从Pandas.DataFrame中的时间戳位置计算速度
[英]Calculate speed from timestamped positions in Pandas.DataFrame
问题描述
I am very new to Pandas but familiar with Numpy and Python. 
我对Pandas很新，但熟悉Numpy和Python。
Supposing I have a `Pandas.DataFrame' of X,Y points (float64) indexed by time (datetime), how can I pythonically calculate speeds from that, providing I already know how to calculate euclidean distances between points? 假设我有一个X的“Pandas.DataFrame”，Y点（float64）由时间（日期时间）索引，我怎么能从中进行pythonically计算速度，假设我已经知道如何计算点之间的欧氏距离？
EDIT: I have just read the help on pandas.Series.diff() , but still I'd like to "replace" the subtraction used on diff by another function, say `euclidean_distance()'. 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，比如说'euclidean_distance（）'。 Is there a way to do that? 有没有办法做到这一点？
DataFrame looks like (index in first column, positions in second): DataFrame看起来像（第一列中的索引，第二列中的位置）：
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]
What I want is some way to get speeds from that, providing the speed of first data sample will always be zero by definition (no known timedelta from a previous sample). 我想要的是从某种方式获得速度，根据定义，第一次数据采样的速度总是为零（没有来自前一个样本的已知时间值）。
Thanks a lot! 非常感谢！
1 个解决方案
解决方案1
4 已采纳 2012-09-08 00:42:02
Would something like this work? 会这样的吗？
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
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