我正在开发一个裁剪 3000 张图像的项目。我已经使用算法创建了 3000 张图像,并正在尝试裁剪它们。目前,它们都位于一个名称遵循相同格式的文件夹中 - “img1”、“img2”、“img3”等。我尝试使用在线工具裁剪所有这些图像,但它不起作用。我正在寻找一种方法,无论是使用免费软件一次裁剪所有图像,还是使用 python 裁剪这些图像并将它们存储在单独的文件夹中。
我已经编写了一段小代码来为你做到这一点。
pip 安装 opencv-python
import cv2 as cv
import os
#First get the list of images in the folder
list_of_original_images = os.listdir("images") #Add the path to the folder
#Create a new directory to store the cropped images
os.mkdir("Cropped_Images")
#Iterate through the image_list
for image_path in list_of_original_images:
image_array = cv.imread(image_path) # Here we load image using opencv
height,width,channel = image_array.shape #Here we get the height,width and color channel
cropped_image = image_array[0:height//2,0:width//2] # Using array slicing we cut some part of the image and save in a new variable
# Write cropped image to Cropped Images folder
cv.write(f"Cropped_Images/{image_path}",cropped_image)
使用 Pillow 库使其变得非常简单:
from PIL import Image;
DirectoryPath = r"C:\Users\...\FolderWithImages";
Images = GetFiles(DirectoryPath);
for Img in Images:
BaseImg = Image.open(Img);
#The arguments to .crop is a rect (X,Y,Width,Height)
CroppedImg = im.crop((0,0,500,500))
CroppedImg.save("Path");
要获取目录中的所有文件,您可以查看此处
BBonless 的示例有效,但对 im.crop 参数存在误解。它不是宽度和高度,而是右坐标和下坐标。仅当您想从左上角裁剪时才有效。
我下面的示例采用顶部、左侧、宽度和高度,并从中计算出右侧和下部位置。
它将提示用户输入源目录并将裁剪后的文件保存在 c: emp 中。使用另一个 fd.askdirectory() 调用,可以很容易地以与输入目录相同的方式提示输出目录。
import glob
import os
from tkinter import filedialog as fd
from PIL import Image
directory_path = fd.askdirectory()
Images = glob.glob(os.path.join(directory_path,"*.jpg"))
for image_file_and_path in Images:
image_filename = os.path.basename(image_file_and_path)
output_filename = os.path.join(r"c:\temp\cropped", image_filename)
uncropped_image = Image.open(image_file_and_path)
# The argument to crop is a box : a 4-tuple defining the left, upper, right, and lower pixel positions.
left = 450
top = 50
width = 2595
height = 3226
CroppedImg = uncropped_image.crop((left, top, width + left, height + top))
CroppedImg.save(output_filename)