Python 3 异常:TypeError:函数缺少 1 个必需的位置参数:'words' [英] Python 3 Exception: TypeError: function missing 1 required positional argument: 'words'

问题描述

我的函数需要一个参数词，但我希望程序在没有参数时不会崩溃.

My function needs one argument words, but i want the program not to crash if no argument.

def get_count(words):
    try:
        consonants_str = "bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ"
        vowels_str = "aeiouAEIOU"
        consonants_ls = []
        vowels_ls = []
        words_ls = []
        result = {"vowels": 0, "consonants": 0}`

        for element in consonants_str:
            consonants_ls.append(element)

        for element in vowels_str:
            vowels_ls.append(element)

        if type(words) != type(None):
            for element in words:
                words_ls.append(element)

            for element in words_ls:
                if element in vowels_ls:
                    result["vowels"] += 1
                if element in consonants_ls:
                    result["consonants"] += 1
                else:
                    continue
            else:
                result["vowels"] = 0
                result["consonants"] = 0

    except TypeError:
        result["vowels"] = 0
        result["consonants"] = 0

    answer = {"vowels": result["vowels"],"consonants": result["consonants"]}

    return answer`

所以如果我用

print(get_count())

我希望程序不会向我显示标题中的错误.例外情况应该在 get_count 的 def 中，因为它应该是一个关闭的文件.我不在同一个文件中执行，所以异常应该独立于其他文件.

I want, that the program doesn't show me an error like the one in the heading. The exception for that should be in the def of get_count because it should be a closed file. I don't execute in the same file, so the Exception should be independent from other files.

我希望你明白我的意思...

I hope you understand what I mean...

感谢您的回答！禁止

推荐答案

当你使用 def 声明你想要一个名为 words 的参数时，你并没有将参数传递给你的函数(words)，像这样调用你的函数或者用你想要的任何字符串输入

You're not passing an argument to your function, when you declared that you want an argument named words with def(words), call your function like this or with any string input you desire

print(get_count('AAAABBBKDKDKDKDA'))

如果你想让程序在没有参数传递时退出，这样做(记住单词现在是一个元组)

if you want the program to exit when no argument is passed, do this (remember words is now a tuple)

import sys

def get_count(*words):
    if not words: # if words is empty, no argument was passed
        sys.exit('You passed in no arguments')
    else:
        pass # do something with the items in words

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