Python argparse require选项，取决于定义的标志 [英] Python argparse requiring option, depending on the defined flags

问题描述

我有一个小的python脚本，它使用 argparse 来让用户定义选项.它为不同的模式使用两个标志，并使用一个参数让用户定义文件.请参见下面的简化示例:

I have a small python script, which uses argparse to let the user define options. It uses two flags for different modes and an argument to let the user define a file. See the simplified example below:

#!/usr/bin/python3

import argparse
from shutil import copyfile

def check_file(f):
    # Mock function: checks if file exists, else "argparse.ArgumentTypeError("file not found")"
    return f

def main():
    aFile = "/tmp/afile.txt"

    parser = argparse.ArgumentParser(description="An example",formatter_class=argparse.RawTextHelpFormatter)
    parser.add_argument("-f", "--file", help="A file, used with method A.", default=aFile, type=check_file)
    parser.add_argument("-a", "--ay", help="Method A, requires file.", action='store_true')
    parser.add_argument("-b", "--be", help="Method B, no file required.", action='store_true')

    args = parser.parse_args()
    f = args.file
    a = args.ay
    b = args.be

    if a:
        copyfile(f, f+".a")
    elif b:
        print("Method B")

if __name__ == "__main__":
    main()

方法A需要该文件.

方法B没有.

如果使用方法A运行脚本，则可以使用默认文件，也可以使用通过 -f /-file 定义的文件.该脚本检查文件是否存在，并且一切正常.

If I run the script with method A, I either use the default file or one that is defined with -f/--file. The script checks if the file exists and everything is fine.

现在，如果我使用方法B运行脚本，则该脚本不需要文件，但是将检查默认选项，并且如果不存在该选项，则argparse函数将引发异常并退出脚本.

Now, if I run the script with method B, it shouldn't require the file, but the default option is checked and if it doesn't exist the argparse function raises the exception and the script exits.

如果定义了 -b 并要求，如果定义了 -a ，我如何配置argparse以使 -f 可选?

How can I configure argparse to make -f optional, if -b is defined and require it, if -a is defined?

我刚刚意识到，使 -f 和 -b 互斥就足够了.但是，如果我仅运行 -b ，则仍将执行 check_file .有办法防止这种情况吗?

edit: I just realized that it would be enough for me to make -f and -b mutually exclusive. But then, if I run -b only, the check_file is executed anyways. Is there a way to prevent that?

#!/usr/bin/python3

import argparse
from shutil import copyfile

def check_file(f):
    # Mock function: checks if file exists, else "argparse.ArgumentTypeError("file not found")"
    print("chk file")
    return f

def main():
    aFile = "/tmp/afile.txt"

    parser = argparse.ArgumentParser(description="An example",formatter_class=argparse.RawTextHelpFormatter)
    group = parser.add_mutually_exclusive_group(required=True)

    group.add_argument("-f", "--file", help="A file, used with method A.", default=aFile, type=check_file)
    parser.add_argument("-a", "--ay", help="Method A, requires file.", action='store_true')
    group.add_argument("-b", "--be", help="Method B, no file required.", action='store_true')

    args = parser.parse_args()
    f = args.file
    a = args.ay
    b = args.be

    if a:
        print("File: "+str(f))
    elif b:
        print("Method B")
        print("file: "+str(f))

if __name__ == "__main__":
    main()

输出:

chk file
Method B
file: /tmp/afile.txt

推荐答案

您可以使用y/be定义子解析器作为子命令，或者为a声明第二个解析器实例.像这样:

You can defined subparser with ay/be as subcommand or alternatively declare a second parser instance for a. Something like:

parser = argparse.ArgumentParser(
    description="An example",
    formatter_class=argparse.RawTextHelpFormatter
)
# ensure either option -a or -b only
group = parser.add_mutually_exclusive_group(required=True)
group.add_argument("-a", "--ay", help="Method A, requires file.",
                   action='store_true')
group.add_argument("-b", "--be", help="Method B, no file required.",
                   action='store_true')
# define a parser for option -a
parser_a = argparse.ArgumentParser()
parser_a.add_argument("-f", "--file", help="A file, used with method A.",
                      type=check_file, required=True)
parser_a.add_argument("-a", "--ay", help="Method A, requires file.",
                      action='store_true')

# first parse - get either -a/-b
args = parser.parse_known_args(sys.argv[1:])
# if -a, use the second parser to ensure -f is in argument
# note parse_known_args return tuple, the first one is the populated namespace
if args[0].ay:
    args = parser_a.parse_args(sys.argv[1:])

这篇关于Python argparse require选项，取决于定义的标志的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！