我在 python 的文件中寻找一些单词。找到每个单词后，我需要从文件中读取接下来的两个单词。我一直在寻找一些解决方案，但我找不到只阅读接下来的单词。
# offsetFile - file pointer
# searchTerms - list of words
for line in offsetFile:
for word in searchTerms:
if word in line:
# here get the next two terms after the word
感谢您的时间。
更新：只需要第一次出现。实际上，在这种情况下，该词只可能出现一次。
文件：
accept 42 2820 access 183 3145 accid 1 4589 algebra 153 16272 algem 4 17439 algol 202 6530
词：['访问'，'代数']
当我遇到“访问”和“代数”时搜索文件，我分别需要 183 3145 和 153 16272 的值。
解决这个问题的一种简单方法是使用生成器读取文件，该生成器一次从文件中生成一个单词。
def words(fileobj):
for line in fileobj:
for word in line.split():
yield word
然后找到您感兴趣的单词并阅读接下来的两个单词：
with open("offsetfile.txt") as wordfile:
wordgen = words(wordfile)
for word in wordgen:
if word in searchterms: # searchterms should be a set() to make this fast
break
else:
word = None # makes sure word is None if the word wasn't found
foundwords = [word, next(wordgen, None), next(wordgen, None)]
现在foundwords[0]是您找到的单词，foundwords[1]是之后的单词，foundwords[2]是之后的第二个单词。如果没有足够的单词，则列表的一个或多个元素将是None.
如果您想强制它仅在一行内匹配，这会稍微复杂一些，但通常您可以将文件视为只是一个单词序列。
def searchTerm(offsetFile, searchTerms):
# remove any found words from this list; if empty we can exit
searchThese = searchTerms[:]
for line in offsetFile:
words_in_line = line.split()
# Use this list comprehension if always two numbers continue a word.
# Else use words_in_line.
for word in [w for i, w in enumerate(words_in_line) if i % 3 == 0]:
# No more words to search.
if not searchThese:
return
# Search remaining words.
if word in searchThese:
searchThese.remove(word)
i = words_in_line.index(word)
print words_in_line[i:i+3]
对于“访问”、“代数”，我得到以下结果：
['访问'，'183'，'3145']
['代数'，'153'，'16272']
如果您只需要检索前两个单词，只需执行以下操作：
offsetFile.readline().split()[:2]
word = '3' #Your word
delim = ',' #Your delim
with open('test_file.txt') as f:
for line in f:
if word in line:
s_line = line.strip().split(delim)
two_words = (s_line[s_line.index(word) + 1],\
s_line[s_line.index(word) + 2])
break