除了nargs，如果你事先知道列表，你可能会想要使用choices:

>>> parser = argparse.ArgumentParser(prog='game.py')
>>> parser.add_argument('move', choices=['rock', 'paper', 'scissors'])
>>> parser.parse_args(['rock'])
Namespace(move='rock')
>>> parser.parse_args(['fire'])
usage: game.py [-h] {rock,paper,scissors}
game.py: error: argument move: invalid choice: 'fire' (choose from 'rock',
'paper', 'scissors')

简短的回答

使用nargs选项或action选项的“append”设置(取决于您希望用户界面如何表现)。

纳尔格斯

parser.add_argument('-l','--list', nargs='+', help='<Required> Set flag', required=True)
# Use like:
# python arg.py -l 1234 2345 3456 4567

Nargs ='+'接受一个或多个参数，Nargs ='*'接受零或多个参数。

附加

parser.add_argument('-l','--list', action='append', help='<Required> Set flag', required=True)
# Use like:
# python arg.py -l 1234 -l 2345 -l 3456 -l 4567

使用append，您可以多次提供该选项来构建列表。

不要使用type=list!!-可能在任何情况下你都不需要在argparse中使用type=list。永远。

长回答

让我们更详细地看看人们可能尝试的一些不同的方法，以及最终的结果。

import argparse

parser = argparse.ArgumentParser()

# By default it will fail with multiple arguments.
parser.add_argument('--default')

# Telling the type to be a list will also fail for multiple arguments,
# but give incorrect results for a single argument.
parser.add_argument('--list-type', type=list)

# This will allow you to provide multiple arguments, but you will get
# a list of lists which is not desired.
parser.add_argument('--list-type-nargs', type=list, nargs='+')

# This is the correct way to handle accepting multiple arguments.
# '+' == 1 or more.
# '*' == 0 or more.
# '?' == 0 or 1.
# An int is an explicit number of arguments to accept.
parser.add_argument('--nargs', nargs='+')

# To make the input integers
parser.add_argument('--nargs-int-type', nargs='+', type=int)

# An alternate way to accept multiple inputs, but you must
# provide the flag once per input. Of course, you can use
# type=int here if you want.
parser.add_argument('--append-action', action='append')

# To show the results of the given option to screen.
for _, value in parser.parse_args()._get_kwargs():
    if value is not None:
        print(value)

以下是你可以期待的输出:

$ python arg.py --default 1234 2345 3456 4567
...
arg.py: error: unrecognized arguments: 2345 3456 4567

$ python arg.py --list-type 1234 2345 3456 4567
...
arg.py: error: unrecognized arguments: 2345 3456 4567

$ # Quotes won't help here... 
$ python arg.py --list-type "1234 2345 3456 4567"
['1', '2', '3', '4', ' ', '2', '3', '4', '5', ' ', '3', '4', '5', '6', ' ', '4', '5', '6', '7']

$ python arg.py --list-type-nargs 1234 2345 3456 4567
[['1', '2', '3', '4'], ['2', '3', '4', '5'], ['3', '4', '5', '6'], ['4', '5', '6', '7']]

$ python arg.py --nargs 1234 2345 3456 4567
['1234', '2345', '3456', '4567']

$ python arg.py --nargs-int-type 1234 2345 3456 4567
[1234, 2345, 3456, 4567]

$ # Negative numbers are handled perfectly fine out of the box.
$ python arg.py --nargs-int-type -1234 2345 -3456 4567
[-1234, 2345, -3456, 4567]

$ python arg.py --append-action 1234 --append-action 2345 --append-action 3456 --append-action 4567
['1234', '2345', '3456', '4567']

外卖:

Use nargs or action='append' nargs can be more straightforward from a user perspective, but it can be unintuitive if there are positional arguments because argparse can't tell what should be a positional argument and what belongs to the nargs; if you have positional arguments then action='append' may end up being a better choice. The above is only true if nargs is given '*', '+', or '?'. If you provide an integer number (such as 4) then there will be no problem mixing options with nargs and positional arguments because argparse will know exactly how many values to expect for the option. Don't use quotes on the command line1 Don't use type=list, as it will return a list of lists This happens because under the hood argparse uses the value of type to coerce each individual given argument you your chosen type, not the aggregate of all arguments. You can use type=int (or whatever) to get a list of ints (or whatever)

他说:我不是说一般情况。我的意思是使用引号将列表传递给argparse不是你想要的。

我认为最优雅的解决方案是传递一个lambda函数给“类型”，正如Chepner所提到的。除此之外，如果你事先不知道列表的分隔符是什么，你也可以传递多个分隔符给re.split:

# python3 test.py -l "abc xyz, 123"

import re
import argparse

parser = argparse.ArgumentParser(description='Process a list.')
parser.add_argument('-l', '--list',
                    type=lambda s: re.split(' |, ', s),
                    required=True,
                    help='comma or space delimited list of characters')

args = parser.parse_args()
print(args.list)


# Output: ['abc', 'xyz', '123']

在argparse的add_argument方法中使用nargs参数

我使用nargs='*'作为add_argument参数。如果我没有传递任何显式参数，我特别使用nargs='*'选项来选择默认值

包括一个代码片段作为示例:

例如:temp_args1.py

请注意:下面的示例代码是用python3编写的。通过改变打印语句的格式，可以在python2中运行

#!/usr/local/bin/python3.6

from argparse import ArgumentParser

description = 'testing for passing multiple arguments and to get list of args'
parser = ArgumentParser(description=description)
parser.add_argument('-i', '--item', action='store', dest='alist',
                    type=str, nargs='*', default=['item1', 'item2', 'item3'],
                    help="Examples: -i item1 item2, -i item3")
opts = parser.parse_args()

print("List of items: {}".format(opts.alist))

注意:我正在收集多个字符串参数，这些参数存储在list - opt . list中 如果需要整数列表，请更改解析器上的类型参数。Add_argument到int

执行结果:

python3.6 temp_agrs1.py -i item5 item6 item7
List of items: ['item5', 'item6', 'item7']

python3.6 temp_agrs1.py -i item10
List of items: ['item10']

python3.6 temp_agrs1.py
List of items: ['item1', 'item2', 'item3']

如果你有一个嵌套列表，其中内部列表有不同的类型和长度，你想要保留类型，例如:

[[1,2]， ["foo"， "bar"]， [3.14， "baz"， 20]]]

然后你可以使用@sam-mason提出的解决方案来解决这个问题，如下所示:

from argparse import ArgumentParser
import json

parser = ArgumentParser()
parser.add_argument('-l', type=json.loads)
parser.parse_args(['-l', '[[1,2],["foo","bar"],[3.14,"baz",20]]'])

这使:

Namespace(l=[[1, 2], ['foo', 'bar'], [3.14, 'baz', 20]])

您可以将列表解析为字符串，并使用eval内置函数将其作为列表读取。在这种情况下，为了确保成功地解析字符串，必须将单引号(或双引号)放入双引号中。

# declare the list arg as a string
parser.add_argument('-l', '--list', type=str)

# parse
args = parser.parse()

# turn the 'list' string argument into a list object
args.list = eval(args.list)
print(list)
print(type(list))

测试:

python list_arg.py --list "[1, 2, 3]"

[1, 2, 3]
<class 'list'>