如何检查python中最长的子字符串

Question

我有一个文本和一个概念清单如下。

concepts = ["data mining", "data", "data source"]
text = "levels and data mining of dna data source methylation"

我希望识别列表中的concepts是否在text中，并将concepts[1:]的所有出现替换为concepts[0]。因此，上述案文的结果应是；

"levels and data mining of dna data mining methylation"

我的代码如下：

concepts = ["data mining", "data", "data source"]
text = "levels and data mining of dna data source methylation"

if any(word in text for word in concepts):
    for terms in concepts[1:]:
        if terms in text:
            text=text.replace(terms,concepts[0])
        text=' '.join(text.split())
    print(text)

然而，我得到的输出为；

levels and data mining mining of dna data mining source methylation

看起来，data这个概念被data mining取代了，这是不正确的。更具体地说，我希望在替换时首先考虑最长的选项。

即使我更改了concepts的顺序，它也不起作用。

concepts = ["data mining", "data source", "data"]
text = "levels and data mining of dna data source methylation"

if any(word in text for word in concepts):
    for terms in concepts[1:]:
        if terms in text:
            text=text.replace(terms,concepts[0])
        text=' '.join(text.split())
    print(text)

我得到了上述代码的如下输出。

levels and data mining mining of dna data mining mining methylation

如果需要，我很乐意提供更多的细节。

Answer 1

这里的问题是您的迭代策略，每次只做一个替换。因为您的替换项包含要替换的术语之一，因此您将在以前的迭代中对已经更改为替换项的内容进行替换。

解决这一问题的一种方法是原子化地完成所有这些替换，使它们都同时发生，并且输出不会影响其他替换的结果。在这方面有几个战略：

1. 您可以将字符串分解为与您的各种术语相匹配的标记，并在事实之后替换它们(并确保没有任何重叠)。
2. 您可以使用一个对多个选项进行原子替换的函数。

#2的一个例子是Python的sub()库的re方法。下面是一个使用它的例子：

import re

concepts = ["data mining", "data source", "data"]
text = "levels and data mining of dna data source methylation"

# Sort targets by descending length, so longer targets that
# might contain shorter ones are found first
targets = sorted(concepts[1:], key=lambda x: len(x), reverse=True)
# Use re.escape to generate version of the targets with special characters escaped
target_re = "|".join(re.escape(item) for item in targets)

result = re.sub(target_re, concepts[0], text)

请注意，这仍然会导致data mining mining与您原来的一组替换，因为它没有现有mining的概念，在data之后。如果您想避免这种情况，可以简单地将要替换的实际项作为替换目标，以便在较短的时间内匹配：

import re

concepts = ["data mining", "data source", "data"]
text = "levels and data mining of dna data source methylation"

# Sort targets by descending length, so longer targets that
# might contain shorter ones are found first
#
# !!!No [1:] !!!
#
targets = sorted(concepts, key=lambda x: len(x), reverse=True)
# Use re.escape to generate version of the targets with special characters escaped
target_re = "|".join(re.escape(item) for item in targets)

result = re.sub(target_re, concepts[0], text)

Answer 2

琥珀的溶液很干净。我写了一个很长的表格版本，有一些评论，在单词中穿行，并展望未来，以检查是否匹配。它应该帮助您使用您原来的代码丢失的概念(检查多个单词匹配并避免双重替换)，因为它只处理相同数量的单词或单个单词匹配的替换，这将不适用于每一个“概念”列表。

concepts = ["data mining", "data source", "data"]
text = "levels and data mining of dna data source methylation"
textSplit = text.split()
finalText = ""
maxX = len(textSplit)
#add a look ahead for mulitwords
for x in range(0, maxX):
    tempSplit = concepts[0].split()
    tempMax = len(tempSplit)
    foundFullMatch = True
    for y in range(0,tempMax):
        if (x + tempMax <= maxX):
            if (textSplit[x+y] != tempSplit[y]):
                foundFullMatch = False
        else:
            foundFullMatch = False
    if (foundFullMatch):
        #skip past it in the loop
        x = x + tempMax
        continue
    else:
        # now start looking at rest of list - make sure is sorted with most words first
        for terms in concepts[1:]:
            tempSplit2 = terms.split()
            tempMax2 = len(tempSplit2)
            foundFullMatch = True
            for y in range(0,tempMax2):
                if (x + tempMax2 <= maxX):
                    if (textSplit[x+y] != tempSplit2[y]):
                        foundFullMatch = False
                else:
                    foundFullMatch = False
            if (foundFullMatch):
                if (tempMax == tempMax2):
                    # found match same number words - replace
                    for y in range(0,tempMax2):
                        textSplit[x+y] = tempSplit[y]
                    x = x + tempMax
                    continue
                else:
                    # found match but not same number of words as concept 0
                    if (tempMax2 == 1):
                        #covers 1 word answer
                        textSplit[x] = concepts[0]
                        continue
print(" ".join(textSplit))