從Python運行Stata並確保沒有錯誤

Question

我知道如何從Python啓動Stata。在這裏，我有小程序

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", "") 

這是工作，或多或少。但它並不能保證Stata程序能夠成功完成。我如何從Stata獲得一些反饋，說「完成成功」？

Answer 1

子過程使用returncode返回成功（零）或底層被叫流程的失敗（非零）的結果。

但是，Stata做的文件不是完全可執行文件，而是作爲batch jobs運行。由於這個原因，Stata.exe將始終返回一個成功代碼，因爲它始終運行，無論代碼輸出如何。因此，請考慮下面的Python讀取位置，並將Stata的日誌輸出到其控制檯以供用戶查看代碼結果。可能甚至要求Python掃描任何日誌文件Stata error code,​​- r(9999)，並且如果在日誌文件中存在，則強制Python向其發送消息。

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, "")