请教个循环判断的问题

[色郎中]

2017040514495198.bin
20170405144956488.txt
201704051450018.ppt
20170405145006232.pas
20170405145006232.txt
201704051450018.txt

目录有上面的文件，想根据后缀名，打印目录下的文件；

用下面注释掉的判断语句，可以打印 列表中的三种文件出来；

而用 if [os.path.splitext(filename)[1] ==pattern[i] for i in range(len(pattern))]: 这句作为判断
PPT文件也打印出来了

如何修改，才能有期望的效果呢？ 谢谢
这样就可以不修改函数判断条件，就可以修改过滤条件了

def filtter_files_pattern(dir,pattern=[]):
    for dirpath,dirnames, filenames in os.walk(dir):
        for filename in filenames:
            #if os.path.splitext(filename)[1] ==pattern[0]or os.path.splitext(filename)[1] == pattern[1] or os.path.splitext(filename)[1] == pattern[2]:
            if [os.path.splitext(filename)[1] ==pattern[i] for i in range(len(pattern))]:              
                print filename       

if __name__ == '__main__':
    pp=['.txt','.pas','.bin']
    filtter_files_pattern('./backFile/',pp)

[色郎中]

引用 2 楼 u012536120 的回复:

像这样子也能实现 用个in

# -*- coding:utf-8 -*-

import os

def filtter_files_pattern(dir,pattern=[]):
    for dirpath,dirnames, filenames in os.walk(dir):
        for filename in filenames:
            #if os.path.splitext(filename)[1] ==pattern[0]or os.path.splitext(filename)[1] == pattern[1] or os.path.splitext(filename)[1] == pattern[2]:
            #zzr=[os.path.splitext(filename)[1] ==pattern[i] for i in range(len(pattern))]
            if os.path.splitext(filename)[1] in pattern:
            	print filename

if __name__ == '__main__':
	pp=['.txt','.pas','.bin']
	filtter_files_pattern('D:/testP/backFile/',pp)

用IN 是达到效果了 ，，，上面的方法也试了，好像还是会把其他类型的文件也打印出来 谢谢 in 的方式可以

[sanGuo_uu]

if list是判断list是否非空的 你那个ppt就是这样子出来的 也就是说，你这个if有问题啊

# -*- coding:utf-8 -*-

import os

def filtter_files_pattern(dir,pattern=[]):
    for dirpath,dirnames, filenames in os.walk(dir):
        for filename in filenames:
            #if os.path.splitext(filename)[1] ==pattern[0]or os.path.splitext(filename)[1] == pattern[1] or os.path.splitext(filename)[1] == pattern[2]:
            zzr=[os.path.splitext(filename)[1] ==pattern[i] for i in range(len(pattern))]
            if zzr:
            	print zzr
                #print filename

if __name__ == '__main__':
	pp=['.txt','.pas','.bin']
	filtter_files_pattern('D:/testP/backFile/',pp)

[sanGuo_uu]

像这样子也能实现 用个in

# -*- coding:utf-8 -*-

import os

def filtter_files_pattern(dir,pattern=[]):
    for dirpath,dirnames, filenames in os.walk(dir):
        for filename in filenames:
            #if os.path.splitext(filename)[1] ==pattern[0]or os.path.splitext(filename)[1] == pattern[1] or os.path.splitext(filename)[1] == pattern[2]:
            #zzr=[os.path.splitext(filename)[1] ==pattern[i] for i in range(len(pattern))]
            if os.path.splitext(filename)[1] in pattern:
            	print filename

if __name__ == '__main__':
	pp=['.txt','.pas','.bin']
	filtter_files_pattern('D:/testP/backFile/',pp)