Question

我正在尝试编写一个程序，将Timelapse照片从其时间戳分组在一起。延时摄影和随机摄影位于一个文件夹中。

例如，如果上一张和当前照片之间的时间戳差异以秒为单位：346、850、13、14、13、14、15、12、12、13、16、11、438。

您可以合理地猜测游戏中时光倒流从13开始到11结束。

现在，我正在尝试一种hacky解决方案，以将百分比差异与上一个差异进行比较。

但是必须存在一个公式/算法，以通过时差将时间戳分组在一起。滚动平均值或其他内容。

我在寻找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

如果您使用itertools.groupby，并使用一个函数，如果延迟满足您对延时摄影区域的标准，则该函数将根据延迟列表返回True，您可以获取每个延迟区域的索引。基本上，我们将功能的True / False输出分组。

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

输出：

间隔拍摄索引2，长度为10

通过时差算法对延时进行分组

1 个答案: