我有一个文件夹“图片-2”,其中有100多个子文件夹,这些子文件夹包括一个图像每个文件夹。def main()打开每个图像,def run(img)接收图像并对其进行处理,但现在我无法将该图像保存在它的子文件夹中。
例如:/ def main c:/def main-2/1/1.png (1是文件夹名,所以我在图-2中有100个文件夹)
如果条件将保存处理过的图像(zero.png)在文件夹图像-2/1/
如何处理100个文件夹,每个文件夹一个图像?
def run(img):
data = img.load()
width, height = img.size
output_img = Image.new("RGB", (100, 100))
Zero=np.zeros(shape=(100, 100),dtype=np.uint8)
for (x, y) in labels:
component = uf.find(labels[(x, y)])
labels[(x, y)] = component
path='C:/Python27/cclabel/Images-2/'
if labels[(x, y)]==0:
Zero[y][x]=int(255)
Zeroth = Image.fromarray(Zero)
for root, dirs in os.walk(path):
print root
print dirs
Zeroth.save(path+'Zero'+'.png','png')
def main():
# Open the image
path="C:/Python27/cclabel/Images-2/"
for root, dirs, files in os.walk(path):
for file_ in files:
img = Image.open(os.path.join(root, file_))
img = img.point(lambda p: p > 190 and 255)
img = img.convert('1')
(labels, output_img) = run(img)
if __name__ == "__main__": main()发布于 2017-12-03 12:30:38
你要给os.walk打两次电话。这是你的问题。这就是我在评论中的意思:
def run(dirname, img):
data = img.load()
width, height = img.size
output_img = Image.new("RGB", (100, 100))
Zero=np.zeros(shape=(100, 100), dtype=np.uint8)
for (x, y) in labels:
component = uf.find(labels[(x, y)])
labels[(x, y)] = component
path = 'C:/Python27/cclabel/Images-2/'
if labels[(x, y)] == 0:
Zero[y][x] = 255
Zeroth = Image.fromarray(Zero)
Zeroth.save(os.path.join(dirname, 'Zero.png'), 'png')
def main():
path = "C:/Python27/cclabel/Images-2/"
for root, dirs, files in os.walk(path):
for file_ in files:
img = Image.open(os.path.join(root, file_))
img = img.point(lambda p: p > 190 and 255)
img = img.convert('1')
(labels, output_img) = run(root, img)
if __name__ == "__main__":
main()https://stackoverflow.com/questions/47616240
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