time.sleep() in a node

asked 2016-04-30 05:00:21 -0500

luketheduke
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updated 2022-02-22 11:23:15 -0500

8729 ●137 ●239 ●256 https://github.com/lucasw

I have a node where I need to wait for a specific amount of time. Can I just use time.sleep() in python or do I need something else? Thanks, luketheduke

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answered 2016-04-30 11:45:42 -0500

spmaniato

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updated 2016-04-30 16:52:36 -0500

You should definitely NOT use time.sleep()

The ROS way of doing this is:

rate = rospy.Rate(1) # 1 Hz
# Do stuff, maybe in a while loop
rate.sleep() # Sleeps for 1/rate sec

Or the similar:

rospy.sleep(1) # Sleeps for 1 sec

In both cases, you'll ofc need to import rospy

Here's the API documentation: http://docs.ros.org/jade/api/rospy/ht...

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I only want to delay once, not repeatedly. How would I implement that?

luketheduke ( 2016-04-30 16:21:39 -0500 )edit

It's the second code snippet above: rospy.sleep(1) # Sleeps for 1 sec Adapt it to how many seconds you want to delay.

spmaniato ( 2016-04-30 16:52:02 -0500 )edit

@luketheduke, if my answer was correct / helpful, please "mark it as correct" by clicking on the checkmark button to its left. It will help people who read this question in the future. Thanks.

spmaniato ( 2016-05-01 14:10:57 -0500 )edit

3

What is the benefit of using rospy.sleep instead of time.sleep? Is it just that it is interruptible?

Rufus ( 2021-05-30 20:39:37 -0500 )edit

1

+1 on @Rufus question comment. It's a very strong NOT, without any explanation.

marko525 ( 2022-04-29 14:22:28 -0500 )edit

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answered 2019-09-27 04:39:37 -0500

Tones
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Another option is using

ros::Duration(1.0).sleep()  // Sleep for one second

while Rate requires the sleep duration in Hz (1/s), Duration requires the sleep time in seconds. Typically, Rate is used inside loops, while Duration is the better choice when the sleeping occurs just once.

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