对于句子中的不同单词，我怎样才能得到一个 output 像 A=B

Question

I'm trying to find different words in 2 different sentence and my sentences inside of Excel's A row and B row.

def UncommonWords(A, B): 

    res = len(test_string.split()) 
    count = {} 

    for word in A.split(): 
        count[word] = count.get(word, 0) + 1

    for word in B.split(): 
        count[word] = count.get(word, 0) + 1

    return [word for word in count if count[word] == 1]  


A = "wu tang clan"
B = "wu tang can"

print(UncommonWords(A, B)) 
print(A,'=',B)

A = "wu tang clan" B = "wu tang can" I have A and B sentences. As you can see 'clan' and 'can' words different. I need an output like:

clan=can

So I dont want to match not unique words. And my A and B sentences always gonna change because i will get my sentences from Excel but my code just return me

wu tang clan=wu tang can

in my code return part give me an output like ['clan'],['can'] but this not seperated. I thought if i can seperate them like that 'A sentence unique word: clan' as dif1 'B sentence unique word: clan' as dif2

i will put dif1 and dif2 to print(dif1,'=',dif2) probably my output will ['clan=can'] and its so fine for me. By the way sorry for my English, thank you for helps.

Edit1:

 Specific examples:

 A = "put returns between two sentence"
 B = "put returns bet ween 2 sentence"

 Output should be like: between two = bet ween 2

 So i have to find unique words in A with in order then find in B, after that output should be as above.

Edit2:

 A = "a breath before surfacing"
 B = "a be before sura"

 Output should be like: breath=be|surfacing=sura

 uncommon_word_in_A[0] ?= uncommon_word_in_B[0]
 uncommon_word_in_A[1] ?= uncommon_word_in_B[1]

Answer 1

The result you get is to be expected since you print your arguments instead of printing the result of the function (noticed the return statement at the end of the function?). You'd want something like:

result = UncommonWords(A, B)
print("{} = {}".format(result[0], result[1])

but this blindly assumes you always have exactly one word of difference between your two strings - which will certainly not be the case for real data.

Also, your code is very inefficient. Using the builtin set type and it's symmetric_difference operation gives the same result in much less time and much less code:

print(set(A).symmetric_difference(set(B)))

Oh and yes, Python's naming conventions are all_lower for variables, functions and modules and CamelCase for classes.

Answer 2

Here is sample code for your reference

 def UncommonWords(A, B): 
    A = A.split()
    B = B.split()
    uncommon = [(A[i],B[i]) for i in range(0,len(A)) if not str(A[i])==str(B[i])]
    return uncommon


A = "wu tang clan"
B = "wu tang can"

uncommon = UncommonWords(A, B) 
print(['='.join([i[0],i[1]]) for i in uncommon])

Hope this helps

Answer 3

def UncommonWords(A, B):
    A_list=A.split()
    B_list=B.split()
    uncommon_word_in_A = []
    uncommon_word_in_B = []
    i = 0
    for word in A_list:
        if word not in B_list:
            uncommon_word_in_A.append(word)


    for word2 in B_list:
        if word2 not in A_list:
            uncommon_word_in_B.append(word2)

    for i in range(0, len(uncommon_word_in_A)):
        print(uncommon_word_in_A[i], end=" ")

    print("=", end=" ")

    for i in range(0, len(uncommon_word_in_B)):
        print(uncommon_word_in_B[i], end=" ")


A = "put returns between two sentence"
B = "put returns bet ween 2 sentence"

UncommonWords(A, B)

Output:

 between two = bet ween 2