提前重试？

[mikumifa]

提前用set_ranges设定可能得结果 也就是把测试提前，没有就重试，从而提高一次的成功率 这种方式是否可行

def calculate_key(ocr, det, pic_url: str) -> str:
    pic_img_bytes = download_img(pic_url)
    pic_img = Image.open(io.BytesIO(pic_img_bytes))
    bg_img = pic_img.crop((0, 0, 344, 344))
    text_img = pic_img.crop((0, 344, 150, 384))
    text_img_bytes = io.BytesIO()
    text_img.save(text_img_bytes, format='PNG')
    ocr.set_ranges(7)
    target = ocr.classification(text_img_bytes.getvalue())
    bg_img_bytes = io.BytesIO()
    bg_img.save(bg_img_bytes, format='PNG')
    bg_det = det.detection(bg_img_bytes.getvalue())
    clicked_path = [None for i in range(len(target))]
    ocr.set_ranges(target)
    for row in bg_det:
        corp = bg_img.crop(row)
        img_byte = io.BytesIO()
        corp.save(img_byte, 'png')
        word = ocr.classification(img_byte.getvalue(), probability=True)
        s = ""
        for i in word['probability']:
            s += word['charsets'][i.index(max(i))]
        if s and s in target and clicked_path[target.index(s)] is None:
            clicked_path[target.index(s)] = row
        else:
            raise ValueError("error")
    res = []
    for path in clicked_path:
        x1, y1, x2, y2 = path
        x = (x1 + x2) / 2
        y = (y1 + y2) / 2
        position = f"{round((x / 333.375) * 10000)}_{round((y / 333.375) * 10000)}"
        res.append(position)

    return ",".join(res)

[mikumifa]

复活