[英]IndexError: string index out of range on my function
我正在尝试创建一个函数,该函数使我可以拆分一个字符串并将每个单词添加到列表中,然后不使用.split()命令而返回该列表中以某个字母开头的单词。 函数的第一部分(将字符串拆分并将每个单词添加到列表中)工作得很好。 问题是当尝试返回该列表中以某个字母开头的值时。 这是我的代码:
def getWordsStartingWith(text, letter):
split_text = [] #This is where each word is appeneded to.
text_addition = "" #This is where the letters from the string are added.
number = 0
gWSW = []
for str in text:
if str == ' ' or str == "'": # Checks to see whether the letter is a space or apostrophy.
split_text.append(text_addition)
text_addition = "" #If there is, then the letters collected so far in text_addition are apended to the list split_text and then cleared from text_addition
else:
text_addition += str #If not, then the letter is added to the string text_addition.
while number < len(split_text)-1:
if split_text[number][0] == letter:
gWSW.append(split_text[number])
number += 1
else:
number += 1
else:
return gWSW
问题在于线
如果split_text [number] [0] ==字母:
如标题中所述返回IndexError。 我很确定它与正在使用的[number]变量有关,但不确定该怎么做。
就像您对问题的评论中指出的那样,您在其中也有很多问题,首先您要删除最后一个单词,可以通过以下方法解决此问题:
else:
text_addition += str #If not, then the letter is added to the string text_addition.
# Avoid dropping last word
if len(text_addition):
split_text.append(text_addition)
while number < len(split_text)-1:
if split_text[number][0] == letter:
然后,我认为您的IndexError问题是在您有两个“空格”时出现的,在这种情况下,您正在添加一个空字符串,并且由于它没有任何char [0],因此是indexError。 您可以使用以下方法解决此问题:
for str in text:
if str == ' ' or str == "'": # Checks to see whether the letter is a space or apostrophy.
if text_addition:
# Here we avoid adding empty strings
split_text.append(text_addition)
text_addition = "" #If there is, then the letters collected so far in text_addition are apended to the list split_text and then cleared from text_addition
else:
text_addition += str #If not, then the letter is added to the string text_addition.
那只是为了回答您的问题。
PD:我在最后一部分所做的一点改进是:
result = []
for str in split_text:
if str.startswith(letter):
result.add(str)
return result
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.