[英]Invert image faster in C#
我正在使用 WinForms。 我的表單中有一個圖片框。 當我在圖片框中打開圖片時,我可以通過單擊按鈕來回反轉顏色,但是我的代碼非常慢。 我怎樣才能提高性能。
private void Button1_Click(object sender, System.EventArgs e)
{
Bitmap pic = new Bitmap(PictureBox1.Image);
for (int y = 0; (y
<= (pic.Height - 1)); y++) {
for (int x = 0; (x
<= (pic.Width - 1)); x++) {
Color inv = pic.GetPixel(x, y);
inv = Color.FromArgb(255, (255 - inv.R), (255 - inv.G), (255 - inv.B));
pic.SetPixel(x, y, inv);
PictureBox1.Image = pic;
}
}
}
每次更改像素時都會設置控件的圖片,這會導致控件重繪自身。 等到你完成圖像:
Bitmap pic = new Bitmap(PictureBox1.Image);
for (int y = 0; (y <= (pic.Height - 1)); y++) {
for (int x = 0; (x <= (pic.Width - 1)); x++) {
Color inv = pic.GetPixel(x, y);
inv = Color.FromArgb(255, (255 - inv.R), (255 - inv.G), (255 - inv.B));
pic.SetPixel(x, y, inv);
}
}
PictureBox1.Image = pic;
如果有人需要 VB.NET 中的類似代碼。
還要注意變量 inv.A 而不是值 255。如果你的圖片框有透明度,你需要這個。
Public Function InvertImageColors(ByVal p As Image) As Image
Dim pic As New Bitmap(p)
For y As Integer = 0 To pic.Height - 1
For x As Integer = 0 To pic.Width - 1
Dim inv As Color = pic.GetPixel(x, y)
inv = Color.FromArgb(inv.A, 255 - inv.R, 255 - inv.G, 255 - inv.B)
pic.SetPixel(x, y, inv)
Next x
Next y
Return pic
End Function
用法:
pic.image = InvertImageColors(pic.image)
它對我有用...
private Image InvertingImage(Image source)
{
//create a blank bitmap the same size as original
Bitmap newBitmap = new Bitmap(source.Width, source.Height);
//get a graphics object from the new image
Graphics g = Graphics.FromImage(newBitmap);
// create the negative color matrix
ColorMatrix colorMatrix = new ColorMatrix(new float[][]
{
new float[] {-1, 0, 0, 0, 0},
new float[] {0, -1, 0, 0, 0},
new float[] {0, 0, -1, 0, 0},
new float[] {0, 0, 0, 1, 0},
new float[] {1, 1, 1, 0, 1}
});
// create some image attributes
ImageAttributes attributes = new ImageAttributes();
attributes.SetColorMatrix(colorMatrix);
g.DrawImage(source, new Rectangle(0, 0, source.Width, source.Height),
0, 0, source.Width, source.Height, GraphicsUnit.Pixel, attributes);
//dispose the Graphics object
g.Dispose();
return newBitmap;
}
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