运行时间序列数据的平均值/频率？

Question

Given a data set with something like:

[2, 3, 4, 5, 6, 7, 8, 9, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 65, 75, 85, 86, 87, 88]

The values are always increasing (in fact it's time), and I want to find out a running average distance between the values. I am in effect trying to determine when the data goes from "1 every second" to "1 every 5 seconds" (or any other value).

I am implementing this in Python, but a solution in any language is most welcome.

The output I am looking for the sample input above, would be something like:

[(2, 1), (10, 5), (55, 10), (85, 1) ]

where, "2" would indicate where the distance between values started out being "1" and, and "10" would indicate where the distance shifted to being "5". (It would have to be exactly there, if the shift was detected a step later, wouldn't matter.)

I am looking for when the average distance between values changes. I realize there will be some kind of trade off between stability of the algorithm and sensitivity to changes in the input.

_{(Is Pandas or NumPy useful for this btw?)}

Answer 1

You could use numpy or pandas like so (the "pandas 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

If you only want to look at the first occurrence of every time step you could also use:

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

However with pandas you would normally use a DatetimeIndex to use the built in time series functionality:

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

From my point of view, for working with time series Pandas is by far the best library available in the Python ecosystem. Not sure what you really want to do, but I would give pandas a try.

Answer 2

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

Output:

[(2, 1), (10, 5), (30, 4), (34, 6), (40, 5), (45, 1), (46, 4), (50, 5)]

Is it what you want ?

Answer 3

How about this generator:

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)]

There's probably a cleaner way to iterate over partition, but using generators should keep it efficient for longer L and differences .

Answer 4

In Python:

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)

Answer 5

I'm a fan of using a window function via islice , it's very useful and I find myself reusing it a lot:

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

Result:

[(2, 1), (10, 5), (30, 4), (34, 6), (40, 5), (45, 1), (46, 4), (50, 5)]