我正在处理存储在FITS文件中的大型传感器数据结构，我使用astropy模块打开和读入这些文件。
为了进行分析，我编写了几个以不同方式处理这些数据的脚本（例如init.py） - 每个脚本创建一个包含多个子图的图像文件。
现在我想创建一个init.py filename arg2 arg3 arg4
脚本，用于初始化诸如文件名和几个参数之类的内容，以便进行分析
analysis1.py我想传递给所有下标，例如#analysis1.py
from init import filename, arg2, arg3, arg4
# import everything else necessary
# do things with the data
# output picture as png-file
#!/bin/bash
# initialization of parameters and data read-in
python init.py filename arg2 arg3 arg4
# analysis scripts running in parallel using the parameters of init.py
python analysis1.py &
python analysis3.py &
使用bash但没有明确地再次输入这些作为参数，如
argv我使用这种方式是因为我通常在CPU机器上工作，并希望尽可能多地使用线程/核心来提高效率。
普通导入当然不起作用，因为import init
from sys import argv
print(argv)
#returns ['mypath/init.py']
没有从上次运行中获取参数
<?php defined('BASEPATH') OR exit('No direct script access allowed');
class Upload_Files extends CI_Controller
{
function __construct() {
parent::__construct();
$this->load->model('file');
}
function index(){
$data = array();
if($this->input->post('fileSubmit') && !empty($_FILES['userFiles']['name'])){
$filesCount = count($_FILES['userFiles']['name']);
for($i = 0; $i < $filesCount; $i++){
$_FILES['userFile']['name'] = $_FILES['userFiles']['name'][$i];
$_FILES['userFile']['type'] = $_FILES['userFiles']['type'][$i];
$_FILES['userFile']['tmp_name'] = $_FILES['userFiles']['tmp_name'][$i];
$_FILES['userFile']['error'] = $_FILES['userFiles']['error'][$i];
$_FILES['userFile']['size'] = $_FILES['userFiles']['size'][$i];
$uploadPath = 'uploads/files/';
$config['upload_path'] = $uploadPath;
$config['allowed_types'] = 'gif|jpg|png';
$this->load->library('upload', $config);
$this->upload->initialize($config);
if($this->upload->do_upload('userFile')){
$fileData = $this->upload->data();
$uploadData[$i]['file_name'] = $fileData['file_name'];
$uploadData[$i]['created'] = date("Y-m-d H:i:s");
$uploadData[$i]['modified'] = date("Y-m-d H:i:s");
}
}
if(!empty($uploadData)){
//Insert file information into the database
$insert = $this->file->insert($uploadData);
$statusMsg = $insert?'Files uploaded successfully.':'Some problem occurred, please try again.';
$this->session->set_flashdata('statusMsg',$statusMsg);
}
}
//Get files data from database
$data['files'] = $this->file->getRows();
//Pass the files data to view
$this->load->view('upload_files', $data);
}
function index2(){
$data = array();
if($this->input->post('fileSubmit2') && !empty($_FILES['userFiles']['name'])){
$filesCount = count($_FILES['userFiles']['name']);
for($i = 0; $i < $filesCount; $i++){
$_FILES['userFile']['name'] = $_FILES['userFiles']['name'][$i];
$_FILES['userFile']['type'] = $_FILES['userFiles']['type'][$i];
$_FILES['userFile']['tmp_name'] = $_FILES['userFiles']['tmp_name'][$i];
$_FILES['userFile']['error'] = $_FILES['userFiles']['error'][$i];
$_FILES['userFile']['size'] = $_FILES['userFiles']['size'][$i];
$uploadPath = 'uploads/index2/';
$config['upload_path'] = $uploadPath;
$config['allowed_types'] = 'gif|jpg|png';
$this->load->library('upload', $config);
$this->upload->initialize($config);
if($this->upload->do_upload('userFile')){
$fileData = $this->upload->data();
$uploadData[$i]['file_name'] = $fileData['file_name'];
$uploadData[$i]['created'] = date("Y-m-d H:i:s");
$uploadData[$i]['modified'] = date("Y-m-d H:i:s");
$uploadData[$i]['divid'] =2;
}
}
if(!empty($uploadData)){
//Insert file information into the database
$insert = $this->file->insert($uploadData);
$statusMsg = $insert?'Files uploaded successfully.':'Some problem occurred, please try again.';
$this->session->set_flashdata('statusMsg',$statusMsg);
}
}
//Get files data from database
$data['files'] = $this->file->getRows2();
//Pass the files data to view
$this->load->view('upload_files', $data);
}
}
我怎么能实现这个目标？
答案 0 :(得分：0)
如果您使用所有输入参数调用bash脚本：
./mybash.sh filename arg2 arg3 arg4
然后你可以编写这样的bash脚本来将输入参数传递给你的所有脚本：
#!/bin/bash
# initialization of parameters and data read-in
python init.py $1 $2 $3 $4
# analysis scripts running in parallel using the parameters of init.py
python analysis1.py $1 $2 $3 $4 &
python analysis3.py $1 $2 $3 $4 &
然后，您的init.py可以访问sys.argv中的参数：
# init.py
from sys import argv
print(argv)
# prints ['mypath/init.py', <filename>, <arg2>, <arg3>, <arg4>]
# everything else
分析脚本相同：
# analysis1.py
from sys import argv
print(argv)
# prints ['mypath/analysis1.py', <filename>, <arg2>, <arg3>, <arg4>]
# everything else
修改 如果目标是让您的init.py将初始输入参数带到bash脚本，更改它们然后将这些更改的值作为参数传递给分析脚本，那么您可以将init.py脚本转储到例如JSON中格式化为stdout，以便您的分析脚本可以在启动时解析它们。
为此，请更改init.py以将初始化参数转储到stdout，例如：
# init.py
import json
from sys import argv
# assign your input parameters
init_filename = argv[1]
init_arg2 = argv[2]
init_arg3 = argv[3]
init_arg4 = argv[4]
# do some initialisation on your input parameters here
# dump your initialised parameters to stdout in json format
print(json.dumps([init_filename, init_arg2, init_arg3, init_arg4]))
然后让您的分析脚本在启动时解析JSON值：
# analysis1.py
import json
from sys import argv
print(json.loads(argv[1]))
# prints [<init_filename>, <init_arg2>, <init_arg3>, <init_arg4>]
# everything else
最后，将init.py脚本中的（JSON）输出分配给变量，并将其传递给bash文件中的分析脚本：
#!/bin/bash
# initialization of parameters and data read-in
PARAMS=$(python init.py $1 $2 $3 $4)
# analysis scripts running in parallel using the parameters dumped by init.py
python analysis1.py ''"$PARAMS"''
python analysis3.py ''"$PARAMS"''
你可以在没有JSON的情况下做到这一点，但是你必须担心你的文件名中的空格和特殊字符以及你转储到stdout的args。
答案 1 :(得分：0)
正如Kendas所说，你必须在某个地方坚持磁盘上的参数。
这将有效：
<强> init.py
import sys
import pickle
def get_parms():
return pickle.load(open("init.dat","rb"))
if __name__ == "__main__":
myparms = sys.argv[:]
pickle.dump(myparms,open("init.dat","wb"))
<强> analysis1.py
import init
myparms = init.get_parms()
print(myparms)
如果你这样运行init.py：
python init.py filename arg1 arg2 arg3
然后analysis1.py将产生此输出：
['(pathname)/init.py', 'filename', 'arg1', 'arg2', 'arg3']
您几乎肯定希望允许analysis1.py接受命令行参数作为来自init的值的覆盖。