Python如何摆脱PyDictionary错误消息

Question

我有以下代码来检查字词中是否有单词。 如果该单词不存在，则对dictionary.meaning的调用将返回None。 问题是它还会发出错误消息“错误：发生以下错误：列表索引超出范围”。 我做了一些研究，似乎我可以使用try：的组合，除了：但无论我尝试了什么，仍然打印出错误信息。 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正常工作？

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

Answer 1

我猜你的try / except块是在错误的块附近，或者你没有正确捕获它，但是没有你的代码很难说。

尝试将try / except放在可能出错的代码部分周围（在本例中为字典检查）。

编辑：

我的错。 错误由PyDictionary库打印。 你应该能够通过做出meaning(word, disable_errors=True)来使它沉默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))