如何高效列出图像序列？Python中的数字序列比较

Question

我有一个包含9张图片的文件夹：

image_0001, image_0002, image_0003
image_0010, image_0011
image_0011-1, image_0011-2, image_0011-3
image_9999

我想以一种高效的方式列出它们，比如这样（9张图片分成4行显示）：

(image_000[1-3], image_00[10-11], image_0011-[1-3], image_9999)

在Python中，有没有办法能简洁地返回一个图片目录，而不需要逐个列出每个文件？

所以，可能的做法是这样的：

列出所有图片，按数字排序，创建一个列表（从头开始依次计数每张图片）。当某张图片缺失时（创建一个新的列表），继续直到原始文件列表结束。这样我就应该得到一些包含连续序列的列表。

我想让这个数字列表更容易阅读和描述。如果我有1000个连续的文件，它可以清晰地列为file[0001-1000]，而不是file['0001','0002','0003'等等...]

编辑1（根据建议）：给定一个扁平化的列表，你会如何推导出glob模式？

编辑2 我正在尝试将问题分解成更小的部分。这里是部分解决方案的一个例子： data1有效，data2返回0010为64，data3（真实数据）不工作：

# Find runs of consecutive numbers using groupby.  The key to the solution
# is differencing with a range so that consecutive numbers all appear in
# same group.
from operator import itemgetter
from itertools import *

data1=[01,02,03,10,11,100,9999]
data2=[0001,0002,0003,0010,0011,0100,9999]
data3=['image_0001','image_0002','image_0003','image_0010','image_0011','image_0011-2','image_0011-3','image_0100','image_9999']

list1 = []
for k, g in groupby(enumerate(data1), lambda (i,x):i-x):
    list1.append(map(itemgetter(1), g))
print 'data1'
print list1

list2 = []
for k, g in groupby(enumerate(data2), lambda (i,x):i-x):
    list2.append(map(itemgetter(1), g))
print '\ndata2'
print list2

返回：

data1
[[1, 2, 3], [10, 11], [100], [9999]]

data2
[[1, 2, 3], [8, 9], [64], [9999]]

Answer 1

def ranges(sorted_list):
    first = None
    for x in sorted_list:
        if first is None:
            first = last = x
        elif x == increment(last):
            last = x
        else:
            yield first, last
            first = last = x
    if first is not None:
        yield first, last

这个 increment 函数留给读者自己去练习。

补充：这里有一个例子，展示如何用整数而不是字符串作为输入来使用它。

def increment(x): return x+1

list(ranges([1,2,3,4,6,7,8,10]))
[(1, 4), (6, 8), (10, 10)]

对于输入中的每个连续范围，你会得到一对数字，表示这个范围的开始和结束。如果某个元素不在任何范围内，那么开始和结束的值是一样的。

Answer 2

好的，我觉得你的问题很有趣，就像一个谜题。我把如何“压缩”数字范围留给你自己去解决（标记为TODO），因为有很多方法可以做到这一点，这取决于你想要的格式，以及你是想要元素数量最少，还是字符串描述长度最短。

这个解决方案使用了一个简单的正则表达式（数字字符串）来把每个字符串分成两类：静态和变量。数据分类后，我用groupby把静态数据收集成最长的匹配组，以达到总结的效果。我在结果中混入了整数索引标记（在matchGrouper中），这样我就可以从所有元素中重新选择变化的部分（在unpack中）。

import re
import glob
from itertools import groupby
from operator import itemgetter

def classifyGroups(iterable, reObj=re.compile('\d+')):
    """Yields successive match lists, where each item in the list is either
    static text content, or a list of matching values.

     * `iterable` is a list of strings, such as glob('images/*')
     * `reObj` is a compiled regular expression that describes the
            variable section of the iterable you want to match and classify
    """
    def classify(text, pos=0):
        """Use a regular expression object to split the text into match and non-match sections"""
        r = []
        for m in reObj.finditer(text, pos):
            m0 = m.start()
            r.append((False, text[pos:m0]))
            pos = m.end()
            r.append((True, text[m0:pos]))
        r.append((False, text[pos:]))
        return r

    def matchGrouper(each):
        """Returns index of matches or origional text for non-matches"""
        return [(i if t else v) for i,(t,v) in enumerate(each)]

    def unpack(k,matches):
        """If the key is an integer, unpack the value array from matches"""
        if isinstance(k, int):
            k = [m[k][1] for m in matches]
        return k

    # classify each item into matches
    matchLists = (classify(t) for t in iterable)

    # group the matches by their static content
    for key, matches in groupby(matchLists, matchGrouper):
        matches = list(matches)
        # Yield a list of content matches.  Each entry is either text
        # from static content, or a list of matches
        yield [unpack(k, matches) for k in key]

最后，我们添加了足够的逻辑来美化输出，并运行一个示例。

def makeResultPretty(res):
    """Formats data somewhat like the question"""
    r = []
    for e in res:
        if isinstance(e, list):
            # TODO: collapse and simplify ranges as desired here
            if len(set(e))<=1:
                # it's a list of the same element
                e = e[0]
            else: 
                # prettify the list
                e = '['+' '.join(e)+']'
        r.append(e)
    return ''.join(r)

fnList = sorted(glob.glob('images/*'))
re_digits = re.compile(r'\d+')
for res in classifyGroups(fnList, re_digits):
    print makeResultPretty(res)

我的图片目录是根据你的示例创建的。你可以用下面的列表替换fnList来进行测试：

fnList = [
 'images/image_0001.jpg',
 'images/image_0002.jpg',
 'images/image_0003.jpg',
 'images/image_0010.jpg',
 'images/image_0011-1.jpg',
 'images/image_0011-2.jpg',
 'images/image_0011-3.jpg',
 'images/image_0011.jpg',
 'images/image_9999.jpg']

当我在这个目录下运行时，我的输出看起来是这样的：

StackOverflow/3926936% python classify.py
images/image_[0001 0002 0003 0010].jpg
images/image_0011-[1 2 3].jpg
images/image_[0011 9999].jpg

Answer 3

这里有一个可以实现你想要的功能的代码，使用了你之前添加的代码作为起点：

#!/usr/bin/env python

import itertools
import re

# This algorithm only works if DATA is sorted.
DATA = ["image_0001", "image_0002", "image_0003",
        "image_0010", "image_0011",
        "image_0011-1", "image_0011-2", "image_0011-3",
        "image_0100", "image_9999"]

def extract_number(name):
    # Match the last number in the name and return it as a string,
    # including leading zeroes (that's important for formatting below).
    return re.findall(r"\d+$", name)[0]

def collapse_group(group):
    if len(group) == 1:
        return group[0][1]  # Unique names collapse to themselves.
    first = extract_number(group[0][1])  # Fetch range
    last = extract_number(group[-1][1])  # of this group.
    # Cheap way to compute the string length of the upper bound,
    # discarding leading zeroes.
    length = len(str(int(last)))
    # Now we have the length of the variable part of the names,
    # the rest is only formatting.
    return "%s[%s-%s]" % (group[0][1][:-length],
        first[-length:], last[-length:])

groups = [collapse_group(tuple(group)) \
    for key, group in itertools.groupby(enumerate(DATA),
        lambda(index, name): index - int(extract_number(name)))]

print groups

这个代码会输出 ['image_000[1-3]', 'image_00[10-11]', 'image_0011-[1-3]', 'image_0100', 'image_9999']，这正是你想要的结果。

历史背景：我最开始回答这个问题时搞错了方向，正如 @Mark Ransom 在下面指出的那样。为了记录历史，我最初的回答是：

你需要用到 glob。试试这个：

import glob
images = glob.glob("image_[0-9]*")

或者，使用你的例子：

images = [glob.glob(pattern) for pattern in ("image_000[1-3]*",
    "image_00[10-11]*", "image_0011-[1-3]*", "image_9999*")]
images = [image for seq in images for image in seq]  # flatten the list