我需要编写的函数的自旋(PIC,x),其中,将拍照和旋转90度的计数器倍顺时针X量。 我只是在一个函数的90度顺时针旋转:
def rotate(pic):
width = getWidth(pic)
height = getHeight(pic)
new = makeEmptyPicture(height,width)
tarX = 0
for x in range(0,width):
tarY = 0
for y in range(0,height):
p = getPixel(pic,x,y)
color = getColor(p)
setColor(getPixel(new,tarY,width-tarX-1),color)
tarY = tarY + 1
tarX = tarX +1
show(new)
return new
..但我不知道我怎么会去上旋转它的时候X量写一个函数。 任何人都知道我能做到这一点?
你可以称之为rotate()
次X量:
def spin(pic, x):
new_pic = duplicatePicture(pic)
for i in range(x):
new_pic = rotate(new_pic)
return new_pic
a_file = pickAFile()
a_pic = makePicture(a_file)
show(spin(a_pic, 3))
但这显然不是最优化的方式,因为你会计算X图像,而不是一个你感兴趣的,我建议你尝试了基本的switch...case
第一(即使这种说法并不存在于Python的办法; ):
xx = (x % 4) # Just in case you want (x=7) to rotate 3 times...
if (xx == 1):
new = makeEmptyPicture(height,width)
tarX = 0
for x in range(0,width):
tarY = 0
for y in range(0,height):
p = getPixel(pic,x,y)
color = getColor(p)
setColor(getPixel(new,tarY,width-tarX-1),color)
tarY = tarY + 1
tarX = tarX +1
return new
elif (xx == 2):
new = makeEmptyPicture(height,width)
# Do it yourself...
return new
elif (xx == 3):
new = makeEmptyPicture(height,width)
# Do it yourself...
return new
else:
return pic
然后,可能是你将能够看到的方式对这些案件合并成一个单一的(但更复杂) double for loop
......玩得开心...