我有一个9张图片的目录：
image_0001, image_0002, image_0003 image_0010, image_0011 image_0011-1, image_0011-2, image_0011-3 image_9999
我希望能够以有效的方式列出它们，例如9个图像的4个条目：
(image_000[1-3], image_00[10-11], image_0011-[1-3], image_9999)
在python中是否有一种方法可以以简短/清晰的方式返回图像目录（不列出每个文件）？
所以，可能是这样的：
列出所有图像，按数字排序，创建一个列表（从开始按顺序计算每个图像）。 缺少图像（创建新列表）时，继续直到原始文件列表完成。 现在我应该有一些包含非破坏序列的列表。
我试图让阅读/描述数字列表变得容易。如果我有1000个连续文件的序列它可以清楚地列为文件[0001-1000]而不是文件['0001'，'0002'，'0003'等......]
Edit1 （基于建议）：给定一个扁平列表，你将如何推导出glob模式？
Edit2 我正试图将问题分解成更小的部分。以下是部分解决方案的示例： data1工作，data2返回0010为64，data3（realworld数据）不起作用：
# 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]]
答案 0 :(得分：6)
以下是您要实现的工作实现，使用您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
这会打印['image_000[1-3]', 'image_00[10-11]', 'image_0011-[1-3]', 'image_0100', 'image_9999']，这就是你想要的。
HISTORY：我最初向后回答了这个问题，正如@Mark Ransom在下面指出的那样。为了历史，我原来的答案是：
您正在寻找glob。尝试：
import glob
images = glob.glob("image_[0-9]*")
或者，使用您的示例：
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
答案 1 :(得分：3)
好的，所以我发现你的问题是一个迷人的难题。我离开了怎么样 “压缩”最多的数字范围（标记为TODO），就像有 取决于你喜欢格式化的方式和不同的方法来实现这一目标 您想要最少数量的元素或最小字符串描述 长度。
此解决方案使用简单的正则表达式（数字字符串）将每个字符串分为两组：静态和变量。数据分类后，我使用groupby来收集数据 将静态数据转换为最长匹配组，以实现汇总效果。我将整数索引sentinals混合到结果中（在matchGrouper中），这样我就可以从所有元素中重新选择不同的部分（在解包中）。
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执行输出的漂亮打印，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您的示例创建的。您可以使用以下列表替换fnList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']
当我针对此目录运行时，我的输出如下：
StackOverflow/3926936% python classify.py
images/image_[0001 0002 0003 0010].jpg
images/image_0011-[1 2 3].jpg
images/image_[0011 9999].jpg
答案 2 :(得分：2)
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
increment功能留给读者练习。
编辑：这里是一个如何使用整数而不是字符串作为输入的示例。
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，您将获得一对指示范围的开始和结束。如果元素不是范围的一部分，则起始值和结束值相同。