Для группировки списков есть функция zip.
| Код |
from operator import itemgetter
...
def get_best_n(img_path, db_dir,n=10,types = ('*.jpg','*.tif','*.png')):
...
return sorted(zip(files_grabbed, score_list), key=itemgetter(1))
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По поводу кода: как-то у Вас все слишком прямолинейно и ресурсоемко
Предлагаю такой вариант:
| Код |
import fnmatch, os, re
from itertools import *
from operator import itemgetter
def locate(pattern="*.*", root=os.curdir, recursive=True):
if isinstance(pattern, str):
pattern = pattern.split(";")
pattern = re.compile("|".join(map(fnmatch.translate, pattern)), re.I).match
root = os.path.abspath(root)
if recursive:
for path, dirs, files in os.walk(root):
for filename in ifilter(pattern, files):
yield os.path.join(path, filename)
else:
for filename in ifilter(pattern, os.listdir(root)):
yield os.path.join(root, filename)
def compare_hist(standard, img_files):
img= cv.LoadImage(standard, cv.CV_LOAD_IMAGE_COLOR)
standard = compute_histogram_rgb(img)
for path in img_files:
img = cv.LoadImage(path, cv.CV_LOAD_IMAGE_COLOR)
hist = compute_histogram_rgb(img)
yield path, cv.CompareHist(standard, hist, cv.CV_COMP_CORREL)
def get_best_n(img_path, db_dir, n=10, types = ('*.jpg','*.tif','*.png')):
score = compare_hist(img_path, locate(types, db_dir))
return sorted(score, key=itemgetter(1))[:n]
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