对于如下这种内容的log.log文件，如何从第9964行开始每跳过30行提取接下来10000行的reward数据：[32m[05-13 17:20:36 MainThread @train.py:373] Total Steps: 122 Reward: 10.252438767788885

可以使用Python的文件读取和处理操作，具体步骤如下：

1. 打开文件并跳过前9963行：

with open('log.log') as f:
    for i in range(9963):
        next(f)

2. 每隔30行读取一次，并提取接下来的10000行数据：

rewards = []
for i in range(333):
    for j in range(10000):
        line = next(f)
        reward = line.split()[-1]
        rewards.append(float(reward))
    # 跳过接下来的29行
    for k in range(29):
        next(f)

3. 最终得到的rewards列表即为所需的reward数据。完整代码如下：

with open('log.log') as f:
    for i in range(9963):
        next(f)

    rewards = []
    for i in range(333):
        for j in range(10000):
            line = next(f)
            reward = line.split()[-1]
            rewards.append(float(reward))
        for k in range(29):
            next(f)