Bash别名自动检测任意命名的文件序列？

Question

我正在寻找一个改变ls结果的bash别名.我经常处理大量文件,不遵循相同的命名约定.关于他们的唯一常见的事情是这个数字是4填充(抱歉不确定正确的说法)并且在扩展之前立即.

例如 - filename_v028_0392.bgeo,test_x34.prerun.0012.simdata,filename_v001_0233.exr

我想将序列列为1元素,这样

filename_v003_0001.geo
filename_v003_0002.geo
filename_v003_0003.geo
filename_v003_0004.geo
filename_v003_0005.geo
filename_v003_0006.geo
filename_v003_0007.geo
filename_v003_0032.geo
filename_v003_0033.geo
filename_v003_0034.geo
filename_v003_0035.geo
filename_v003_0036.geo
testxxtest.0057.exr
testxxtest.0058.exr
testxxtest.0059.exr
testxxtest.0060.exr
testxxtest.0061.exr
testxxtest.0062.exr
testxxtest.0063.exr

将显示为某些行

[seq]filename_v003_####.geo (1-7)
[seq]filename_v003_####.geo (32-36)
[seq]testxxtest.####.exr (57-63)

同时仍然列出未更改的非序列.

我真的不知道从哪里开始接近这个.我知道有相当数量的python,但不确定这是否真的是最好的方法.任何帮助将不胜感激!

谢谢

Answer 1

我得到了一个 python 2.7 脚本，它通过解决折叠仅按序列号更改的几行的更普遍的问题来解决您的问题

import re

def do_compress(old_ints, ints):
    """
    whether the ints of the current entry is the continuation of the previous
    entry
    returns a list of the indexes to compress, or [] or False when the current
    line is not part of an indexed sequence
    """
    return len(old_ints) == len(ints) and \
        [i for o, n, i in zip(old_ints, ints, xrange(len(ints))) if n - o == 1]

def basic_format(file_start, file_stop):
    return "[seq]{} .. {}".format(file_start, file_stop)


def compress(files, do_compress=do_compress, seq_format=basic_format):
    p = None
    old_ints = ()
    old_indexes = ()

    seq_and_files_list = [] 
        # list of file names or dictionaries that represent sequences:
        #   {start, stop, start_f, stop_f}

    for f in files:
        ints = ()
        indexes = ()

        m = p is not None and p.match(f) # False, None, or a valid match
        if m:
            ints = [int(x) for x in m.groups()]
            indexes = do_compress(old_ints, ints)

        # state variations
        if not indexes: # end of sequence or no current sequence
            p = re.compile( \
                '(\d+)'.join(re.escape(x) for x in re.split('\d+',f)) + '$')
            m = p.match(f)
            old_ints = [int(x) for x in m.groups()]
            old_indexes = ()
            seq_and_files_list.append(f)

        elif indexes == old_indexes: # the sequence continues
            seq_and_files_list[-1]['stop'] = old_ints = ints
            seq_and_files_list[-1]['stop_f'] = f
            old_indexes = indexes

        elif old_indexes == (): # sequence started on previous filename
            start_f = seq_and_files_list.pop()
            s = {'start': old_ints, 'stop': ints, \
                'start_f': start_f, 'stop_f': f}
            seq_and_files_list.append(s)

            old_ints = ints
            old_indexes = indexes

        else: # end of sequence, but still matches previous pattern
            old_ints = ints
            old_indexes = ()
            seq_and_files_list.append(f)

    return [ isinstance(f, dict) and seq_format(f['start_f'], f['stop_f']) or f 
        for f in seq_and_files_list ]


if __name__ == "__main__":
    import sys
    if len(sys.argv) == 1:
        import os
        lst = sorted(os.listdir('.'))
    elif sys.argv[1] in ("-h", "--help"):
        print """USAGE: {} [FILE ...]
compress the listing of the current directory, or the content of the files by
collapsing identical lines, except for a sequence number
"""
        sys.exit(0)
    else:
        import string
        lst = [string.rstrip(l, '\r\n') for f in sys.argv[1:] for l in open(f)])
    for x in compress(lst):
        print x

也就是说，根据您的数据：

bernard $ ./ls_sequence_compression.py given_data
[seq]filename_v003_0001.geo .. filename_v003_0007.geo
[seq]filename_v003_0032.geo .. filename_v003_0036.geo
[seq]testxxtest.0057.exr .. testxxtest.0063.exr

它基于与非数字文本匹配的连续两行中存在的整数之间的差异。这允许在用作序列基础的字段发生变化时处理非统一输入......

下面是一个输入示例：

01 - test8.txt
01 - test9.txt
01 - test10.txt
02 - test11.txt
02 - test12.txt
03 - test13.txt
04 - test13.txt
05 - test13.txt
06
07
08
09
10

这使：

[seq]01 - test8.txt .. 01 - test10.txt
[seq]02 - test11.txt .. 02 - test12.txt
[seq]03 - test13.txt .. 05 - test13.txt
[seq]06 .. 10

有什么意见欢迎留言！

哈...我附近忘记了：没有参数，这个脚本输出当前目录的折叠内容。