Hi, everyone, I am new to the computer vision task, but I am trying to learn something from this competition.
\ni met some puzzles in this notebook, These are the following two questions:
():\n\n x = np.zeros((Config.image_size, Config.image_size, Config.in_chans), dtype=np.uint8)\n t = self.df.iloc[idx]\n st_id = (t[])\n label = t[:].values.astype(np.int64)\n\n\n \n i (, , ):\n :\n p = \n img = Image.(p).convert()\n img = np.array(img)\n x[..., i] = img.astype(np.uint8)\n :\n ()\n \n\n \n i (, , ):\n :\n p = \n img = Image.(p).convert()\n img = np.array(img)\n x[..., i+] = img.astype(np.uint8)\n :\n ()\n \n\n \n axt2 = glob()\n axt2 = (axt2)\n\n step = (axt2) / \n st = (axt2)/ - *step\n end = (axt2)+\n\n i, j (np.arange(st, end, step)):\n :\n p = axt2[(, ((j-).()))]\n img = Image.(p).convert()\n img = np.array(img)\n x[..., i+] = img.astype(np.uint8)\n :\n ()\n \n\n np.(x)>\n\n self.transform :\n x = self.transform(image=x)[]\n\n x = x.transpose(, , )\n\n x, label\n epoch (, Config.epochs + ):\n ()\n model.train()\n total_loss = \n tqdm(train_dataloader, leave=) pbar:\n optimizer.zero_grad()\n idx, (x, t) (pbar):\n x = x.to(Config.device)\n t = t.to(Config.device)\n\n autocast:\n loss = \n y = model(x)\n col (Config.n_labels):\n pred = y[:,col*:col*+]\n gt = t[:,col]\n loss = loss + criterion(pred, gt) / Config.n_labels\nThank you very much for your help!
", "messages": [ { "id": "2997348", "postDate": "09/24/2024 13:23:17", "content": "Hi, everyone, I am new to the computer vision task, but I am trying to learn something from this competition.
\ni met some puzzles in this notebook, These are the following two questions:
():\n\n x = np.zeros((Config.image_size, Config.image_size, Config.in_chans), dtype=np.uint8)\n t = self.df.iloc[idx]\n st_id = (t[])\n label = t[:].values.astype(np.int64)\n\n\n \n i (, , ):\n :\n p = \n img = Image.(p).convert()\n img = np.array(img)\n x[..., i] = img.astype(np.uint8)\n :\n ()\n \n\n \n i (, , ):\n :\n p = \n img = Image.(p).convert()\n img = np.array(img)\n x[..., i+] = img.astype(np.uint8)\n :\n ()\n \n\n \n axt2 = glob()\n axt2 = (axt2)\n\n step = (axt2) / \n st = (axt2)/ - *step\n end = (axt2)+\n\n i, j (np.arange(st, end, step)):\n :\n p = axt2[(, ((j-).()))]\n img = Image.(p).convert()\n img = np.array(img)\n x[..., i+] = img.astype(np.uint8)\n :\n ()\n \n\n np.(x)>\n\n self.transform :\n x = self.transform(image=x)[]\n\n x = x.transpose(, , )\n\n x, label\n epoch (, Config.epochs + ):\n ()\n model.train()\n total_loss = \n tqdm(train_dataloader, leave=) pbar:\n optimizer.zero_grad()\n idx, (x, t) (pbar):\n x = x.to(Config.device)\n t = t.to(Config.device)\n\n autocast:\n loss = \n y = model(x)\n col (Config.n_labels):\n pred = y[:,col*:col*+]\n gt = t[:,col]\n loss = loss + criterion(pred, gt) / Config.n_labels\nThank you very much for your help!
", "rawMarkdown": "Hi, everyone, I am new to the computer vision task, but I am trying to learn something from this competition.\ni met some puzzles in this notebook, These are the following two questions:\n1. do you think that the three types of MRI images (Sagittal T1, Sagittal T2/STIR, and Axial T2) in different channels of the same image? \n```python\n def __getitem__(self, idx):\n\n x = np.zeros((Config.image_size, Config.image_size, Config.in_chans), dtype=np.uint8)\n t = self.df.iloc[idx]\n st_id = int(t['study_id'])\n label = t[1:].values.astype(np.int64)\n \n\n # Sagittal T1:\n for i in range(0, 10, 1):\n try:\n p = f'../data/cvt_png/{st_id}/Sagittal T1/{i:03d}.png'\n img = Image.open(p).convert('L')\n img = np.array(img)\n x[..., i] = img.astype(np.uint8)\n except:\n print(f'failed to load on {st_id}, Sagittal T1')\n pass\n\n # Sagittal T2/STIR:\n for i in range(0, 10, 1):\n try:\n p = f'../data/cvt_png/{st_id}/Sagittal T2_STIR/{i:03d}.png'\n img = Image.open(p).convert('L')\n img = np.array(img)\n x[..., i+10] = img.astype(np.uint8)\n except:\n print(f'failed to load on {st_id}, Sagittal T2/STIR')\n pass\n \n # Axial T2\n axt2 = glob(f'../data/cvt_png/{st_id}/Axial T2/*.png')\n axt2 = sorted(axt2)\n \n step = len(axt2) / 10.0\n st = len(axt2)/2.0 - 4.0*step\n end = len(axt2)+0.0001\n \n for i, j in enumerate(np.arange(st, end, step)):\n try:\n p = axt2[max(0, int((j-0.5001).round()))]\n img = Image.open(p).convert('L')\n img = np.array(img)\n x[..., i+20] = img.astype(np.uint8)\n except:\n print(f'failed to load on {st_id}, Sagittal T2/STIR')\n pass \n\n assert np.sum(x)>0\n\n if self.transform is not None:\n x = self.transform(image=x)['image']\n\n x = x.transpose(2, 0, 1)\n\n return x, label\n```\n2. why is it possible to learn and infer 75 prediction labels at once? Because the labels' shape is (25, )\n```python\n for epoch in range(1, Config.epochs + 1):\n print(f'start epoch {epoch}')\n model.train()\n total_loss = 0\n with tqdm(train_dataloader, leave=True) as pbar:\n optimizer.zero_grad()\n for idx, (x, t) in enumerate(pbar):\n x = x.to(Config.device)\n t = t.to(Config.device)\n\n with autocast:\n loss = 0\n y = model(x)\n for col in range(Config.n_labels):\n pred = y[:,col*3:col*3+3]\n gt = t[:,col]\n loss = loss + criterion(pred, gt) / Config.n_labels\n\n\n```\n\nThank you very much for your help!", "votes": null }, { "id": "2997842", "postDate": "09/25/2024 00:29:48", "content": "Hi @alannikos and welcome, I don't quite get the first question but as for the second part:
\ny[:, col*3:col*3+3] slices out the predictions for each label from a range of 3 logits. gt = t[:, col] grabs the corresponding ground truth (gt) for the same label.Now you pass everything to BCE, I believe, and it works just fine. 3 logits and one corresponding labels (0, 1 or 2).
\nHope it makes sense.
", "rawMarkdown": "Hi @alannikos and welcome, I don't quite get the first question but as for the second part:\n\n1. `y[:, col*3:col*3+3]` slices out the predictions for each label from a range of 3 logits. \n So it is just fancy ways to reshape (75, ) into 25 by 3\n2. `gt = t[:, col]` grabs the corresponding ground truth (`gt`) for the same label.\n\nNow you pass everything to BCE, I believe, and it works just fine. 3 logits and one corresponding labels (0, 1 or 2).\n\nHope it makes sense.", "votes": null }, { "id": "2999088", "postDate": "09/26/2024 10:50:49", "content": "Thank you for your answer! I upvoted the answer. My first question is that during the establishment of this dataset, it seems that 30 png images were loaded into one sample, that is, in_channels=30. However, I wonder if the general image data is 3channels, is it because it is medical image data?
", "rawMarkdown": "Thank you for your answer! I upvoted the answer. My first question is that during the establishment of this dataset, it seems that 30 png images were loaded into one sample, that is, in_channels=30. However, I wonder if the general image data is 3channels, is it because it is medical image data?", "votes": null }, { "id": "2999917", "postDate": "09/27/2024 04:20:22", "content": "Yes, it is because the DICOM slices are gray scale. There are variable number of slices so the code is taking an average guess of 30 which may contain the relevant data.
", "rawMarkdown": "Yes, it is because the DICOM slices are gray scale. There are variable number of slices so the code is taking an average guess of 30 which may contain the relevant data.", "votes": null } ], "comments": [ { "id": 2997842, "author_name": "sergiosaharovskiy", "author_url": "", "post_date": "09/25/2024 00:29:48", "content": "Hi @alannikos and welcome, I don't quite get the first question but as for the second part:
\ny[:, col*3:col*3+3] slices out the predictions for each label from a range of 3 logits. gt = t[:, col] grabs the corresponding ground truth (gt) for the same label.Now you pass everything to BCE, I believe, and it works just fine. 3 logits and one corresponding labels (0, 1 or 2).
\nHope it makes sense.
", "votes": null, "replies": [ { "id": 2999088, "author_name": "alannikos", "author_url": "", "post_date": "09/26/2024 10:50:49", "content": "Thank you for your answer! I upvoted the answer. My first question is that during the establishment of this dataset, it seems that 30 png images were loaded into one sample, that is, in_channels=30. However, I wonder if the general image data is 3channels, is it because it is medical image data?
", "votes": null, "replies": [ { "id": 2999917, "author_name": "coderrkj", "author_url": "", "post_date": "09/27/2024 04:20:22", "content": "Yes, it is because the DICOM slices are gray scale. There are variable number of slices so the code is taking an average guess of 30 which may contain the relevant data.
", "votes": null, "replies": [] } ] } ] } ], "raw_markdown_by_id": { "2997348": "Hi, everyone, I am new to the computer vision task, but I am trying to learn something from this competition.\ni met some puzzles in this notebook, These are the following two questions:\n1. do you think that the three types of MRI images (Sagittal T1, Sagittal T2/STIR, and Axial T2) in different channels of the same image? \n```python\n def __getitem__(self, idx):\n\n x = np.zeros((Config.image_size, Config.image_size, Config.in_chans), dtype=np.uint8)\n t = self.df.iloc[idx]\n st_id = int(t['study_id'])\n label = t[1:].values.astype(np.int64)\n \n\n # Sagittal T1:\n for i in range(0, 10, 1):\n try:\n p = f'../data/cvt_png/{st_id}/Sagittal T1/{i:03d}.png'\n img = Image.open(p).convert('L')\n img = np.array(img)\n x[..., i] = img.astype(np.uint8)\n except:\n print(f'failed to load on {st_id}, Sagittal T1')\n pass\n\n # Sagittal T2/STIR:\n for i in range(0, 10, 1):\n try:\n p = f'../data/cvt_png/{st_id}/Sagittal T2_STIR/{i:03d}.png'\n img = Image.open(p).convert('L')\n img = np.array(img)\n x[..., i+10] = img.astype(np.uint8)\n except:\n print(f'failed to load on {st_id}, Sagittal T2/STIR')\n pass\n \n # Axial T2\n axt2 = glob(f'../data/cvt_png/{st_id}/Axial T2/*.png')\n axt2 = sorted(axt2)\n \n step = len(axt2) / 10.0\n st = len(axt2)/2.0 - 4.0*step\n end = len(axt2)+0.0001\n \n for i, j in enumerate(np.arange(st, end, step)):\n try:\n p = axt2[max(0, int((j-0.5001).round()))]\n img = Image.open(p).convert('L')\n img = np.array(img)\n x[..., i+20] = img.astype(np.uint8)\n except:\n print(f'failed to load on {st_id}, Sagittal T2/STIR')\n pass \n\n assert np.sum(x)>0\n\n if self.transform is not None:\n x = self.transform(image=x)['image']\n\n x = x.transpose(2, 0, 1)\n\n return x, label\n```\n2. why is it possible to learn and infer 75 prediction labels at once? Because the labels' shape is (25, )\n```python\n for epoch in range(1, Config.epochs + 1):\n print(f'start epoch {epoch}')\n model.train()\n total_loss = 0\n with tqdm(train_dataloader, leave=True) as pbar:\n optimizer.zero_grad()\n for idx, (x, t) in enumerate(pbar):\n x = x.to(Config.device)\n t = t.to(Config.device)\n\n with autocast:\n loss = 0\n y = model(x)\n for col in range(Config.n_labels):\n pred = y[:,col*3:col*3+3]\n gt = t[:,col]\n loss = loss + criterion(pred, gt) / Config.n_labels\n\n\n```\n\nThank you very much for your help!", "2997842": "Hi @alannikos and welcome, I don't quite get the first question but as for the second part:\n\n1. `y[:, col*3:col*3+3]` slices out the predictions for each label from a range of 3 logits. \n So it is just fancy ways to reshape (75, ) into 25 by 3\n2. `gt = t[:, col]` grabs the corresponding ground truth (`gt`) for the same label.\n\nNow you pass everything to BCE, I believe, and it works just fine. 3 logits and one corresponding labels (0, 1 or 2).\n\nHope it makes sense.", "2999088": "Thank you for your answer! I upvoted the answer. My first question is that during the establishment of this dataset, it seems that 30 png images were loaded into one sample, that is, in_channels=30. However, I wonder if the general image data is 3channels, is it because it is medical image data?", "2999917": "Yes, it is because the DICOM slices are gray scale. There are variable number of slices so the code is taking an average guess of 30 which may contain the relevant data." }, "source": "meta" }