给定一个数据集，例如：
[2, 3, 4, 5, 6, 7, 8, 9, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 65, 75, 85, 86, 87, 88]
这些值总是在增加（实际上是时间），我想找出这些值之间的运行平均距离。我实际上是在尝试确定数据何时从“每秒 1”变为“每 5 秒 1”（或任何其他值）。
我在 Python 中实现这个，但任何语言的解决方案都是最受欢迎的。
我正在寻找上面的示例输入的输出将类似于：
[(2, 1), (10, 5), (55, 10), (85, 1) ]
其中，“2”表示值之间的距离从“1”开始的位置，“10”表示距离变为“5”的位置。（它必须准确地存在，如果稍后检测到转移，那没关系。）
我正在寻找值之间的平均距离何时发生变化。我意识到在算法的稳定性和对输入变化的敏感性之间会有某种权衡。
您可以像这样使用 numpy 或 pandas（“pandas 版本”）：
In [256]: s = pd.Series([2, 3, 4, 5, 6, 7, 8, 9, 10, 15, 20, 25, 30, 35,
40, 45, 50, 55, 65, 75, 85, 86, 87, 88])
In [257]: df = pd.DataFrame({'time': s,
'time_diff': s.diff().shift(-1)}).set_index('time')
In [258]: df[df.time_diff - df.time_diff.shift(1) != 0].dropna()
Out[258]:
time_diff
time
2 1
10 5
55 10
85 1
如果您只想查看每个时间步的第一次出现，您还可以使用：
In [259]: df.drop_duplicates().dropna() # set take_last=True if you want the last
Out[259]:
time_diff
time
2 1
10 5
55 10
但是对于 pandas，您通常会使用 aDatetimeIndex来使用内置的时间序列功能：
In [44]: a = [2, 3, 4, 5, 6, 7, 8, 9, 10, 15, 20, 25, 30, 35,
40, 45, 50, 55, 65, 75, 85, 86, 87, 88]
In [45]: start_time = datetime.datetime.now()
In [46]: times = [start_time + datetime.timedelta(seconds=int(x)) for x in a]
In [47]: idx = pd.DatetimeIndex(times)
In [48]: df = pd.DataFrame({'data1': np.random.rand(idx.size),
'data2': np.random.rand(idx.size)},
index=idx)
In [49]: df.resample('5S') # resample to 5 Seconds
Out[49]:
data1 data2
2012-11-28 07:36:35 0.417282 0.477837
2012-11-28 07:36:40 0.536367 0.451494
2012-11-28 07:36:45 0.902018 0.457873
2012-11-28 07:36:50 0.452151 0.625526
2012-11-28 07:36:55 0.816028 0.170319
2012-11-28 07:37:00 0.169264 0.723092
2012-11-28 07:37:05 0.809279 0.794459
2012-11-28 07:37:10 0.652836 0.615056
2012-11-28 07:37:15 0.508318 0.147178
2012-11-28 07:37:20 0.261157 0.509014
2012-11-28 07:37:25 0.609685 0.324375
2012-11-28 07:37:30 NaN NaN
2012-11-28 07:37:35 0.736370 0.551477
2012-11-28 07:37:40 NaN NaN
2012-11-28 07:37:45 0.839960 0.118619
2012-11-28 07:37:50 NaN NaN
2012-11-28 07:37:55 0.697292 0.394946
2012-11-28 07:38:00 0.351824 0.420454
从我的角度来看，对于时间序列，Pandas 是迄今为止 Python 生态系统中可用的最好的库。不知道你真的想做什么，但我会试试 pandas。
在 Python 中：
a = [2, 3, 4, 5, 6, 7, 8, 9, 10, 15, 20, 25, 30, 34, 40, 45, 46, 50, 55]
# zip() creates tuples of two consecutive values
# (it zips lists of different length by truncating longer list(s))
# then tuples with first value and difference are placed in 'diff' list
diff = [(x, y-x) for x, y in zip(a, a[1:])]
# now pick only elements with changed difference
result = []
for pair in diff:
if not len(result) or result[-1][1]!=pair[1]: # -1 to take last element
result.append(pair)
a = [2, 3, 4, 5, 6, 7, 8, 9, 10, 15, 20, 25, 30, 34, 40, 45, 46, 50, 55]
ans = [(a[0], a[1]-a[0])]
for i in range(1, len(a)-1):
if a[i+1] - a[i] - a[i] + a[i-1] is not 0:
ans.append((a[i], a[i+1]-a[i]))
print ans
输出：
[(2, 1), (10, 5), (30, 4), (34, 6), (40, 5), (45, 1), (46, 4), (50, 5)]
是你想要的吗？
这个生成器怎么样：
L = [2, 3, 4, 5, 6, 7, 8, 9, 10, 15, 20, 25, 30, 34, 40, 45, 46, 50, 55]
def differences_gen(L, differences):
previous = L[0]
differences = iter(differences + [None])
next_diff = next(differences)
for i, n in enumerate(L[1:]):
current_diff = n - previous
while next_diff is not None and current_diff >= next_diff:
yield (previous, next_diff)
next_diff = next(differences)
previous = n
list(differences_gen(L, [1,5]))
# [(2, 1), (10, 5)]
可能有一种更简洁的方法来迭代分区，但是使用生成器应该可以让它在更长的时间内保持高效，L并且differences.
我喜欢通过 使用窗口函数islice，它非常有用，我发现自己经常重复使用它：
from itertools import islice
def window(seq, n=2):
"""
Returns a sliding window (of width n) over data from the iterable
s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...
"""
it = iter(seq)
result = tuple(islice(it, n))
if len(result) == n:
yield result
for elem in it:
result = result[1:] + (elem,)
yield result
# Main code:
last_diff = None
results = []
for v1, v2 in window(a):
diff = abs(v1 - v2)
if diff != last_diff:
results.append((v1, diff))
last_diff = diff
结果：
[(2, 1), (10, 5), (30, 4), (34, 6), (40, 5), (45, 1), (46, 4), (50, 5)]