时差算法分组时差

Question

我试着写一个程序，让蒂莫西丝的照片从他们的时间戳组合在一起。时间流逝照片和随机照片在一个文件夹中。

例如，如果上一张照片与当前照片之间的时间戳秒差为: 346、850、13、14、13、14、15、12、12、13、16、11、438。

你可以合理地猜测时间从13开始到11结束。

现在，我正在尝试一种比较与前一种方案的百分比差异的方法。

，但是必须有一个公式/algo来根据时间差将时间戳组合在一起。滚动意味着什么。

我在考虑一个简单的解决方案吗？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

Answer 1

如果使用itertools.groupby，使用返回True的函数，如果延迟符合时间流逝照片区域的条件，则可以根据延迟列表获得每个此类区域的索引。基本上，我们对该函数的真/假输出进行分组。

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

产出：

时间流逝指数2，长度10