你可以这样使用numpy或panda（“panda version”）：

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点的第一次出现，还可以用途：

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

然而，对于Pandas，您通常会使用DatetimeIndex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

在我看来，对于处理时间序列来说，Pandas是Python生态系统中最好的库，不知道你到底想做什么，但我会给予Pandas一个尝试。

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)]

这是你想要的吗？

这个发电机怎么样：

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)]

• 可能有一种更干净的方法来迭代分区，但是使用生成器应该可以在更长的L和differences时间内保持它的效率。*

我喜欢通过islice使用窗口函数，它非常有用，我发现自己经常重用它：

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)]