我有一个9图像的目录:
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,就像这样(9个图像有4个条目):
(image_000[1-3], image_00[10-11], image_0011-[1-3], image_9999)
在python中有一种方法,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,你将如何得出glob模式？
Edit2 我试图将问题分解成更小的部分.以下是解决方案的一部分示例: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
收益:
data1
[[1, 2, 3], [10, 11], [100], [9999]]
data2
[[1, 2, 3], [8, 9], [64], [9999]]
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
打印['image_000[1-3]', 'image_00[10-11]', 'image_0011-[1-3]', 'image_0100', 'image_9999'],这是你想要的.
历史:我最初回答了这个问题,正如@Mark Ransom在下面指出的那样.为了历史,我最初的答案是:
你正在寻找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