Python-比较两个列表时出现循环问题

Question

I have a little problem, I am trying to compare 2 lists with words in it to establish a similarity percentage but here is the thing, if I have the same word 2 times in each lists, I get a falsied percentage.

First I made this little script :

data1 = ['test', 'super', 'class', 'test', 'boom']
data2 = ['test', 'super', 'class', 'test', 'boom']
res = 0
nb = (len(data1) + len(data2)) / 2
if data1 and data2 and nb != 0:
    for id1, item1 in enumerate(data1):
        for id2, item2 in enumerate(data2):
            if item1 == item2:
                res += 1 - abs(id1 - id2) / nb
    print(res / nb * 100)

The problem is that if i have 2 time the same word in the lists the percentage will be greater than 100%. So to counter that, i added a 'break' just after the line 'res += 1 - abs(id1 - id2) / nb' but the percentage is still falsified.

I hope you've understand my problem, thanks you for your help !

Answer 1

You can use difflib.SequenceMatcher instead to compare the similarity of two lists. Try this :

from difflib import SequenceMatcher as sm
data1 = ['test', 'super', 'class', 'test', 'boom']
data2 = ['test', 'super', 'class', 'test', 'boom']
matching_percentage = sm(None, data1, data2).ratio() * 100

Output :

100.0

Answer 2

data1 = ['test', 'super', 'class', 'test', 'boom']
data2 = ['test', 'super', 'class', 'test', 'boom']
from collections import defaultdict

dic1 =defaultdict(int)
dic2=defaultdict(int)

for i in data1:
    dic1[i]+=1

for i in data2:
    dic2[i]+=1

count = 0

for i in dic1:
    if i in dic2.keys():
        count+=abs(dic2[i]-dic1[i])


result =( (1-count/(len(data1)+len(data2))) *100)

output

100.0

Answer 3

Try this code:

data1 = ['test', 'super', 'class', 'class', 'test', 'boom']
data2 = ['test', 'super', 'class', 'class', 'test', 'boom']
res = 0
nb = (len(data1) + len(data2)) / 2.0

def pos_iter(index, sz):
    yield index
    i1 = index - 1
    i2 = index + 1
    while i1 >=0 and i2 < sz:
        if i1 >= 0:
            yield i1
            i1 -=1
        if i2 < sz:
            yield i2
            i2 += 1
if data1 and data2 and nb != 0:
    for id1, item1 in enumerate(data1):
        for id2 in pos_iter(id1, len(data2)):
            item2 = data2[id2]
            if item1 == item2:
                res += max(0, 1 - abs(id1 - id2) / nb)
                break
    print(res / nb * 100)

Problem with your code is, that you look for matching word in second data2 always from beginning. Which will give you invalid values, if words repeat. You need to search always "around" position of word in data1 , because u want to find the closest one.

Also you need break you've added, otherwise text with all the same words will go way above 1.0. Your nb variable needs to be double (or python2 will round division result). And you should make sure 1 - abs(id1 - id2) / nb is greater than zero, hence i've added max(0, ...) .