如何用循环实现图像3x3网格分割的坐标生成代码？

我正在编写代码将一张图像沿3x3网格模式分割成多个小图像。目前我为每个“网格单元”显式分配四个角的坐标，以便用于PIL.Image.crop()方法。请问是否有简便的方式用循环实现这段代码？

原代码 #

from math import floor
import matplotlib.pyplot as plt

height = 75
width = 75

numRows = 5
numCols = 5

numVertLines = numRows+1
numHorLines  = numCols+1

xcoords = []
ycoords = []

grid4coords = []

colIncrement = floor(width/numCols)
rowIncrement = floor(height/numRows)
print("Incremementing columns by", colIncrement, "pixels")
print("Incrementing rows by", rowIncrement, "pixels")

for i in range(0, (height+1), rowIncrement):    #start from (0,0), increment by rowIncrement until reached 
    xcoords.append(i)

for j in range(0, (width+1), colIncrement): #start from (0,0), increment by colIncrement until reached right
    ycoords.append(j)
    
print("xcoords:",xcoords)
print("ycoords",ycoords)
################################## row 1 grid coords
TL11 = [xcoords[0], ycoords[0]]
TR11 = [xcoords[1], ycoords[0]]
BL11 = [xcoords[0], ycoords[1]]
BR11 = [xcoords[1], ycoords[1]]

TL12 = [xcoords[1], ycoords[0]]
TR12 = [xcoords[2], ycoords[0]]
BL12 = [xcoords[1], ycoords[1]]
BR12 = [xcoords[2], ycoords[1]]

TL13 = [xcoords[2], ycoords[0]]
TR13 = [xcoords[3], ycoords[0]]
BL13 = [xcoords[2], ycoords[1]]
BR13 = [xcoords[3], ycoords[1]]


################################ row 2 grid coords
TL21 = [xcoords[0], ycoords[1]]
TR21 = [xcoords[1], ycoords[1]]
BL21 = [xcoords[0], ycoords[2]]
BR21 = [xcoords[1], ycoords[2]]

TL22 = [xcoords[1], ycoords[1]]
TR22 = [xcoords[2], ycoords[1]]
BL22 = [xcoords[1], ycoords[2]]
BR22 = [xcoords[2], ycoords[2]]

TL23 = [xcoords[2], ycoords[1]]
TR23 = [xcoords[3], ycoords[1]]
BL23 = [xcoords[2], ycoords[2]]
BR23 = [xcoords[3], ycoords[2]]

################################ row 3 grid coords
TL31 = [xcoords[0], ycoords[2]]
TR31 = [xcoords[1], ycoords[2]]
BL31 = [xcoords[0], ycoords[3]]
BR31 = [xcoords[1], ycoords[3]]

TL32 = [xcoords[1], ycoords[2]]
TR32 = [xcoords[2], ycoords[2]]
BL32 = [xcoords[1], ycoords[3]]
BR32 = [xcoords[2], ycoords[3]]

TL33 = [xcoords[2], ycoords[2]]
TR33 = [xcoords[3], ycoords[2]]
BL33 = [xcoords[2], ycoords[3]]
BR33 = [xcoords[3], ycoords[3]]

当然可以用嵌套循环简化代码，避免手动重复定义每个网格的坐标。而且PIL.Image.crop()实际需要的是一个元组(左, 上, 右, 下)，对应你代码里的TL.x, TL.y, BR.x, BR.y，可以直接在循环中生成这个参数，或者按需求保存四个角的坐标。

优化代码示例 #

from math import floor
import matplotlib.pyplot as plt

height = 75
width = 75

numRows = 5
numCols = 5

colIncrement = floor(width / numCols)
rowIncrement = floor(height / numRows)
print("Incremementing columns by", colIncrement, "pixels")
print("Incrementing rows by", rowIncrement, "pixels")

# 用range直接生成坐标点列表，替代原循环追加
xcoords = list(range(0, height + 1, rowIncrement))
ycoords = list(range(0, width + 1, colIncrement))

print("xcoords:", xcoords)
print("ycoords", ycoords)

# 嵌套循环生成所有网格的坐标信息
grid_coords = []
# 遍历行索引（每个网格对应行区间[i, i+1]）
for i in range(numRows):
    row_grids = []
    # 遍历列索引（每个网格对应列区间[j, j+1]）
    for j in range(numCols):
        # 计算当前网格的四个角坐标
        TL = [xcoords[i], ycoords[j]]
        TR = [xcoords[i+1], ycoords[j]]
        BL = [xcoords[i], ycoords[j+1]]
        BR = [xcoords[i+1], ycoords[j+1]]
        # 把坐标和crop所需参数存入字典，方便后续调用
        row_grids.append({
            'TL': TL,
            'TR': TR,
            'BL': BL,
            'BR': BR,
            'crop_box': (TL[0], TL[1], BR[0], BR[1])
        })
    grid_coords.append(row_grids)

# 打印验证前3行前3列的网格（和原代码输出对应）
for row_idx in range(3):
    print(f"################################## row {row_idx+1} grid coords")
    for col_idx in range(3):
        grid = grid_coords[row_idx][col_idx]
        print(f"TL{row_idx+1}{col_idx+1} = {grid['TL']}")
        print(f"TR{row_idx+1}{col_idx+1} = {grid['TR']}")
        print(f"BL{row_idx+1}{col_idx+1} = {grid['BL']}")
        print(f"BR{row_idx+1}{col_idx+1} = {grid['BR']}")

代码说明 #

1. 用list(range(...))替代原代码中的循环追加，更简洁地生成坐标点列表。
2. 外层循环遍历行起始索引i，内层遍历列起始索引j，通过i与i+1、j与j+1的组合，直接获取当前网格的边界坐标，无需手动逐个定义。
3. 将每个网格的四个角坐标和crop_box参数存入字典，后续可以直接提取crop_box传给PIL.Image.crop()，也可以单独调用四个角的坐标。
4. 最后打印的部分和原代码输出对应，用于验证循环生成的坐标是否正确。

内容的提问来源于stack exchange，提问作者nickreitz