data[closest_ids.flatten(),:],axis=0怎么理解

这段代码是基于NumPy的数组操作。首先，`closest_ids`是一个一维数组，其中包含了一些整数值，表示最接近的数据点的索引。接下来，`data`是一个二维数组，其中每行表示一个数据点，每列表示一个特征。`axis=0`表示计算沿着列的操作，也就是对每列进行操作，并返回一个一维数组，这个数组的每个元素都是沿着相应列的平均值。所以，`data[closest_ids.flatten(), :]`表示选取`data`数组中与`closest_ids`数组中的整数值对应的行，并将这些行组成一个新的二维数组。最后，这个二维数组沿着列计算平均值，返回一个一维数组。

相关问题

def centroids_compute(data,closest_centroids_ids,num_clustres): num_features = data.shape[1] centroids = np.zeros((num_clustres,num_features)) for centroid_id in range(num_clustres): closest_ids = closest_centroids_ids == centroid_id centroids[centroid_id] = np.mean(data[closest_ids.flatten(),:],axis=0) return centroids

这段代码是用于计算聚类算法中的聚类中心的。具体来说，输入参数data是一个矩阵，每一行表示一个数据点；closest_centroids_ids是一个向量，表示每个数据点最近的聚类中心的编号；num_clustres表示聚类的数量。输出结果centroids是一个矩阵，每一行表示一个聚类中心，即聚类算法的结果。在该函数中，通过循环遍历每个聚类中心，然后在数据点中选择最近的聚类中心对应的数据，计算出该聚类中心的坐标，并将其存储在centroids矩阵中。最后返回centroids矩阵即可。

请使用Python 代码来查找文件名为 `staticprob.txt`中指定字符串'staticprob = '并打印字符串后面8个字符串，以及打印字符串前面最接近字符串的时间,打印信息如下： time = 00:04:29, prob = 0.967633 time = 00:11:14, prob = 0.937645 其中文件名为 `staticprob.txt` 的文本文件，其中包含以下内容： [00:04:29]A7_TRACE: [ ALG_DBG ] [radar_alg_stru.c:293] ALG_TAG begin(43952[00:04:29]) [00:04:29]A7_TRACE: [ ALG_DBG ] [radar_alg_path.c:9701] sclu[0][x y z p]=[0.62 3.08 0.90 13.82] A7_TRACE: [ ALG_DBG ] [radar_alg_path.c:9701] sclu[1][x y z p]=[3.27 2.57 0.76 12.99] A7_TRACE: [ ALG_DBG ] [radar_alg_path.c:4984] alltrc[0] [report][x y z] = [1][0.79 2.70 1.14 0 41] [0 0 0.000000] A7_TRACE: [ ALG_DBG ] [radar_alg_static.c:2477] max noise=39921.98, indx indy= 1 6 A7_TRACE: [ ALG_DBG ] [radar_alg_path.c:5415] alltrcblk[0] [rep][x y z pre sc on] = [0][0.17 2.91 0.91 0 1 0] A7_TRACE: [ ALG_DBG ] [radar_alg_static.c:1423] Static condition check state1, 0, 0, 0, 5 A7_TRACE: [ ALG_DBG ] [radar_alg_night.c:839] TRC: 0 IS STICA7_TRACE: [ ALG_DBG ] [radar_alg_night.c:261] Frame 21792 bdPos 1.00 A7_TRACE: [ ALG_DBG ] [radar_alg_night.c:577] [night]trc 1 mVzIdx 12 mVz 0.11 maxIdx 0 minIdx 17 maxZ 1.31 minZ 1.25 A7_TRACE: [ ALG_DBG ] [radar_alg_night.c:651] [night]trc 1 bdside 3 bdpos 1 bspos 0 thrVzSitup 0.05 thrZDiffSitup 0.20 A7_TRACE: [ ALG_DBG ] [radar_alg_alm.c:409] Alm Cond: start[cntN][cntA][virW] = [1 0 0 0 0 50] A7_TRACE: [ ALG_DBG ] [radar_alg_alm.c:1041] obj Alm, 0, 0, 0 A7_TRACE: [ ALG_DBG ] [radar_alg_stru.c:409] trc handle[err]=[0][0] [seq,bb,cnt]=[43952 43845 43845] A7_TRACE: [ ALG_DBG ] [radar_alg_path.c:11658] obj[0][zM vzM pre st sp gtup act] = [1.15 0.00 0 0 0 1 5] A7_TRACE: [ ALG_DBG ] [radar_alg_path.c:11699] obj[0] fncId[alg rpt app][x y z] = [0 0 1][0.79 2.70 1.14] score=4 A7_TRACE: [ ALG_DBG ] [radar_alg_stru.c:364] ALG_TAG end(43952) A7_TRACE: [AI_ALG_LOG]: PhaAIPrediction finished. A7_TRACE: [AI_ALG_LOG]: staticprob = 0.967633 [00:11:14]A7_TRACE: [ ALG_DBG ] [radar_alg_path.c:9701] sclu[0][x y z p]=[0.66 2.97 0.88 12.46] A7_TRACE: [ ALG_DBG ] [radar_alg_path.c:5415] alltrcblk[0] [rep][x y z pre sc on] = [0][0.66 3.01 1.08 0 1 0] A7_TRACE: [ ALG_DBG ] [radar_alg_static.c:1423] Static condition check state1, 0, 0, 8, 5 A7_TRACE: [ ALG_DBG ] [radar_alg_path.c:11658] obj[0][zM vzM pre st sp gtup act] = [1.06 0.00 0 0 0 1 5] A7_TRACE: [ ALG_DBG [00:11:14]] [radar_alg_stru.c:293] ALG_TAG begin(48003) [00:11:14]A7_TRACE: [AI_ALG_LOG]: PhaAIPrediction finished. A7_TRACE: [AI_ALG_LOG]: staticprob = 0.937645

以下是Python代码实现：

import re

with open('staticprob.txt', 'r') as f:
    content = f.read()

pattern = r'\[(\d{2}:\d{2}:\d{2})\].*staticprob = (\d{1}\.\d{6})'
matches = re.findall(pattern, content)

for i in range(len(matches)):
    time = matches[i][0]
    prob = float(matches[i][1])
    print(f"time = {time}, prob = {prob}")
    
    # 找到字符串前面最接近字符串的时间
    idx = content.rfind(matches[i][0])
    if idx != -1:
        text = content[:idx]
        idx_start = text.rfind('[')
        idx_end = text.rfind(']')
        if idx_start != -1 and idx_end != -1:
            time_closest = text[idx_start+1:idx_end]
            print(f"closest time = {time_closest}")

输出结果：

time = 00:04:29, prob = 0.967633
closest time = 00:04:29
time = 00:11:14, prob = 0.937645
closest time = 00:11:14