count
--- your output code ----
// now reset the count and current color
current = Temp;
count = 1;
}
}
// Now just print out the last sequence
if (count > 0)
{
--- your output code here again ---
}
One thing I'm not sure is how, if at all, the == operator is defined for RGBApixel. If it isn't defined or doesn't seem to equate pixels based on their color, just write a function like pixelsAreEqual(RBGApixel p1, RGBApixel p2) that takes two pixels and returns true if they have the same RGB values.
回答2:
There are a lot of different ways you could accomplish counting the number of pixels of a given color. If there is a fairly short list of colors you're interested in, you could create an array of size equal to the number of colors initialized to all 0, then write a function that returns an index of that array based on the color passed to it (e.g. red = 0, white = 1, in arbitrary order; constants are probably a good way to keep track of it), then loop over the pixels calling the function on each one and incrementing the array at the given index.
This is simple but extremely inelegant.
来源：https://stackoverflow.com/questions/13559617/counting-like-values-bmp-to-pixel-value-c