从 Python 运行 Stata 并确保没有错误

Question

我知道如何从 Python 启动 Stata。这里我有个小程序

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", "")

这或多或少是有效的。但这并不能保证 Stata 程序会成功完成。如何从 Stata 获得一些反馈给 Python，说“已成功完成”？

Answer 1

子进程使用returncode返回底层调用进程的成功（零）或失败（非零）结果。

然而，Stata do 文件并不完全是可执行文件，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，它都会始终运行。因此，请考虑以下 Python 读取 Stata 的日志并将其输出到其控制台以供用户查看代码结果的情况。甚至可能让 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 向其发送消息。

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, "")