python pandas：如何计算导数/梯度 [英] python pandas: how to calculate derivative/gradient

问题描述

鉴于我有以下两个向量：

Given that I have the following two vectors:

In [99]: time_index
Out[99]: 
[1484942413,
 1484942712,
 1484943012,
 1484943312,
 1484943612,
 1484943912,
 1484944212,
 1484944511,
 1484944811,
 1484945110]

In [100]: bytes_in
Out[100]: 
[1293981210388,
 1293981379944,
 1293981549960,
 1293981720866,
 1293981890968,
 1293982062261,
 1293982227492,
 1293982391244,
 1293982556526,
 1293982722320]

其中 bytes_in 是仅增量计数器，而 time_index 是

Where bytes_in is an incremental only counter, and time_index is a list to unix timestamps (epoch).

目标：我要计算的是比特率。

Objective: What I would like to calculate is the bitrate.

这意味着我将构建一个数据框，例如

That means that I will build a data frame like

In [101]: timeline = pandas.to_datetime(time_index, unit="s")

In [102]: recv = pandas.Series(bytes_in, timeline).resample("300S").mean().ffill().apply(lambda i: i*8)

In [103]: recv
Out[103]: 
2017-01-20 20:00:00    10351849683104
2017-01-20 20:05:00    10351851039552
2017-01-20 20:10:00    10351852399680
2017-01-20 20:15:00    10351853766928
2017-01-20 20:20:00    10351855127744
2017-01-20 20:25:00    10351856498088
2017-01-20 20:30:00    10351857819936
2017-01-20 20:35:00    10351859129952
2017-01-20 20:40:00    10351860452208
2017-01-20 20:45:00    10351861778560
Freq: 300S, dtype: int64

问题：现在，很奇怪，手动计算梯度会给我：

Question: Now, what is strange, calculating the gradient manually gives me :

In [104]: (bytes_in[1]-bytes_in[0])*8/300
Out[104]: 4521.493333333333

这是正确的值..

同时用熊猫计算梯度给了我

while calculating the gradient with pandas gives me

In [124]: recv.diff()
Out[124]: 
2017-01-20 20:00:00          NaN
2017-01-20 20:05:00    1356448.0
2017-01-20 20:10:00    1360128.0
2017-01-20 20:15:00    1367248.0
2017-01-20 20:20:00    1360816.0
2017-01-20 20:25:00    1370344.0
2017-01-20 20:30:00    1321848.0
2017-01-20 20:35:00    1310016.0
2017-01-20 20:40:00    1322256.0
2017-01-20 20:45:00    1326352.0
Freq: 300S, dtype: float64

与上面不同， 1356448.0与4521.493333333333

请问我做错了什么吗？

推荐答案

pd.Series.diff（）只接受差异。

这会为您提供答案

recv.diff() / recv.index.to_series().diff().dt.total_seconds()

2017-01-20 20:00:00            NaN
2017-01-20 20:05:00    4521.493333
2017-01-20 20:10:00    4533.760000
2017-01-20 20:15:00    4557.493333
2017-01-20 20:20:00    4536.053333
2017-01-20 20:25:00    4567.813333
2017-01-20 20:30:00    4406.160000
2017-01-20 20:35:00    4366.720000
2017-01-20 20:40:00    4407.520000
2017-01-20 20:45:00    4421.173333
Freq: 300S, dtype: float64

---

您也可以使用 numpy.gradient 传递 bytes_in 和您期望的增量。这不会使长度减少一，而是对边缘进行假设。

---

You could also use numpy.gradient passing the bytes_in and the delta you expect to have. This will not reduce the length by one, instead making assumptions about the edges.

np.gradient(bytes_in, 300) * 8

array([ 4521.49333333,  4527.62666667,  4545.62666667,  4546.77333333,
        4551.93333333,  4486.98666667,  4386.44      ,  4387.12      ,
        4414.34666667,  4421.17333333])

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