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python - Python如何摆脱PyDictionary错误消息

问题描述

我有以下代码来检查字典中是否有单词。如果单词不存在，则对 dictionary.meaning 的调用返回 None。问题是它还吐出一条错误消息“错误：发生以下错误：列表索引超出范围”。我做了一些研究，看来我可以使用 try:, except: 的组合，但无论我尝试了什么，错误消息仍然会打印出来。这是一个显示问题的测试用例。如何在不显示索引错误的情况下使此代码工作？

代码：

    def is_word(word):
        from PyDictionary import PyDictionary
        dictionary=PyDictionary()
        rtn = (dictionary.meaning(word))
        if rtn == None:
           return(False)
        else:
           return (True)

    my_list = ["no", "act", "amp", "xibber", "xyz"]

    for word in my_list:
        result = is_word(word)
        if result == True:       
           print(word, "is in the dictionary")
        else:
           print(word, "is NOT in the dictionary")

输出：

no is in the dictionary
act is in the dictionary
amp is in the dictionary
Error: The Following Error occured: list index out of range
xibber is NOT in the dictionary
Error: The Following Error occured: list index out of range
xyz is NOT in the dictionary

标签： pythonpython-3.xdictionary

解决方案

我猜你的 try/except 块是在错误的块周围，或者你没有正确地捕捉到它，但是如果没有你的代码就很难判断。

尝试将 try/except 放在可能出错的代码部分（在这种情况下是字典检查）。

编辑：

我的错。该错误正在由PyDictionary库打印。您应该能够通过执行使其静音meaning(word, disable_errors=True)。

def is_word(word):
    from PyDictionary import PyDictionary

    dictionary = PyDictionary()

    try:
        output = dictionary.meaning(word, disable_errors=True)
    except:
        return False
    else:
        return bool(output)

my_list = ["no", "act", "amp", "xibber", "xyz"]

for word in my_list:
    result = is_word(word)
    if result:       
       print("{} is in the dictionary".format(word))
    else:
       print("{} is NOT in the dictionary".format(word))

第二次编辑：使用https://github.com/tasdikrahman/vocabulary。

from vocabulary.vocabulary import Vocabulary
vb = Vocabulary()

my_list = ["no", "act", "amp", "xibber", "xyz"]

for word in my_list:
    if vb.meaning(word):
       print("{} is in the dictionary".format(word))
    else:
       print("{} is NOT in the dictionary".format(word))

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