小编coo*_*kie的帖子

给定n个点的列表,如何使用numpy生成包含从每个点到每个其他点的距离的矩阵?

嘿家伙所以我试图在python中重写以下matlab代码:

repmat(points, 1, length(points)) - repmat(points', length(points),1);
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points 是一个包含几个点的弧度值的数组.

上面的代码给我一个像这样的矩阵输出:

 0   1   2   0   1   2   0   1   2
-1   0   1  -1   0   1  -1   0   1
-2  -1   0  -2  -1   0  -2  -1   0
 0   1   2   0   1   2   0   1   2
-1   0   1  -1   0   1  -1   0   1
-2  -1   0  -2  -1   0  -2  -1   0
 0   1   2   0   1   2   0   1   2
-1   0   1  -1   0   1  -1 …
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python matlab numpy matrix

4
推荐指数
1
解决办法
166
查看次数

使用“localhost”配置侦听器会导致无法检索有关代理的元数据

我正在尝试使用 docker 容器设置单个 Kafka 代理,并且我正在使用此处的图像。目标是在 docker 中运行 Kafka 代理,并从主机运行生产者/消费者。

在配置KAFKA_LISTENERSKAFKA_ADVERTISED_LISTENERS属性时,我意识到如果以下配置到位,我将无法检索代理元数据

KAFKA_BROKER_ID: 0
KAFKA_INTER_BROKER_LISTENER_NAME: INTERNAL
KAFKA_LISTENER_SECURITY_PROTOCOL_MAP: INTERNAL:PLAINTEXT,EXTERNAL:PLAINTEXT
# KAFKA_ADVERTISED_LISTENERS must be EQUAL or SUBSET of KAFKA_LISTENERS
KAFKA_LISTENERS: INTERNAL://kafka0:29092, EXTERNAL://localhost:9092
# INTERNAL://kafka0:29092 is specified because of the KAFKA_INTER_BROKER_LISTENER_NAME configuration.
KAFKA_ADVERTISED_LISTENERS: INTERNAL://kafka0:29092, EXTERNAL://localhost:9092
KAFKA_ZOOKEEPER_CONNECT: zookeeper:2181
KAFKA_AUTO_CREATE_TOPICS_ENABLE: 'false'
KAFKA_MESSAGE_MAX_BYTES: '200000'
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> kafkacat -b localhost:9092 -L
ERROR: Failed to acquire metadata: Local: Broker transport failure
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但是在配置中替换EXTERNAL://localhost:9092EXTERNAL://kafka0:9092EXTERNAL://:9092(默认接口)修复了它。KAFKA_LISTENERS

>kafkacat -b localhost:9092 -L
Metadata …
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apache-kafka docker docker-compose

4
推荐指数
1
解决办法
4432
查看次数

使用sklearn对弧度距离矩阵进行DBSCAN?

我希望对几个时间戳(以分钟为单位)进行聚类。所以到目前为止我所做的是:

1) 将点转换为弧度

#points containing time value in minutes
points = [100, 200, 600, 659, 700]

def convert_to_radian(x):
    return((x / (24 * 60)) * 2 * pi)

rad_function = np.vectorize(convert_to_radian)
points_rad = rad_function(points)
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2) 生成距离矩阵

#generate distance matrix from each point
dist = points_rad[None,:] - points_rad[:, None]
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3)分配每个点的最短距离

dist[((dist > pi) & (dist <= (2*pi)))] = dist[((dist > pi) & (dist <= (2*pi)))] -(2*pi)
dist[((dist > (-2*pi)) & (dist <= (-1*pi)))] = dist[((dist > (-2*pi)) & (dist <= (-1*pi)))] + …
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python numpy data-mining scipy scikit-learn

3
推荐指数
1
解决办法
2102
查看次数