请帮我编写一个脚本间隔5秒启动一个ros1的launch文件循环执行100次文件名字叫imulaunch成功启动后有‘’四元数‘’打印信息时为成功启动否则为失败并分别记录成功失败的次数

以下是Python脚本示例：

import subprocess
import time

launch_file = "imu.launch"
success_count = 0
failure_count = 0

for i in range(100):
    process = subprocess.Popen(['roslaunch', launch_file])
    time.sleep(5)
    process.terminate()
    output = process.communicate()[0]
    if "四元数" in output:
        print("Successfully launched {}".format(launch_file))
        success_count += 1
    else:
        print("Failed to launch {}".format(launch_file))
        failure_count += 1

print("Number of successful launches: {}".format(success_count))
print("Number of failed launches: {}".format(failure_count))

脚本使用subprocess模块启动roslaunch命令，并在每次启动后等待5秒，然后关闭进程并检查输出是否包含“四元数”字样来判断是否成功启动。最后输出成功和失败的次数。请注意，该脚本需要在ROS环境下运行。