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python - 从 Pandas.DataFrame 中带时间戳的位置计算速度

转载 作者:太空狗 更新时间:2023-10-30 01:16:36 24 4
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我对 Pandas 很陌生,但熟悉 Numpy 和 Python。

假设我有一个按时间(日期时间)索引的 X、Y 点 (float64) 的“Pandas.DataFrame”,如果我已经知道如何计算点之间的欧几里德距离,我如何通过 python 计算速度?

编辑:我刚刚阅读了关于pandas.Series.diff() 的帮助,但我仍然想用另一个函数“替换” diff 上使用的减法,比如 `euclidean_distance( )'。有办法吗?

DataFrame 看起来像(索引在第一列,位置在第二列):

2009-08-07 16:16:44    [37.800185, -122.426361]
2009-08-07 16:16:48 [37.800214, -122.426153]
2009-08-07 16:16:49 [37.800222, -122.426118]
2009-08-07 16:16:52 [37.800197, -122.426072]
2009-08-07 16:17:32 [37.800214, -122.425903]
2009-08-07 16:17:34 [37.800236, -122.425826]
2009-08-07 16:17:40 [37.800282, -122.425534]
2009-08-07 16:17:44 [37.800307, -122.425315]
2009-08-07 16:17:46 [37.800324, -122.425207]
2009-08-07 16:17:47 [37.800331, -122.425153]
2009-08-07 16:17:49 [37.800343, -122.425047]
2009-08-07 16:17:50 [37.800355, -122.424994]
2009-08-07 16:17:51 [37.800362, -122.424942]
2009-08-07 16:17:54 [37.800378, -122.424796]
2009-08-07 16:17:56 [37.800357, -122.424764]

我想要的是从中获得速度的某种方法,前提是第一个数据样本的速度根据定义始终为零(上一个样本没有已知的时间增量)。

非常感谢!

最佳答案

这样的东西行得通吗?

In [99]: df
Out[99]:
X Y
2009-08-07 00:00:00 -0.900602 -1.107547
2009-08-07 01:00:00 0.398914 1.545534
2009-08-07 02:00:00 -0.429100 2.052242
2009-08-07 03:00:00 0.857940 -0.348118
2009-08-07 04:00:00 0.394655 -1.578197
2009-08-07 05:00:00 -0.240995 -1.474097
2009-08-07 06:00:00 0.619148 -0.040635
2009-08-07 07:00:00 -1.403177 -0.187540
2009-08-07 08:00:00 -0.360626 -0.399728
2009-08-07 09:00:00 0.179741 -2.709712

In [100]: df['Time'] = df.index.asi8

In [101]: dist = df.diff().fillna(0.)

In [102]: dist['Dist'] = np.sqrt(dist.X**2 + dist.Y**2)

In [103]: dist['Speed'] = dist.Dist / (dist.Time / 1e9)

In [104]: dist
Out[104]:
X Y Time Dist Speed
2009-08-07 00:00:00 0.000000 0.000000 0.000000e+00 0.000000 NaN
2009-08-07 01:00:00 1.299516 2.653081 3.600000e+12 2.954248 0.000821
2009-08-07 02:00:00 -0.828013 0.506708 3.600000e+12 0.970752 0.000270
2009-08-07 03:00:00 1.287040 -2.400360 3.600000e+12 2.723637 0.000757
2009-08-07 04:00:00 -0.463285 -1.230079 3.600000e+12 1.314430 0.000365
2009-08-07 05:00:00 -0.635650 0.104100 3.600000e+12 0.644118 0.000179
2009-08-07 06:00:00 0.860143 1.433462 3.600000e+12 1.671724 0.000464
2009-08-07 07:00:00 -2.022324 -0.146906 3.600000e+12 2.027653 0.000563
2009-08-07 08:00:00 1.042550 -0.212188 3.600000e+12 1.063924 0.000296
2009-08-07 09:00:00 0.540367 -2.309984 3.600000e+12 2.372345 0.000659

关于python - 从 Pandas.DataFrame 中带时间戳的位置计算速度,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12326641/

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