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Calculate speed with temporary positions in Pandas.DataFrame

I am very new to Pandas, but familiar with Numpy and Python.

Suppose I have a `Pandas.DataFrame 'of X, Y points (float64) indexed by time (datetime), how can I pythonically calculate the speeds from this, providing that I already know how to calculate Euclidean distances between points?

EDIT: I just read the help of pandas.Series.diff() , but still I would like to “replace” the subtraction used for diff with another function, for example `euclidean_distance (). Is there any way to do this?

The DataFrame looks like (index in the first column, position in the second):

 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]

I want some way to get the speed from this, since the speed of the first data sample will always be zero by definition (the timedelta from the previous sample is not known).

Thanks a lot!

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1 answer

Something like this work?

 In [99]: df Out[99]: XY 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]: XY 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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Source: https://habr.com/ru/post/1433072/

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