python – 时间序列重新取样

在一些情况下,数据的采样率从0Hz到超过200Hz从第二到第二变化,所提供的数据集中的最高采样率是每秒约50个样本.

当采集样本时,它们总是均匀分布在第二个样本上

time                   x
2012-12-06 21:12:40    128.75909883327378
2012-12-06 21:12:40     32.799224301545976
2012-12-06 21:12:40     98.932953779777989
2012-12-06 21:12:43    132.07033814856786
2012-12-06 21:12:43    132.07033814856786
2012-12-06 21:12:43     65.71691352191452
2012-12-06 21:12:44    117.1350194748169
2012-12-06 21:12:45     13.095622561808861
2012-12-06 21:12:47     61.295242676059246
2012-12-06 21:12:48     94.774064119961352
2012-12-06 21:12:49     80.169378222553533
2012-12-06 21:12:49     80.291142695702533
2012-12-06 21:12:49    136.55650749231367
2012-12-06 21:12:49    127.29790925838365

应该

time                        x
2012-12-06 21:12:40 000ms   128.75909883327378
2012-12-06 21:12:40 333ms    32.799224301545976
2012-12-06 21:12:40 666ms    98.932953779777989
2012-12-06 21:12:43 000ms   132.07033814856786
2012-12-06 21:12:43 333ms   132.07033814856786
2012-12-06 21:12:43 666ms    65.71691352191452
2012-12-06 21:12:44 000ms   117.1350194748169
2012-12-06 21:12:45 000ms    13.095622561808861
2012-12-06 21:12:47 000ms    61.295242676059246
2012-12-06 21:12:48 000ms    94.774064119961352
2012-12-06 21:12:49 000ms    80.169378222553533
2012-12-06 21:12:49 250ms    80.291142695702533
2012-12-06 21:12:49 500ms   136.55650749231367
2012-12-06 21:12:49 750ms   127.29790925838365

是否有一种简单的方法来使用熊猫时间序列重新采样功能,还是有一些内置于numpy或scipy中的东西会起作用？

解决方法:

我认为没有内置的pandas或numpy方法/功能来做到这一点.

但是,我赞成使用python生成器：

def repeats(lst):
    i_0 = None
    n = -1 # will still work if lst starts with None
    for i in lst:
        if i == i_0:
            n += 1
        else:
            n = 0
        yield n
        i_0 = i
# list(repeats([1,1,1,2,2,3])) == [0,1,2,0,1,0]

然后你可以把这个generator into a numpy array：

import numpy as np
df['rep'] = np.array(list(repeats(df['time'])))

重复重复：

from collections import Counter
count = Counter(df['time'])
df['count'] = df['time'].apply(lambda x: count[x])

并进行计算(这是计算中最昂贵的部分)：

df['time2'] = df.apply(lambda row: (row['time'] 
                                 + datetime.timedelta(0, 1) # 1s
                                     * row['rep'] 
                                     / row['count']),
                 axis=1)

注意：要删除计算列,请使用del df [‘rep’]和del df [‘count’].

实现它的一种“内置”方式可能会使用shift两次完成,但我认为这会有些麻烦……

python – 时间序列重新取样

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