Buckets:
| #!/usr/bin/env python3 | |
| """exp73b — 直接验证 THINGS-EEG 的「测试概念 vs 训练概念」是否重叠。 | |
| ## 为什么需要这个 | |
| exp73 做的是 **200 选 1 图像解码**。一个必须先回答的前提问题： | |
| > 200 张测试图的**概念**，是否出现在 1654 个训练概念里？ | |
| - 若**重叠**：解码器可能靠"同概念、同语义"取巧，数字是虚高的。 | |
| - 若**不相交（concept-disjoint）**：数字就是标准协议下的值，可与文献比较。 | |
| exp73 内置了一个**免标签**的 ViT 特征近邻探针来回答这个问题， | |
| 但那是**间接**证据。这个脚本给出**直接**证据：从桶内图像清单取概念名。 | |
| ## 证据来源（桶内，无需下载 .pt 标签） | |
| hf buckets ls -R hf://buckets/illusion-brain/training-data/eeg/things-eeg/Image_set/training_images/ | |
| hf buckets ls -R hf://buckets/illusion-brain/training-data/eeg/things-eeg/Image_set/test_images/ | |
| 结构（实测）： | |
| training_images/<NNNNN_concept>/<concept>_NN{b,s}.jpg → 1654 概念 × 10 图 | |
| test_images/<NNNNN_concept>/<concept>_NN{b,s}.jpg → 200 概念 × 1 图 | |
| 行序即特征行序：ViT-H-14_features_train.pt 的第 i 行 ↔ 训练清单第 i 行。 | |
| ## 用法 | |
| hf buckets ls -R <train_dir> > train_images.txt | |
| hf buckets ls -R <test_dir> > test_images.txt | |
| python exp73b_concept_overlap.py train_images.txt test_images.txt | |
| """ | |
| from __future__ import annotations | |
| import re | |
| import sys | |
| from pathlib import Path | |
| PAT = re.compile(r"Image_set/(?:training_images|test_images)/(\d+)_(.+?)/") | |
| def parse(path): | |
| ids, names = [], [] | |
| unmatched = [] | |
| with open(path, encoding="utf-8", errors="replace") as fh: | |
| for line in fh: | |
| m = PAT.search(line) | |
| if m: | |
| ids.append(int(m.group(1))) | |
| names.append(m.group(2)) | |
| elif line.strip() and not line.startswith(("Hint:", "---", "\x1b")): | |
| unmatched.append(line.strip()[:110]) | |
| return ids, names, unmatched | |
| def main(train_txt, test_txt): | |
| tr_ids, tr_names, tr_un = parse(train_txt) | |
| te_ids, te_names, te_un = parse(test_txt) | |
| print(f"训练行 {len(tr_names)} 测试行 {len(te_names)}") | |
| if tr_un: | |
| print(f" ! 未匹配的训练行 {len(tr_un)}，样例: {tr_un[:2]}") | |
| if te_un: | |
| print(f" ! 未匹配的测试行 {len(te_un)}，样例: {te_un[:2]}") | |
| groups = [] | |
| consistent = True | |
| for i in range(0, len(tr_names), 10): | |
| g = tr_names[i:i + 10] | |
| if len(set(g)) != 1: | |
| consistent = False | |
| g = [f"<不一致:{sorted(set(g))}>"] | |
| groups.append(g[0]) | |
| print(f"训练概念组 {len(groups)}（应为 1654），每组 10 张且组内同名: {consistent}") | |
| print(f"训练概念字母序: {groups == sorted(groups)}") | |
| print(f"测试概念 {len(te_names)}，互不相同: {len(set(te_names)) == len(te_names)}") | |
| print(f"测试概念字母序: {te_names == sorted(te_names)}") | |
| overlap = sorted(set(te_names) & set(groups)) | |
| print("\n" + "=" * 72) | |
| print(f"概念重叠：{len(overlap)} / {len(te_names)}") | |
| if overlap: | |
| print(" 重叠样例:", overlap[:10]) | |
| print(" ⇒ 测试概念**部分**出现在训练概念里 ⇒ 解码数字可能含同概念泄漏") | |
| else: | |
| print(" ⇒ 测试概念与训练概念**完全不相交**（concept-disjoint）") | |
| print(" ⇒ exp73 的 200 选 1 数字是**新概念**泛化，不是同概念取巧") | |
| print("=" * 72) | |
| print(f"训练概念 id 范围: {min(tr_ids):05d}–{max(tr_ids):05d}（{len(set(tr_ids))} 个）") | |
| print(f"测试概念 id 范围: {min(te_ids):05d}–{max(te_ids):05d}（{len(set(te_ids))} 个）") | |
| print(f"id 区间有交集: {bool(set(tr_ids) & set(te_ids))}" | |
| " （id 是各自分割内的编号，不是全局概念 id）") | |
| return 0 if not overlap else 1 | |
| if __name__ == "__main__": | |
| if len(sys.argv) != 3: | |
| print(__doc__) | |
| sys.exit(2) | |
| sys.exit(main(sys.argv[1], sys.argv[2])) | |
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