You can use a set of algorithms to solve this problem. Of course, since the problem is complicated, the solution is also complex.

• First break both your string and phrases into words and record the indices. Now you have both of them as words.

• Loop through words of phrases and words of string, and check if you have an exactly similar word. If you do, then you “may” have a possible output within a close window of string. Here, you can create a window, and see if your phrase is in the window, “exactly” or “partially”. For “partially” you can use Levenshtein, and for “exact” you can use KMP.

I did not code the whole algorithm, will take you some time to do so. Partially, I did though, and you can do the rest.

import re, string, collections


def _matches(s, p):
    return [(match.group(0), match.start(), match.end()) for match in re.finditer(p, s)]


def get_lps(p):
    lps, j = [0] * len(p), 0
    for i in range(1, len(p)):
        while j and p[i] != p[j]:
            j = lps[j - 1]
        if p[i] == p[j]:
            j += 1
        lps[i] = j
    return lps


def kmp(s, p):
    lps, indices, j = get_lps(p), [], 0
    for i in range(len(s)):
        while j and s[i] != p[j]:
            j = lps[j - 1]
        if s[i] == p[j]:
            j += 1
        if j == len(p):
            indices.append(i - len(p) + 1)
            j = lps[j - 1]
    return indices


def levenshtein_edit_distance(A, B):
    dp = [[0] * (len(B) + 1) for _ in range(len(A) + 1)]
    for i in range(len(A) + 1):
        dp[i][0] = i
    for j in range(len(B) + 1):
        dp[0][j] = j

    for i in range(1, len(A) + 1):
        for j in range(1, len(B) + 1):
            if A[i - 1] == B[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]
            else:
                dp[i][j] = 1 + min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1])

    return dp[-1][-1]


phrase_list = ['Gregor Samse', 'in his bed into', 'horrible creature']

phs = list(set(phrase_list))
s = ' ' * 50 + 
    """
One morning, when Gregor Samsa woke from troubled dreams, he found himself transformed in his bed into a horrible vermin. 
He lay on his armour-like back, and if he lifted his head a little he could see his brown belly, slightly domed and divided by arches into stiff sections. The bedding was hardly able to cover it and seemed ready to slide off any moment. 
His many legs, pitifully thin compared with the size of the rest of him, waved about helplessly as he looked...""" 
+ ' ' * 50
p = r'(?:b[^rnst,?:;'"!_-]+?b)'

s_matches = _matches(s, p)
print(s_matches)

d = {ph: _matches(ph, p) for ph in phs}
print(d)


res = []
possibles = collections.defaultdict(list)
for k, val in d.items():
    for (word_a, st_a, end_a) in val:
        for (word_s, st_s, end_s) in s_matches:
            if word_a == word_s:
                search_window = s[st_s - 10:end_s + 10]
                found = kmp(search_window, word_a)[0]
                print(st_s, found, search_window)
                possibles[k] += [(st_s, found, search_window)]
                # break
print(possibles)

Levenshtein Distance

Knuth–Morris–Pratt algorithm

Similarity

Note that Levenshtein calculates the distance of two strings.

• If their distance is 0, it is an exact match.
• If their distance is close, it is a close match.
• If their distance is not close, it is not a match.

def levenshtein_edit_distance(A, B):
    dp = [[0] * (len(B) + 1) for _ in range(len(A) + 1)]
    for i in range(len(A) + 1):
        dp[i][0] = i
    for j in range(len(B) + 1):
        dp[0][j] = j

    for i in range(1, len(A) + 1):
        for j in range(1, len(B) + 1):
            if A[i - 1] == B[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]
            else:
                dp[i][j] = 1 + min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1])

    return dp[-1][-1]


phrase = 'horrible creature'
s = 'horrible vermin'
dist = levenshtein_edit_distance(phrase, s)
if levenshtein_edit_distance(phrase, s) < len(phrase) // 2:
    print(f"'{s}' and '{phrase}' are similar matches with a character distance of {dist}.")

Prints

‘horrible vermin’ and ‘horrible creature’ are similar matches with a
character distance of 7.