我承担巨蟒编程类,并且我们正在定义一个镜点镜像图像,然后使用嵌套的for循环从一个侧面复制像素到其他工作。 例如,垂直镜像的图像将使用以下代码:
def mirrorVertical(source):
mirrorPoint = getWidth(source) / 2
width = getWidth(source)
for y in range(0,getHeight(source)):
for x in range(0,mirrorPoint):
leftPixel = getPixel(source,x,y)
rightPixel = getPixel(source,width - x - 1,y)
color = getColor(leftPixel)
setColor(rightPixel,color)
我目前工作的一个分配问题,问我们,以便左上方被反射到右下方对角镜上的图像。 我发现到目前为止,只有每一个例子,答案适用于正方形图像,我需要能够将其应用到任何图像,最好通过定义一个对角镜点。 我一直在试图定义使用AY = MX + B风格式的镜面点,但我不知道如何告诉Python作出这样的一条线。 任何不特定于方形图像帮助将不胜感激!
注:因为我是全新的在这里,我不能发表图片呢,但是从左下到右上斜镜点将运行。 在上面留下三角形的图像将被反射到右下方。
您可以把左上角与非方阵像这样的右下:
height = getHeight(source)
width = getWidth(source)
for i in range(height - 1):
for j in range(int(width * float(height - i) / height)):
# Swap pixel i,j with j,i
这适用于沿对角线镜像。 你似乎在暗示你可能要多带些任意位置镜像。 在这种情况下,你将需要决定如何在没有上镜线的另一侧的匹配像素的像素填充。
你提到你在分配工作,所以你可能需要明确地做循环,但请注意,如果你把你的数据变成numpy的数组,可以轻松(和更有效)达到你想要的numpy的功能的组合是什么fliplr
, flipud
,并transpose
。
还要注意的是,在你的代码示例(镜像左/右),您要复制左侧像素的右侧,反之则不行,所以你实际上并没有镜像图像。
只是分享对角镜像的一种方式, 左下到右上 :
学生负责相适应,如果他需要从“右上到左下的”镜像或使用相反的对角线。
# !! Works only with squared pictures... !!
def diagMirrorPicture(picture):
height = getHeight(picture)
width = getWidth(picture)
if (height != width):
printNow("Error: The input image is not squared. Found [" + \
str(width) + " x " + str(height) + "] !")
return None
newPicture = makeEmptyPicture(width, height)
mirrorPt = 0
for x in range(0, width, 1):
for y in range(mirrorPt, height, 1):
sourcePixel = getPixel(picture, x, y)
color = getColor(sourcePixel)
# Copy bottom-left as is
targetPixel = getPixel(newPicture, x, y)
setColor(targetPixel, color)
# Mirror bottom-left to top right
targetPixel = getPixel(newPicture, y, x)
# ^^^^ (simply invert x and y)
setColor(targetPixel, color)
# Here we shift the mirror point
mirrorPt += 1
return newPicture
file = pickAFile()
picture = makePicture(file)
picture = diagMirrorPicture(picture)
if (picture):
writePictureTo(picture, "/home/mirror-diag2.png")
show(picture)
注意:这表现为,如果我们操作的垂直镜,用于独立地每条像素线,沿着垂直轴通过由mirrorPt
点。
输出( 由安东尼塔皮埃斯绘画 ):
................. ................. ..................
以下是实验性只是为了好玩...
# Draw point, with check if the point is in the image area
def drawPoint(pic, col, x, y):
if (x >= 0) and (x < getWidth(pic)) and (y >= 0) and (y < getHeight(pic)):
px = getPixel(pic, x, y)
setColor(px, col)
# Draw line segment given two points
# From Bresenham's line algorithm :
# http://en.wikipedia.org/wiki/Bresenham%27s_line_algorithm
def drawLine(pic, col, x0, y0, x1, y1):
dx = abs(x1-x0)
dy = abs(y1-y0)
sx = sy = 0
#sx = 1 if x0 < x1 else -1
#sy = 1 if y0 < y1 else -1
if (x0 < x1):
sx = 1
else:
sx = -1
if (y0 < y1):
sy = 1
else:
sy = -1
err = dx - dy
while (True):
drawPoint(pic, col, x0, y0)
if (x0 == x1) and (y0 == y1):
break
e2 = 2 * err
if (e2 > -dy):
err = err - dy
x0 = x0 + sx
if (x0 == x1) and (y0 == y1):
drawPoint(pic, col, x0, y0)
break
if (e2 < dx):
err = err + dx
y0 = y0 + sy
# Works only with squared cropped areas :
# i.e. in [(x0, y0), (x1, y1)], abs(x1-x0) must be equal to abs(y1-y0)
#
# USAGE :
# * To get bottom reflected to top use x0 > x1
# * To get top reflected to bottom use x0 < x1
def diagCropAndMirrorPicture(pic, startPt, endPt):
w = getWidth(pic)
h = getHeight(pic)
if (startPt[0] < 0) or (startPt[0] >= w) or \
(startPt[1] < 0) or (startPt[1] >= h) or \
(endPt[0] < 0) or (endPt[0] >= w) or \
(endPt[1] < 0) or (endPt[1] >= h):
printNow("Error: The input points must be in the image range !")
return None
new_w = abs(startPt[0] - endPt[0])
new_h = abs(startPt[1] - endPt[1])
if (new_w != new_h):
printNow("Error: The input points do not form a square !")
return None
printNow("Given: (" + str(startPt[0]) + ", " + str(endPt[0]) + ") and (" \
+ str(startPt[1]) + ", " + str(endPt[1]) + ")")
newPicture = makeEmptyPicture(new_w, new_h)
if (startPt[0] < endPt[0]):
offsetX = startPt[0]
switchX = False
switchTB = True
else:
offsetX = endPt[0]
switchX = True
switchTB = False
if (startPt[1] < endPt[1]):
offsetY = startPt[1]
switchY = False
else:
offsetY = endPt[1]
switchY = True
# (switchX XOR switchY)
changeDiag = (switchX != switchY)
mirror_pt = 0
for x in range(0, new_w, 1):
for y in range(mirror_pt, new_h, 1):
#for y in range(0, new_h, 1):
oldX = x
oldY = y
if (switchTB):
sourcePixel = getPixel(picture, offsetX+new_w-1- oldX, offsetY+new_h-1- oldY)
else:
sourcePixel = getPixel(picture, offsetX+oldX, offsetY+oldY)
color = getColor(sourcePixel)
if (changeDiag):
newX = new_w-1 - x
newY = new_h-1 - y
#printNow("Change Diag !")
else:
newX = x
newY = y
# Copied half
if (switchTB):
targetPixel = getPixel(newPicture, new_w-1- x, new_h-1- y)
else:
targetPixel = getPixel(newPicture, x, y)
setColor(targetPixel, color)
# Mirror half (simply invert x and y)
if (switchTB):
targetPixel = getPixel(newPicture, new_h-1- newY, new_w-1- newX)
else:
targetPixel = getPixel(newPicture, newY, newX)
setColor(targetPixel, color)
# Here we shift the mirror point
if (not changeDiag):
mirror_pt += 1
return newPicture
file = pickAFile()
pic = makePicture(file)
picture = makePicture(file)
# Draw working area
drawLine(pic, white, 30, 60, 150, 180)
drawLine(pic, white, 30, 180, 150, 60)
drawLine(pic, black, 30, 60, 30, 180)
drawLine(pic, black, 30, 60, 150, 60)
drawLine(pic, black, 150, 60, 150, 180)
drawLine(pic, black, 30, 180, 150, 180)
show(pic)
writePictureTo(pic, "D:\\pic.png")
# Build cropped and mirrored areas
pic1 = diagCropAndMirrorPicture(picture, (150, 60), (30, 180))
pic2 = diagCropAndMirrorPicture(picture, (30, 180), (150, 60))
pic3 = diagCropAndMirrorPicture(picture, (150, 180), (30, 60))
pic4 = diagCropAndMirrorPicture(picture, (30, 60), (150, 180))
# Show cropped and mirrored areas
if (pic1):
writePictureTo(pic1, "D:\\pic1.png")
show(pic1)
if (pic2):
writePictureTo(pic2, "D:\\pic2.png")
show(pic2)
if (pic3):
writePictureTo(pic3, "D:\\pic3.png")
show(pic3)
if (pic4):
writePictureTo(pic4, "D:\\pic4.png")
show(pic4)
.................................................. .... .................................................. ....
....... .......... .......... .......... .......
我假设你想根据45度线,而不是矩形对角线镜像。
你必须创建一个新的形象,它的宽度是原来的高度,它的高度是原来的宽度。
如果你的坐标系的原点是左下方,在原来的复制点(X,Y)的(Y,X)在新的图像位置。 如果它是另一个角落,你必须要考虑多一点;)