Question

我知道如何从Python启动Stata。这里我有一个小程序

def dostata(dofile, *params):
    ## Launch a do-file, given the fullpath to the do-file
    ## and a list of parameters.       
    cmd = ["C:\Program Files (x86)\Stata13\StataMP-64.exe", "do", dofile]
    for param in params:
        cmd.append(param)
    a = subprocess.Popen(cmd, shell=True)

path = "C:/My/do/file/dir/"
filename = "try.do"

dostata(path + filename, model, "1", "")

这或多或少都有效。但它并不能保证Stata计划能够顺利完成。我怎样才能从Stata到Python获得一些反馈意见＆＃34;成功完成＆＃34;？

Answer 1

子流程使用returncode返回基础被调用进程的成功（零）或失败（非零）结果。

但是，Stata do文件不是完全可执行文件，而是以batch jobs运行。因此，Stata.exe将始终返回成功代码，因为无论.do代码输出如何，它始终会运行。因此，请考虑下面Python读取的位置，并将Stata的日志输出到其控制台，以便用户查看代码结果。甚至可能会使Python扫描日志文件中的任何Stata error code，r(1) - r(9999)，如果存在于日志文件中，则强制Python发送消息。

import os, subprocess

# CURRENT DIRECTORY
cd = os.path.dirname(os.path.abspath(__file__))

def openlog(filename):
    with open(filename, 'r') as txt:
        for line in txt:
            print(line.strip())

def dostata(dofile, logfile, *params):
    ## Launch a do-file, given the fullpath to the do-file
    ## and a list of parameters.       
   cmd = ["C:\Program Files (x86)\Stata13\StataMP-64.exe", "/b", "do", dofile]
   for param in params:
       cmd.append(param)
   a = subprocess.Popen(cmd, shell=True)

   print('STATA OUTPUT:\n')
   openlog(os.path.join(cd, logfile))


path = "C:/My/do/file/dir/"
filename = "try.do"
logname = "try.log"

result = dostata(os.path.join(path, filename), logname, "")

从Python运行Stata并确保没有错误

1 个答案: