Question

is it possible to sort through a list and do something depending on your preferred order or pre determined choice?

so i have a list of files containing definitions, if i re.search the string and return a match, i would like to only print the string + highest definition

i started to explore the below idea but i cannot seem to get it to print only the highest -currently it prints all 3

#!/usr/bin/python3
import re
definitions = ['1080p', '720p', 'SD'] # order (highest to lowest)
some_files = ['movie_a_1080p.mp4', 'movie_b_720p.mp4','movie_c_SD.mp4'] 

for each in some_files:
    for defs in definitions:
        match = re.search(defs, each, re.M | re.I)
        if match:
            if match.group() == '1080p':
                print('1080p', each)
                break
            else:
                if match.group() == '720p':
                    print('720p', each)
                    break
                else:
                if match.group() == 'SD':
                    print('SD', each)
                    break

any help would be awesome

Answer 1

由于您的definitions正常，您可以遍历列表，并在找到匹配项时返回文件。

def best_quality(l):
    definitions = ['1080p', '720p', 'SD'] # order (highest to lowest)
    for definition in definitions:
        for file in some_files:
            if definition in file: return file

some_files = ['movie_c_SD.mp4', 'movie_a_1080p.mp4', 'movie_b_720p.mp4']
print best_quality(some_files)
>>> movie_a_1080p.mp4

Answer 2

如果您只想要一个结果（最高）并且考虑到您的列表从最高到最低排序，您可以在找到第一个结果后返回：

import re
definitions = ['1080p', '720p', 'SD'] # order (highest to lowest)
some_files = ['movie_a_1080p.mp4', 'movie_b_720p.mp4', 'movie_c_SD.mp4']

def find():
    for each in some_files:
        for defs in definitions:
            match = re.search(defs, each, re.M | re.I)
            if match:
                return match.group(0)


print(find())

Answer 3

如果我理解你要做什么，这应该有效：

def best(some_files):
    for definition in ('1080p', '720p', 'SD'):
        match = next((file for file in some_files if definition in file), None)
        if match is not None:
            return match

print(best(['movie_a_1080p.mp4', 'movie_b_720p.mp4', 'movie_c_SD.mp4']))  # movie_a_1080p.mp4
print(best(['movie_b_720p.mp4', 'movie_c_SD.mp4', 'movie_a_1080p.mp4']))  # movie_a_1080p.mp4
print(best(['movie_b_720p.mp4', 'movie_c_SD.mp4']))  # movie_b_720p.mp4
print(best(['movie_d.mp4', 'movie_c_SD.mp4']))  # movie_c_SD.mp4
print(best(['movie_d.mp4']))  # None

您的方法的主要问题是break仅突破最近的循环。 @ alfasin的答案通过函数return来修复此问题。

如果你愿意，我的答案也可以在没有函数的情况下使用，因为它只有一个循环可以突破：

for definition in ('1080p', '720p', 'SD'):
    match = next((file for file in some_files if definition in file), None)
    if match is not None:
        print('Found the best one: {}'.format(match))
        break

sort a list by prefered order

3 个答案: