6

我写了一个示例订阅者。我想将从 rospy.Subscriber 获得的数据提供给另一个变量,以便稍后在程序中使用它进行处理。目前我可以看到订阅者正在运行,因为当我使用 rospy.loginfo() 函数时,我可以看到正在打印的订阅值。虽然我不知道如何将这些数据存储到另一个变量中。我尝试使用赋值运算符“=”将其直接分配给变量,但出现错误。

我尝试使用 rospy.loginfo 编写回调函数来打印来自订阅对象的位置数据。我订阅了 JointState,它包含标题、位置、速度和努力数组。使用 rospy.loginfo 我可以验证订阅者是否正在订阅。但是当我试图将它直接分配给一个变量时,我得到了一个错误。

我正在显示来自回调函数的 loginfo,如下所示

def callback(data):
   rospy.loginfo(data.position)
   global listen
    listen = rospy.Subscriber("joint_states", JointState, 
    callback)
    rospy.spin()

这很好用。但是当我稍微修改代码以分配订阅值时,我得到以下错误,即

   listen1 = rospy.Subscriber("joint_states", JointState, 
   callback=None)
   listen = listen1.position
   #rospy.loginfo(listen)
   print(listen)
   rospy.spin()```

The error is as follows, 
 ```listen = listen1.position
    AttributeError: 'Subscriber' object has no attribute 'position'

编辑:这是我在程序中定义的节点,

    #rospy.loginfo(msg.data)
    global tactile_states
    tactile_states = data.data

def joint_callback(data):
    #rospy.loginfo(data.position)
    global g_joint_states 
    global g_position
    global g_pos1
    g_joint_states = data
    #for i in len(data.position):
        #g_position[i] = data.position[i]
    g_position = data.position
    if len(data.position) > 0:
        print("jointstate more than 0")
        g_pos1 = data.position[0]
    #print(g_position)


def joint_modifier(*args):
    #choice describes what the node is supposed to do whether act as publisher or subscribe to joint states or tactile sensors
    rospy.init_node('joint_listener_publisher', anonymous=True)
    pub1 = rospy.Publisher('joint_states', JointState, queue_size = 10)
    if(len(args)>1):
        choice = args[0]
        joint_name = args[1]
        position = args[2]
    else:
        choice = args[0]
    if (choice == 1):
        rate = rospy.Rate(1)
        robot_configuration = JointState()
        robot_configuration.header = Header()
        robot_configuration.name = [joint_name]
        robot_configuration.position = [position]
        robot_configuration.velocity = [10]
        robot_configuration.effort = [100]
        while not rospy.is_shutdown():
            robot_configuration.header.stamp = rospy.Time.now()
            rospy.loginfo(robot_configuration)
            break
        pub1.publish(robot_configuration)
        rospy.sleep(2)
    if (choice == 2):
        #rospy.Timer(rospy.Duration(2), joint_modifier)
        listen = rospy.Subscriber("joint_states", JointState, joint_callback)
        rospy.spin()
    if (choice == 3):
        #rospy.Timer(rospy.Duration(2), joint_modifier)
        tactile_sub = rospy.Subscriber("/sr_tactile/touch/ff", Float64, tactile_callback)
        rospy.spin()

这就是我在程序主体内调用节点的方式,

           joint_modifier(2)
           print("printing g_position")
           print(g_position)#to check the format of g_position
           print("printed g _position")
           leg_1 = Leg_attribute(g_position[0], g_position[1], g_position[2], velocity1 = 10, velocity2 = 10, velocity3 = 10, effort1 = 100, effort2 = 100, effort3 = 100, acceleration=1)

以这种方式调用时,程序会卡在joint_modifier(2)该函数具有rospy.spin().


4

1 回答 1

3

您使用的样式不是很标准。我假设你已经看过 ROS wiki 上的示例,我已经对其进行了修改以演示下面的标准用法。

主要是解决您发布的代码,您需要listen在回调之外设置全局范围。这是存储data您想要的,而不是订阅者对象。rospy.spin() 永远不会进入回调,只有主节点函数/部分。不经常使用的订阅者对象listen1不返回任何内容,也不存储它获取的数据。也就是说,您需要 Subscriber() 有一个非 None 回调。它更像是一个bind, 给 thedatacallback不是从订阅者返回它。这就是listen1 (Subscriber)没有属性position (JointState)的原因。

import rospy
from sensor_msgs.msg import JointState

# Subscribers
#     joint_sub (sensor_msgs/JointState): "joint_states"

# This is where you store all your data you recieve
g_joint_states = None
g_positions = None
g_pos1 = None

def timer_callback(event): # Type rospy.TimerEvent
    print('timer_cb (' + str(event.current_real) + '): g_positions is')
    print(str(None) if g_positions is None else str(g_positions))

def joint_callback(data): # data of type JointState
    # Each subscriber gets 1 callback, and the callback either
    # stores information and/or computes something and/or publishes
    # It _does not!_ return anything
    global g_joint_states, g_positions, g_pos1
    rospy.loginfo(data.position)
    g_joint_states = data
    g_positions = data.position
    if len(data.position) > 0:
        g_pos1 = data.position[0]
    print(g_positions)

# In your main function, only! here do you subscribe to topics
def joint_logger_node():
    # Init ROS
    rospy.init_node('joint_logger_node', anonymous=True)

    # Subscribers
    # Each subscriber has the topic, topic type, AND the callback!
    rospy.Subscriber('joint_states', JointState, joint_callback)
    # Rarely need to hold onto the object with a variable: 
    #     joint_sub = rospy.Subscriber(...)
    rospy.Timer(rospy.Duration(2), timer_callback)

    # spin() simply keeps python from exiting until this node is stopped
    # This is an infinite loop, the only code that gets ran are callbacks
    rospy.spin()
    # NO CODE GOES AFTER THIS, NONE! USE TIMER CALLBACKS!
    # unless you need to clean up resource allocation, close(), etc when program dies

if __name__ == '__main__':
    joint_logger_node()

编辑 1:Subscriber()、spin() 和 _callback(s) 的作用似乎有些混乱。在 Python 中它有点模糊,但是有一个主程序管理所有节点,并在它们之间发送节点。在每个节点中,我们向该节点存在的主程序注册,以及它有哪些发布者和订阅者。通过注册,这意味着我们告诉主程序,“嘿,我想要那个主题!”;在您的情况下,对于您的(未声明的)joint_sub 订阅者,“嘿,我想要JointStatejoint_states主题的所有消息!” 主程序每次(从某个地方的某个发布者)获得一个joint_states JointState新消息时,都会将其发送给该订阅者。订阅者使用回调处理、处理和处理 msg(数据):当(!)我收到一条消息时,运行回调。

所以主程序joint_states JointState从某个发布者那里收到一条新的消息。然后它,因为我们向它注册了一个订阅者,所以将它发送到这个节点。rospy.spin() 是一个等待该数据的无限循环。这就是它的作用(主要是):

def rospy.spin():
    while rospy.ok():
        for new_msg in get_new_messages from master():
            if I have a subscriber to new_msg:
                my_subscriber.callback(new_msg)

rospy.spin() 是您的回调、joint_callback(和/或 timer_callback 等)实际被调用和执行的地方。它在有数据时运行。

更根本的是,我认为由于这种混乱,您的程序结构存在缺陷;您的功能不会像您认为的那样做。这就是你应该如何制作你的节点。

  1. 将执行 NN 的数学部分(所有真正的非 ros 代码)放入一个单独的模块中,并创建一个函数来运行它。
  2. 如果只想在收到数据时运行它,请在回调中运行它。如果要发布结果,请在回调中发布。
  3. 不要调用主函数!if __init__ == '__main__': my_main_function()应该是它被调用的唯一地方,这将调用你的代码。我再说一遍:声明订阅者/发布者/初始化/计时器/参数的主函数仅在 中运行if __init__ ...,并且此函数运行您的代码。要让它运行您的代码,请将您的代码放在回调中。计时器回调对此很方便。

我希望这个代码示例能澄清:

import rospy
from std_msgs.msg import Header
from sensor_msgs.msg import JointState
import my_nn as nn # nn.run(data)

# Subscribers
#     joint_sub (sensor_msgs/JointState): "joint_states"

# Publishers
#     joint_pub (sensor_msgs/JointState): "target_joint_states"

joint_pub = None

def joint_callback(data): # data of type JointState
    pub_msg = JointState() # Make a new msg to publish results
    pub_msg.header = Header()
    pub_msg.name = data.name
    pub_msg.velocity = [10] * len(data.name)
    pub_msg.effort = [100] * len(data.name)
    # This next line might not be quite right for what you want to do,
    # But basically, run the "real code" on the data, and get the
    # result to publish back out
    pub_msg.position = nn.run(data.position) # Run NN on data, store results
    joint_pub.publish(pub_msg) # Send it when ready!

if __name__ == '__main__':
    # Init ROS
    rospy.init_node('joint_logger_node', anonymous=True)
    # Subscribers
    rospy.Subscriber('joint_states', JointState, joint_callback)
    # Publishers
    joint_pub = rospy.Publisher('target_joint_states', JointState, queue_size = 10)
    # Spin
    rospy.spin()
    # No more code! This is not a function to call, but its
    # own program! This is an executable! Run your code in
    # a callback!

请注意,我们设计为 ros 节点的 python 模块没有要调用的函数。它具有定义的回调结构和它们之间共享的全局数据,所有这些都在__init__.

于 2019-07-30T18:17:07.150 回答