“sys.argv"是什么意思? [英] What does 'sys.argv' mean?

问题描述

我正在从代码中学习，但我对其中的一行代码感到困惑:

I am learning from code, and I am get confused by one of its lines which is:

things = [float(arg) for arg in sys.argv[1:]]
Omega_a, Omega_b, Delta_a, Delta_b, \
init_pop_a, init_pop_b, tstep, tfinal = things

我在网上搜索并试图理解 sys.arg 的含义，以下是我的理解:

I have searched online and tried to understand what sys.arg means, and here is my understanding:

所以 sys.argv[0] 是文件名，而 sys.argv[1:] 是其余的参数，应该由用户给出.我不确定我的理解是否正确，如果是，那么我不明白为什么不能这样:

So sys.argv[0] is the file name, and sys.argv[1:] is the rest of the parameters which should given by users. I am not sure am I understood it right, and if it is, then I don't understand why cant it be like:

Omega_a = input() 
Omega_b = input()
etc...

这两种给定参数的方式有什么区别?

What's the difference between these two ways of giving parameters?

此外，如果我运行代码(按 F5)，Python shell 会出现如下错误:

Also, if I run the code (press F5), the Python shell give me an error like:

Traceback (most recent call last):

File "C:\Users\testcode.py", line 55, in <module>
init_pop_a, init_pop_b, tstep, tfinal = things
ValueError: need more than 0 values to unpack

在它给我一个错误之前，我什至没有机会给出参数 (sys.argv[1:]).于是上网查了一下.看起来我需要在 cmd 中运行这段代码，这让我更加困惑，为什么要这样做以及我应该如何放入 cmd 才能运行它?

I wasn't even given a chance to give parameters (sys.argv[1:]) before it gave me an error. So I searched online. It looks like I need to run this code in cmd which confused me more, why should it and how should I put into cmd in order to run it?

推荐答案

不同的是，sys.argv(命令行)参数是在程序运行前(启动时)给出的:

The difference is, that sys.argv (command line) parameters are given before the program is running (while starting it):

python testcode.py arg1 arg2 arg3 arg4 and so on ...

这将导致您的变量为:

Omega_a = 'arg1'
Omega_b = 'arg2'
Delta_a = 'arg3'
Delta_b = 'arg4'
init_pop_a = 'and'
init_pop_b = 'so'
tstep = 'on'
tfinal = '...'

虽然 input() 是在程序运行时给出的.

While the input()s are given when the program is running.

由于您没有使用参数启动程序，因此会出现错误，因为没有足够的(恰好 0 个)参数可以解压到变量中.

As you do not start the program with parameters it gives you the error, because there are not enough (exactly 0) parameters to be unpacked into the variables.

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