ROS从python节点发布数组

时间:2015-07-12 16:21:01

标签: python arrays ros

我是ros + python的新手,我正试图从python ros节点发布一维数组。我使用Int32MultiArray但我无法理解多阵列中布局的概念。有谁可以向我解释一下?或者是否有其他方式发布数组?

感谢。

#!/usr/bin/env python

import roslib
roslib.load_manifest('test_drone')
import numpy
import rospy
import sys
import serial
from std_msgs.msg import String,Int32,Int32MultiArray,MultiArrayLayout,MultiArrayDimension
from rospy.numpy_msg import numpy_msg
from rospy_tutorials.msg import Floats

#port = "dev/ttyS0"
#baud = 115200

#ser = serial.Serial()
#ser.port = port
#ser.baudrate = baud



################################################################################################


def main(args):
   pub=rospy.Publisher('sonar_vals',Int32MultiArray,queue_size = 10)
   rospy.init_node('ca_serial')
   r = rospy.Rate(0.2)
   while not rospy.is_shutdown():
      print "LOOP running"
      a = [3250,2682,6832,2296,8865,7796,6955,8236]
      pub.publish(a)
      r.sleep();



     # try:
         #data_raw = ser.readline()
     # except e:
        # print e
      #sd = data_raw.split(',')   
      #a = numpy.array([sd[0],sd[1],sd[2],sd[3],sd[4],sd[5],sd[6],sd[7],sd[8],sd[9]],dtype=numpy.float32)
      #if sd[0] == 777:
      #   pub.publish(a)
     # else:
       #  print 'Invalid Data'




if __name__ == '__main__':
    import sys, getopt
    main(sys.argv)

3 个答案:

答案 0 :(得分:2)

*MultiArray消息在您的情况下有点矫枉过正。我认为如果为此创建自己的简单消息类型IntList会更简单(参见this tutorial on how to create custom messages)。 IntList.msg - 文件如下所示:

int32[] data

要发布包含此消息的列表,请使用以下代码段:

a = IntList()
a.data = [3250,2682,6832,2296,8865,7796,6955,8236]
pub.publish(a)

请注意,您无法直接发布列表,但必须实例化IntList对象并填充此对象的data成员(这适用于所有消息类型,即使您只是想发布一个单整数!)。

答案 1 :(得分:1)

如果您确实想使用MultiArray,请尝试:

array = []
my_array_for_publishing = Int32MultiArray(data=array)

答案 2 :(得分:0)

您可以对python使用以下方法:     pub = rospy.Publisher('chatter2', Float64MultiArray, queue_size=10) data_to_send = Float64MultiArray() # the data to be sent, initialise the array data_to_send.data = array # assign the array with the value you want to send pub.publish(data_to_send)