Question

尝试匹配并标记基于字符的n-gram。字符串

txt = "how does this work"

与列表中的n-gram相匹配

ngrams = ["ow ", "his", "s w"]

并标有<> - 但是，仅当没有先前打开的报价时。我正在寻找这个字符串的输出是h<ow >does t<his w>ork（注意2 ^-nd部分中的双匹配，但只有1对预期的引号）。

我试过的for循环并没有产生所需的输出：

switch = False

for i in txt:
    if i in "".join(ngrams) and switch == False:
        txt = txt.replace(i, "<" + i)
        switch = True
    if i not in "".join(ngrams) and switch == True:
        txt = txt.replace(i, ">" + i)
        switch = False

print(txt)

非常感谢任何帮助。

Answer 1

这应该有效：

else

Answer 2

此解决方案使用str.find方法在txt字符串中查找ngram的所有副本，将每个副本的索引保存到indices集，以便我们可以轻松处理重叠匹配。

然后我们将txt，char by char复制到result列表，在需要的地方插入尖括号。此策略比使用多个.replace调用插入尖括号更有效，因为每个.replace调用都需要重建整个字符串。

我稍微扩展了你的数据，以说明我的代码处理了ngram的多个副本。

txt = "how does this work now chisolm"
ngrams = ["ow ", "his", "s w"]
print(txt)
print(ngrams)

# Search for all copies of each ngram in txt
# saving the indices where the ngrams occur
indices = set()
for s in ngrams:
    slen = len(s)
    lo = 0
    while True:
        i = txt.find(s, lo)
        if i == -1:
            break
        lo = i + slen 
        print(s, i)
        indices.update(range(i, lo-1))

print(indices)

# Copy the txt to result, inserting angle brackets
# to show matches
switch = True
result = []
for i, u in enumerate(txt):
    if switch:
        if i in indices:
            result.append('<')
            switch = False
        result.append(u)
    else:
        result.append(u)
        if i not in indices:
            result.append('>')
            switch = True

print(''.join(result))

<强>输出

how does this work now chisolm
['ow ', 'his', 's w']
ow  1
ow  20
his 10
his 24
s w 12
{1, 2, 10, 11, 12, 13, 20, 21, 24, 25}
h<ow >does t<his w>ork n<ow >c<his>olm

如果要合并相邻的组，我们可以使用str.replace方法轻松完成。但为了使其正常工作，我们需要预处理原始数据，将所有空白行转换为单个空格。一种简单的方法是拆分数据并重新加入。

txt = "how  does this\nwork  now chisolm hisow"
ngrams = ["ow", "his", "work"]

#Convert all whitespace to single spaces
txt = ' '.join(txt.split())

print(txt)
print(ngrams)

# Search for all copies of each ngram in txt
# saving the indices where the ngrams occur
indices = set()
for s in ngrams:
    slen = len(s)
    lo = 0
    while True:
        i = txt.find(s, lo)
        if i == -1:
            break
        lo = i + slen 
        print(s, i)
        indices.update(range(i, lo-1))

print(indices)

# Copy the txt to result, inserting angle brackets
# to show matches
switch = True
result = []
for i, u in enumerate(txt):
    if switch:
        if i in indices:
            result.append('<')
            switch = False
        result.append(u)
    else:
        result.append(u)
        if i not in indices:
            result.append('>')
            switch = True

# Convert the list to a single string
output = ''.join(result)

# Merge adjacent groups
output = output.replace('> <', ' ').replace('><', '')
print(output)

<强>输出

how does this work now chisolm hisow
['ow', 'his', 'work']
ow 1
ow 20
ow 34
his 10
his 24
his 31
work 14
{32, 1, 34, 10, 11, 14, 15, 16, 20, 24, 25, 31}
h<ow> does t<his work> n<ow> c<his>olm <hisow>

Answer 3

这是一个只有一个for循环的方法。我把它计时了，它和这个问题的其他答案一样快。我认为它更清楚一点，尽管那可能是因为我写了它。

我遍历n-gram中第一个字符的索引，如果匹配，我会使用一堆if-else子句来查看是否应该添加<或>在这种情况下。我从原始output添加到字符串txt的末尾，所以我并没有真正插入字符串的中间。

txt = "how does this work"
ngrams = set(["ow ", "his", "s w"])
n = 3
prev = -n
output = ''
shift = 0
open = False
for i in xrange(len(txt) - n + 1):
    ngram = txt[i:i + n]
    if ngram in ngrams:
        if i - prev > n:
            if open:
                output += txt[prev:prev + n] + '>' + txt[prev + n:i] + '<'
            elif not open:
                if prev > 0:
                    output += txt[prev + n:i] + '<'
                else:
                    output += txt[:i] + '<'
                open = True
        else:
            output += txt[prev:i]
        prev = i
if open:
    output += txt[prev:prev + n] + '>' + txt[prev + n:]
print output

Python：替换字符串，从列表中匹配

3 个答案: