如何在numpy数组指定列中查找数值序列并提取对应特征

该方案支持任意长度的查询序列，无额外依赖，兼容性适配老旧版本Numpy：

import numpy as np

def find_matching_sequences(label_arr, query_seq):
    n = len(label_arr)
    k = len(query_seq)
    # 边界处理：查询序列长度超过原数组直接返回空
    if k > n:
        return np.array([]), []
    # 构造所有长度为k的滑动窗口索引
    window_idx = np.arange(k) + np.arange(n - k + 1)[:, None]
    # 取出所有窗口的label值与查询序列全匹配比对
    matches = (label_arr[window_idx] == query_seq).all(axis=1)
    # 提取匹配的起始索引
    start_indices = np.where(matches)[0]
    # 生成完整匹配的索引列表
    full_indices = [list(range(s, s + k)) for s in start_indices]
    return start_indices, full_indices

# 你的场景调用示例
v = np.array([[121,1],
         [131,1],
         [113,1],
         [131,1],
         [223,1],
         [242,1],
         [212,1],
         [131,1],
         [113,1],
         [131,1]])
label_col = v[:, 0]
sequence = [131,113,131]
start_indices, full_indices = find_matching_sequences(label_col, sequence)
# 输出结果：[1 7]，[[1, 2, 3], [7, 8, 9]]
print("匹配起始索引：", start_indices)
print("完整匹配索引列表：", full_indices)

# 提取对应feature的示例
feature_arr = ['a','b','c','d','e','f','g','h','i','j']
matched_features = [[feature_arr[i] for i in idx_list] for idx_list in full_indices]
# 输出结果：[['b','c','d'], ['h','i','j']]
print("匹配的feature序列：", matched_features)

若你使用的Numpy版本过老不支持上述广播写法，可替换为np.lib.stride_tricks.as_strided构造滑动窗口，逻辑完全一致。

如果允许使用pandas，可通过滚动窗口实现，逻辑更直观：

import pandas as pd

df = pd.DataFrame(v, columns=['label', 'other'])
df['feature'] = ['a','b','c','d','e','f','g','h','i','j']
k = len(sequence)
# 滚动窗口全匹配判断
match_mask = df['label'].rolling(k).apply(lambda x: (x == sequence).all()).fillna(0).astype(bool)
# 匹配起始索引 = 匹配位置往前推k-1位
start_indices = df.index[match_mask] - k + 1

内容的提问来源于stack exchange，提问作者Kunal Shah