Question

我正在寻找最清晰，最Pythonic且最快的方法来检查字符串是否包含列表中的单词

这是我到目前为止提出的内容

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

此代码可以很好地查找所有组合。这是一个嵌套的for循环。乍一看还不清楚。

我想知道是否有更好的方法。

Answer 1

将any()与链式列表理解结合使用会更Python化：

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

但是这只会返回true / false；它不会识别在哪个子列表中找到哪个词。您可以使用此列表理解来打印特定的匹配项：

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。

Answer 2

我正在寻找最清晰，最Pythonic且最快的方法来检查字符串是否包含列表中的单词

首先，我将列表弄平

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，分割字符串：

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]

那是非常蟒蛇的，对速度或可靠性不是很确定

Answer 3

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())

尽管set需要一些空间，但可以提高效率。谢谢彼得·伍德。

Answer 4

使用itertools

import itertools
merged = list(itertools.chain.from_iterable(allStrings))
[print(x) for x in testtt.split() if x in merged]

最清晰，Python式，可靠且最快的方法来检查字符串是否包含列表中的单词

4 个答案: