我想编写一个程序,组延时照片的同时从他们的时间戳。该延时照片的和随机的照片的是一个文件夹中。
例如,如果在先前和当前照片之间秒时间戳差值是:346,850,13,14,13,14,15,12,12,13,16,11,438。
你可以做一个合理的猜测的延时开始在13和11结束。
现在,我想一个哈克的解决方案比较与前一个的百分比差异。
但是,必须有一个式/ ALGO到组由为TimeDifference时间戳一起。滚动平均值或东西。
我是不是在寻找一个简单的解决方案?谢谢!
DEF cat_algo(文件夹):
# Get a list with all the CR2 files in the folder we are processing
file_list = folder_to_file_list(folder)
# Extract the timestamp out of the CR2 file into a sorted dictionary
cr2_timestamp = collections.OrderedDict()
for file in file_list:
cr2_timestamp[file] = return_date_from_raw(file)
print str(file) + " - METADATA TIMESTAMP: " + \
str(return_date_from_raw(file))
# Loop over the dictionary to compare the timestamps and create a new dictionary with a suspected group number per shot
# Make sure we know that there is no first file yet using this (can be refractored)
item_count = 1
group_count = 0
cr2_category = collections.OrderedDict()
# get item and the next item out of the sorted dictionary
for item, nextitem in zip(cr2_timestamp.items(), cr2_timestamp.items()[1::]):
# if not the first CR2 file
if item_count >= 2:
current_date_stamp = item[1]
next_date_stamp = nextitem[1]
delta_previous = current_date_stamp - previous_date_stamp
delta_next = next_date_stamp - current_date_stamp
try:
difference_score = int(delta_next.total_seconds() /
delta_previous.total_seconds() * 100)
print "diffscore: " + str(difference_score)
except ZeroDivisionError:
print "zde"
if delta_previous > datetime.timedelta(minutes=5):
# if difference_score < 20:
print item[0] + " - hit - " + str(delta_previous)
group_count += 1
cr2_category[item[0]] = group_count
else:
cr2_category[item[0]] = group_count
# create a algo to come up with percentage difference and use this to label timelapses.
print int(delta_previous.total_seconds())
print int(delta_next.total_seconds())
# Calculations done, make the current date stamp the previous datestamp for the next iteration
previous_date_stamp = current_date_stamp
# If time difference with previous over X make a dict with name:number, in the end everything which has the
# same number 5+ times in a row can be assumed as a timelapse.
else:
# If it is the first date stamp, assign it the current one to be used in the next loop
previous_date_stamp = item[1]
# To help make sure this is not the first image in the sequence.
item_count += 1
print cr2_category
如果你使用itertools.groupby,使用返回True如果延迟满足你的标准延时照片的区域,根据延迟的名单上的功能,你可以得到每个这样的区域的折射率。基本上,我们分组对功能的真/假输出。
from itertools import groupby
# time differences given in original post
data = [346, 850, 13, 14, 13, 14, 15, 12, 12, 13, 16, 11, 438]
MAX_DELAY = 25 # timelapse regions will have a delay no larger than this
MIN_LENGTH = 3 # timelapse regions will have at least this many photos
index = 0
for timelapse, g in groupby(data, lambda x: x <= MAX_DELAY):
length = len(list(g))
if (timelapse and length > MIN_LENGTH):
print ('timelapse index {}, length {}'.format(index, length))
index += length
输出:
延时索引2,长度10