How can I check multiple words in string, by some conditions?

Question

I need a wise advice from Stack Overflow again. I'm not sure the title is properly showing what I am wondering right now.

The thing is this.

there is two groups of words, and I need to know if a string has one(or more) word in group A while it also has a word in group B. Like... this.

Group_A = ['nice','car','by','shop']
Group_B = ['no','thing','great']

t_string_A = 'there is a car over there'
t_string_B = 'no one is in a car'

t_string_A has 'car' from Group_A, while nothing from Group_B, so it must return... I don't know, let's say 0 while t_string_B has 'car' from Group_A, and 'no' from Group_B, so it should return 1

Actually I was doing this job by somewhat... primitive way. Like bunch of sets of codes like

if 'nice' in t_string_A and 'no' in t_string_A:
    return 1

But as you know, as the length of Group A or B increases, I should make too many sets of those sets. And this is certainly not efficient.

I appreciate your help and attention :D Thanks in advance!

Answer 1

you could work with sets:

Group_A = set(('nice','car','by','shop'))
Group_B = set(('no','thing','great'))

t_string_A = 'there is a car over there'
t_string_B = 'no one is in a car'

set_A = set(t_string_A.split())
set_B = set(t_string_B.split())

def test(string):
    s = set(string.split())
    if Group_A & set_A and Group_B & set_A:
        return 1
    else:
        return 0

what should be the result if there are no words from Group_A and Group_B?

depending on your phrases the test may be more efficient this way:

def test(string):
    s = string.split()
    if any(word in Group_A for word in s) and any(word in Group_B for word in s):
        return 1
    else:
        return 0

Answer 2

You can use itertools.product to generate all possible pairs of words from the give groups. You then iterate through the list of strings, and if a pair is present in the string, the result is True, otherwise result is False.

import itertools as it

Group_A = ['저는', '저희는', '우리는']
Group_B = ['입니다','라고 합니다']

strings = [ '저는 학생입니다.', '저희는 회사원들 입니다.' , '이 것이 현실 입니다.', '우리는 배고파요.' , '우리는 밴디스트라고 합니다.']

#Get all possible combinations of words from the group
z = list(it.product(Group_A, Group_B))

results = []

#Run through the list of string
for s in strings:
    flag = False
    for item in z:
        #If the word is present in the string, flag is True
        if item[0] in s and item[1] in s:
            flag = True
            break
    #Append result to results string
    results.append(flag)

print(results)

The result will then look like

[True, True, False, False, True]

In addition for the inputs below

Group_A = ['thing']
Group_B = ['car']
strings = ['there is a thing in a car', 'Nothing is in a car','Something happens to my car']

The values will be [True, True, True]

Answer 3

Group_A = ['nice','car','by','shop']
Group_B = ['no','thing','great']

from collections import defaultdict

group_a=defaultdict(int)
group_b=defaultdict(int)

for i in Group_A:
    group_a[i]=1

for i in Group_B:
    group_b[i]=1

t_string_A = 'there is a car over there'
t_string_B = 'no one is in a car'

def fun2(string):
    l=[]
    past=0
    for i in range(len(string)):
        if string[i]==' ':
            if string[past:i]!='':
                l.append(string[past:i])
            past=i+1
    return l

def fun(string,dic):
    for i in fun2(string):
   # for i in string.split():
        try:
            if dic[i]:
                return 1
        except:
            pass
    return 0

if fun(t_string_A,group_a)==fun(t_string_B,group_b):
    print(1)
else:
    print(0)

Answer 4

This can be solved efficiently as variations on the Aho Corasick algorithm

It is an efficient dictionary matching algorithm that locates patterns within text simultaneously in O(p + q + r), with p = length of patterns, q = length of text, r = length of returned matches.

You may want to run two separate state machines simultaneously, and you would need to modify them so they terminate on the first match.

I took a stab at the modifications, starting with this python implementation

class AhoNode(object):
    def __init__(self):
        self.goto = {}
        self.is_match = False
        self.fail = None

def aho_create_forest(patterns):
    root = AhoNode()
    for path in patterns:
        node = root
        for symbol in path:
            node = node.goto.setdefault(symbol, AhoNode())
        node.is_match = True
    return root

def aho_create_statemachine(patterns):
    root = aho_create_forest(patterns)
    queue = []
    for node in root.goto.itervalues():
        queue.append(node)
        node.fail = root
    while queue:
        rnode = queue.pop(0)
        for key, unode in rnode.goto.iteritems():
            queue.append(unode)
            fnode = rnode.fail
            while fnode is not None and key not in fnode.goto:
                fnode = fnode.fail
            unode.fail = fnode.goto[key] if fnode else root
            unode.is_match = unode.is_match or unode.fail.is_match
    return root

def aho_any_match(s, root):
    node = root
    for i, c in enumerate(s):
        while node is not None and c not in node.goto:
            node = node.fail
        if node is None:
            node = root
            continue
        node = node.goto[c]
        if node.out:
            return True
    return False

def all_any_matcher(*pattern_lists):
    ''' Returns an efficient matcher function that takes a string
    and returns True if at least one pattern from each pattern list
    is found in it.
    '''
    machines = [aho_create_statemachine(patterns) for patterns in pattern_lists]

    def matcher(text):
        return all(aho_any_match(text, m) for m in machines)
    return matcher

and to use it

patterns_a = ['nice','car','by','shop']
patterns_b = ['no','thing','great']

matcher = all_any_matcher(patterns_a, patterns_b)

text_1 = 'there is a car over there'
text_2 = 'no one is in a car'
for text in (text_1, text_2):
    print '%r - %s' % (text, matcher(text))

This displays

'there is a car over there' - False
'no one is in a car' - True

Answer 5

You can iterate over the words and see if any of them is in the string:

from typing import List

def has_word(string: str, words: List[str]) -> bool:
    for word in words:
        if word in string:
            return True
    return False

This function can be modified easily to have has_all_words too.