Question

我的代码接受了句子并在该句中找到了一个单词。

如果单词在句子中，则需要说明找到了单词以及单词所在的位置。

如果单词不在句子中，则应显示错误消息。

我有这个：

print("Please insert your sentence without punctuation")
sentence=(input())
variable1='sentence'
print("Which word would you like to find in your sentence?")
word=input()
variable2='word'
if 'word'=='COUNTRY':
    'variable3'==5
    'variable4'==17
if word in sentence:
    print([word], "is in positions", [variable3], "and", [variable4]);
else:
    print("Your word is not in the sentence!")

Answer 1

Python序列提供index方法。它为您提供元素的索引，或者如果元素不在序列中则引发错误。在字符串上，它允许您查找子字符串。

>>> 'hello world'.index('world')
6
>>> 'hello world'.index('word')
ValueError: substring not found

基本上，您必须为句子和要搜索的单词添加输入。就是这样。

print("Insert sentence without punctuation...")
sentence=input()  # get input, store it to name `sentence`
print("Insert word to find...")
word=input()
try:
    idx = sentence.index(word)
except ValueError:  # so it wasn't in the sentence after all...
    print('Word', word, 'not in sentence', repr(sentence))
else:  # if we get here, IndexError was NOT thrown
    print('Word', word, 'first occurs at position', idx)

这里有一些警告，例如'fooworldbar'也会匹配。正确处理这些事情取决于人们究竟想要什么。我猜你真的想要 word 职位。

如果您需要“n单词”含义的位置，则必须将句子转换为单词列表。 str.split这样做。然后，您可以再次使用index。此外，如果您想要所有位置，则必须重复调用索引。

print("Insert sentence without punctuation...")
sentence = input()  # get input, store it to name `sentence`
words = sentence.split()  # split at whitespace, creating a list of words
print("Insert word to find...")
word=input()
positions, idx = [], -1
while idx < len(words):
    try:
        idx = words.index(word, idx+1)
    except ValueError:  # so it wasn't in the rest of the sentence after all...
        break
    else:  # if we get here, IndexError was NOT thrown
        positions.append(idx)  # store the index to list of positions
# if we are here, we have searched through the whole string
if positions:  # positions is not an empty list, so we have found some
    print('Word', word, 'appears at positions', ', '.join(str(pos) for pos in positions))
else:
    print('Word', word, 'is not in the sentence')

Answer 2

我想在提供的代码中处理一些误解。

首先，

print("Please insert your sentence without punctuation")
sentence=(input())

更简单

sentence = input("Please insert your sentence without punctuation")

现在我有一个名为sentence的变量，不应该混淆字符串'sentence'

同样我们可以说

word = input("Which word would you like to find in your sentence?")

再次给出另一个变量word，不要混淆字符串'word'

假设我们有参数，

sentence = "Has this got an elephant in?"

我们会搜索'elephant'

这个词

发布的代码尝试使用in，但这会发生：

>>> "elephant" in sentence
True
>>> "ele" in sentence
True
>>> "giraffe" in sentence
False
>>>

关闭。但还不够近。由于我们在'ele'中找到了'elephant'，所以它并不是在寻找整个单词。

如果您split将句子翻译成单词，如其他答案所示，则可以搜索整个单词和找到位置。（查找拆分;您可以选择除默认' '之外的其他字符。）

words = sentence.split()
word = 'ele'
words.index(word)

如果单词不存在，您将收到错误：

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 'ele' is not in list

我会把错误处理留给你。

Answer 3

您可以使用re模块：

import re

sentence = input('Sentence: ')
word =  input('Word: ')

## convert word in regular expression for search method.
regex_word = r'(' + word + ')(?=\s|$)'

## initialize search var.
search = re.search(regex_word, sentence)

if search:
    while search:
        match_pos = search.span()
        sentence = sentence[:match_pos[0]] + sentence[match_pos[1]:]

        print(word + ' is in position ' + str(match_pos[0]))
        search = re.search(regex_word, sentence)
else:
    print(word + ' is not present in this sentence')

需要句子分析方面的帮助

3 个答案: