Question

我有一个全部转换为整数的时间戳列表。如何将它们分成2秒间隔的单独列表？

d = [1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256401, 1499256401, 1499256401, 1499256401, 1499256402, 1499256402, 1499256403, 1499256403, 1499256404, 1499256404, 1499256405, 1499256405, 1499256405, 1499256405, 1499256406, 1499256406, 1499256407, 1499256407, 1499256408, 1499256408, 1499256408, 1499256408, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409]

我想将它们分成2秒的间隔，这样我就可以打印出每个间隔的结果。

预期输出：

[1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256401, 1499256401, 1499256401, 1499256401, 1499256402, 1499256402]
[1499256403, 1499256403, 1499256404, 1499256404, 1499256405, 1499256405, 1499256405, 1499256405]
[1499256406, 1499256406, 1499256407, 1499256407, 1499256408, 1499256408, 1499256408, 1499256408]
[1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409]

感谢。

Answer 1

您可以使用简单的for循环对最大差异为2的项目进行分组：

lst = [[]]
start = d[0]
for x in d:
    if x-start <= 2:
        lst[-1].append(x)
    else:
        lst.append([x])
        start = x

如果元素未排序，您可以在分组前调用sorted进行预测。

对于代码重用（导入及所有这些），您可以创建一个grouper类，然后使用itertools.groupby进行分组：

from itertools import groupby

class Grouper(object):
    def __init__(self, start=None, diff=2):
        self.diff = diff
        self.start = start
        self._count = 0

    def __call__(self, x):
        if x-self.start > self.diff:
            self._count += 1
            self.start = x
        return self._count

然后分组现在简化为：

lst = [list(g) for _, g in groupby(d, Grouper(start=d[0]))]

[[1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256401, 1499256401, 1499256401, 1499256401, 1499256402, 1499256402], 
 [1499256403, 1499256403, 1499256404, 1499256404, 1499256405, 1499256405, 1499256405, 1499256405], 
 [1499256406, 1499256406, 1499256407, 1499256407, 1499256408, 1499256408, 1499256408, 1499256408], 
 [1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409]]

Answer 2

您可以尝试这一点 - 确定d的最小值和最大值的范围，然后检查每个元素是否在2秒(x, x+1, x+2)内的时间间隔内：

for x in range(min(d), max(d)+1, 3):
    print ([item for item in d if item in (x, x+1, x+2) ])

返回：

[1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256400, 1499256401, 1499256401, 1499256401, 1499256401, 1499256402, 1499256402]
[1499256403, 1499256403, 1499256404, 1499256404, 1499256405, 1499256405, 1499256405, 1499256405]
[1499256406, 1499256406, 1499256407, 1499256407, 1499256408, 1499256408, 1499256408, 1499256408]
[1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409, 1499256409]

在python中按时间间隔2秒分割时间戳列表

2 个答案: