python - Compare if two items from os.listdir are similar? -

ignore question. different actual question needed ask. people answered question, i'm sorry. in future, though.

read new thread here: opening files found os.listdir() , comparing lines inside?

basically, i'm running os.listdir() listing of files, , trying compare if 2 different files have similar names. how go this?

basically, code this:

config_dir = "/etc/netctl/"  profiles = os.listdir(config_dir) in profiles:     if os.path.isfile(config_dir + i):         if in i:             print "true"     else:         pass 

i'm not sure use check similarities in names, though. however, know "if in i" checking same word... don't know how go saving last one...

i tried:

i2 = "" profiles = os.listdir(config_dir) in profiles:     if os.path.isfile(config_dir + i):         if i2 == "":             i2 =             print i2         elif i2 == i:             continue         if i2 in i:             print "true"     else:         pass 

i think might overthinking this, though. output of os.listdir:

['hooks', 'interfaces', 'examples', 'ddwrt', 'momandkids_wifiz', 'backups', 'momandkids'] 

the files ddwrt momandkids_wifiz , momandkids. basically, want detect names "momandkids" , "momandkids_wifiz" similar, , return true.

this should it:

from difflib import sequencematcher glob import glob os import path  config_dir = '/etc/netctl' min_ratio = 0.90 # 90%  profiles = dict((i, {'full_path': v, 'matches': [], 'file_name': path.splitext(path.split(v)[-1])[0]}) (i, v) in enumerate(glob(config_dir + '/*.*')))  k, v in profiles.items():     sm = sequencematcher(a=v['file_name'], b='')     k, v in profiles.items():         if k == k or k in v['matches']:             continue         sm.set_seq2(v['file_name'])         if sm.ratio() > min_ratio:             v['matches'].append(k)             v['matches'].append(k)  # display output k, v in profiles.items():     print k, v