Python如何摆脱PyDictionary错误消息 [英] Python how to get rid of PyDictionary error messages

问题描述

我有以下代码来检查字典中是否有单词.如果该词不存在，则对dictionary.意义的调用将返回None.问题在于它还会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:的组合，但:但是:无论我尝试了什么，错误消息仍然会打印出来.这是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?

I have the following code to check if a word is in the dictionary. If the word does not exist the call to dictionary.meaning returns None. The problem is that it also spits out an error message "Error: The Following Error occured: list index out of range". I did some research and it appeared that I could use a combination of try:, except: but no matter what I tried the error message is still printed out. Here is a test case that shows the problem. How can I make this code work without displaying the index error?

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

推荐答案

我猜测您的try/except块位于错误的块周围，或者您没有正确捕获它，但是如果没有代码，很难分辨出来.

I'm guessing your try/except block was around the wrong block, or you weren't catching it properly, but it's tough to tell without your code.

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

Try putting the try/except around the section of code that would be erroring (the dictionary check in this case).

我的错误.错误已由 PyDictionary 库.您应该可以通过执行意义(单词，disable_errors = True)使它静音.

My mistake. The error is getting printed by the PyDictionary library. You should be able to silence it by doing 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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