for循环中的意外输出-Python

Question

I have this list:

t=[['universitario de deportes'],['lancaster'],['universitario de'],['juan aurich'],['muni'],['juan']]

I want to reorder the list according to the jaccard distance. If I reorder t the expected ouput should be:

[['universitario de deportes'],['universitario de'],['lancaster'],['juan aurich'],['juan'],['muni']]

The code of the jackard distance is working OK, but the rest of the code doesn't give the expected output.The code is below:

def jack(a,b):
    x=a.split()
    y=b.split()
    k=float(len(set(x)&set(y)))/float(len((set(x) | set(y))))
    return k
t=[['universitario de deportes'],['lancaster'],['universitario de'],['juan aurich'],['muni'],['juan']]

import copy as cp


b=cp.deepcopy(t)

c=[]

while (len(b)>0):
    c.append(b[0][0])
    d=b[0][0]
    del b[0]
    for m in range (0 , len(b)+1):
        if m > len(b):
            break
            if jack(d,b[m][0])>0.3:
                c.append(b[m][0])
                del b[m]

Unfortunately, the unexpected output is the same list :

print c
['universitario de deportes', 'lancaster', 'universitario de', 'juan aurich', 'muni', 'juan']

EDIT:

I tried to correct my code but it didn't work too but I got a little closer to the expected output:

t=[['universitario de deportes'],['lancaster'],['universitario de'],['juan aurich'],['muni'],['juan']]

import copy as cp


b=cp.deepcopy(t)

c=[]

while (len(b)>0):
    c.append(b[0][0])
    d=b[0][0]
    del b[0]
    for m in range(0,len(b)-1):
        if jack(d,b[m][0])>0.3:
            c.append(b[m][0])
            del b[m]

The "close" output is:

['universitario de deportes', 'universitario de', 'lancaster', 'juan aurich', 'muni', 'juan']

Second edit:

Finally, I came up with a solution that has quite fast computational. Currently, I'll use the code to order 60 thousands names. The code is below:

t=['universitario de deportes','lancaster','lancaste','juan aurich','lancaster','juan','universitario','juan franco']

import copy as cp


b=cp.deepcopy(t)

c=[]

while (len(b)>0):
    c.append(b[0])
    e=b[0]
    del b[0]
    for val in b:
        if jack(e,val)>0.3:
            c.append(val)
            b.remove(val)

print c
['universitario de deportes', 'universitario', 'lancaster', 'lancaster', 'lancaste', 'juan aurich', 'juan', 'juan franco'

Answer 1

Firstly, not sure why you've got everything in single-item lists, so I suggest flattening it out first:

t = [l[0] for l in t]

This gets rid of the extra zero indices everywhere, and means you only need shallow copies (as strings are immutable).

Secondly, the last three lines of your code never run:

if m > len(b):
    break # nothing after this will happen
    if jack(d,b[m][0])>0.3:
       c.append(b[m][0])
       del b[m]

I think what you want is:

out = [] # this will be the sorted list
for index, val1 in enumerate(t): # work through each item in the original list
    if val1 not in out: # if we haven't already put this item in the new list
        out.append(val1) # put this item in the new list
    for val2 in t[index+1:]: # search the rest of the list
        if val2 not in out: # if we haven't already put this item in the new list
            jack(val1, val2) > 0.3: # and the new item is close to the current item
                out.append(val2) # add the new item too

This gives me

out == ['universitario de deportes', 'universitario de', 
      'lancaster', 'juan aurich', 'juan', 'muni']

I would generally recommend using better variable names than a , b , c , etc..