我正在寻找最清晰、最快捷的方法来检查字符串是否包含列表中的单词。
这就是我到目前为止想出来的
introStrings = ['introduction:' , 'case:' , 'introduction' , 'case' ]
backgroundStrins = ['literature:' , 'background:', 'Related:' , 'literature' , 'background', 'related' ]
methodStrings = [ 'methods:' , 'method:', 'techniques:', 'methodology:' , 'methods' , 'method', 'techniques', 'methodology' ]
resultStrings = [ 'results:', 'result:', 'experimental:', 'experiments:', 'experiment:', 'results', 'result', 'experimental', 'experiments', 'experiment']
discussioStrings = [ 'discussion:' , 'Limitations:' , 'discussion' , 'limitations']
conclusionStrings = ['conclusion:' , 'conclusions:', 'concluding:' , 'conclusion' , 'conclusions', 'concluding' ]
allStrings = [ introStrings, backgroundStrins, methodStrings, resultStrings, discussioStrings, conclusionStrings ]
testtt = 'this may thod be in techniques ever material and methods'
for item in allStrings:
for word in testtt.split():
if word in item:
print('yes')
break这段代码非常适合所有组合。这是一个嵌套的for循环。乍一看还不太清楚。
我想知道是否有更好的方法。
发布于 2019-06-21 22:26:54
我能得到的是使用chain和any
resultStrings = [
"results:",
"result:",
"experimental:",
"experiments:",
"experiment:",
"results",
"result",
"experimental",
"experiments",
"experiment",
]
conclusionStrings = [
"conclusion:",
"conclusions:",
"concluding:",
"conclusion",
"conclusions",
"concluding",
]
allStrings = [resultStrings, conclusionStrings]
testtt = "this may thod be in techniques ever material and methods"
from itertools import chain
string_set = set(chain(*allStrings))
any(i in string_set for i in testtt.split())虽然set需要一定的空间,但它可以提高效率。谢谢彼得伍德.
发布于 2019-06-21 22:17:13
如果将any()与链式列表理解结合使用,则会更具有Pythonic风格:
print any(word in sublist for word in testtt.split() for sublist in allStrings)但是,这只会返回true/false;它不会标识在哪个子列表中找到了哪个单词。您可以使用此列表理解打印特定的匹配:
print [(word,sublist) for word in testtt.split() for sublist in allStrings if word in sublist]通过多次计算testtt.split(),您的代码有点浪费。
发布于 2019-06-21 22:22:07
我正在寻找最清晰、最快捷的方法来检查字符串是否包含列表中的单词。
首先,我会把名单弄平
all_strings = [*intro, *back, *methods, ...] # You get the idea(或者,使用嵌套列表理解)
all_strings = [word for list in [intro, back, ...] for word in list] # if you're into that接下来,拆分字符串:
string_words = a_string.split()最后,只需查找单词:
found = [w for w in string_words if w in all_strings]那是一种很重的电子琴,对速度和可靠性不太确定。
https://stackoverflow.com/questions/56710904
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