最清晰的，Pythonic的，可靠的，和最快的方法来检查一个字符串是否包含来自列表的单词

Question

我正在寻找最清晰、最通俗、最快速的方法来检查字符串是否包含列表中的单词

到目前为止我就是这么想的

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的组合。这是一个嵌套for循环。乍一看还不太清楚。你知道吗

我想知道有没有更好的办法。你知道吗

Answer 1

我可以使用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需要一些空间，但它可以提高效率。谢谢彼得·伍德。

Answer 2

I am looking for the most clear, Pythonic, and fastest way to check if a string contains words from a list of lists

首先，我要把名单弄平

all_strings = [*intro, *back, *methods, ...] # You get the idea

（或者，使用嵌套列表）

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]

这是很霸道，不太确定的速度或可靠性

Answer 3

将any()与链式列表结合使用会更具python风格：

print any(word in sublist for word in testtt.split() for sublist in allStrings)

然而，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。您可以使用以下列表打印特定匹配项：

print [(word,sublist) for word in testtt.split() for sublist in allStrings if word in sublist]

多次计算testtt.split()会有点浪费代码。你知道吗