[英]I am having a trouble with an algorithm for assinging syllable boundaries in Python
could you help me please? 请问你能帮帮我吗?
I am trying to desing an algorithm that assings the syllable boundaries for any given string of consonanats and vowels 'CVCCVCCVCC'. 我正在尝试设计一种算法,该算法可对任何给定的辅音和元音“ CVCCVCCVCC”字符串的音节边界进行评估。 These strings are stored in a list that may run into hundred of thousands or even million of itmes in length. 这些字符串存储在一个列表中,该列表的长度可能达到数十万甚至上百万个。 However, it seems that I did not understand handling lists properly and that is why I am getting this trouble. 但是,似乎我不了解如何正确地处理列表,这就是我遇到此麻烦的原因。 After composing the code, I click "Run" the module, and Python shell restarts, raising no syntax error. 编写代码后,单击“运行”模块,然后Python Shell重新启动,不会引发语法错误。 However, when I test the code, it returns an empty list. 但是,当我测试代码时,它返回一个空列表。 The code is here. 代码在这里。 Could you please tell me what is the wrong with my algorithm? 您能告诉我我的算法有什么问题吗? I am getting really disappointed. 我真的很失望。
Thanks in advance for kind help! 在此先感谢您的帮助!
def assing_boundaries(L):
""" (List of str -> List of str)
Assings the syllable boundaries for a given string of Syllable Templates, with the output meeting the two conditions given below:
Condition1: item[2:] != 'CC'
Condition2: item[:-2] !='C.C'
>>> assing_boundaries(['CVCCVCC', 'CVCCV:C'])
['CVC.CVCC', 'CVC.CV:C']
"""
boundary = []
i = 0
for item in L:
for char in item:
for i in range(len(item) - 1):
if item[i] == 'CC':
char = 'C.C'
i = i + 1
boundary = boundary.append(item)
return boundary
if item[i] == 'CC'
, but item[i]
can only ever be a single character. 您可以测试if item[i] == 'CC'
,但item[i]
只能是单个字符。 boundary = boundary.append(item)
is wrong, because .append
returns None
. boundary = boundary.append(item)
是错误的,因为.append
返回None
。 Just do boundary.append(item)
. 只需执行boundary.append(item)
。 Simpler yet, why not just do return [item.replace('CC', 'C.C') for item in L]
instead? 更简单的是,为什么不只return [item.replace('CC', 'C.C') for item in L]
呢? You can tweak the list comprehension to filter the list as needed. 您可以调整列表理解以根据需要过滤列表。
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