Python 以numpy为单位的聚合时间序列

我试图将一个变量聚合到时间段上，但我不确定如何以Pythonic的方式实现这一点。我的计划是每十秒钟汇总一次。 我的时间数据是第一列，从午夜开始以秒表示。第二列是我要聚合的变量。大多数条目是零，但其他条目是我想要求和的实际数字

因此，我们的想法是从第1列开始，每隔10秒对第2列求和。 对于第一列，矩阵向量每10秒有一次时间，对于第二列，聚合变量有一次时间

我的第一个想法是在第二列做一个while循环，当我们在同一个时间段内时，对所有的数字求和。并将其放置在for循环中，每增加10秒。这似乎是难以置信的沉重和缓慢

非常感谢您的建议。多谢各位

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---

考虑以下方法：

import pandas as pd

df = pd.read_csv(filename, header=None, names=['ts','val'])

In [31]: df.groupby(df['ts']//10*10)['val'].sum()
Out[31]:
ts
34200.0    1806.0
34210.0     907.0
34220.0     100.0
34230.0    -201.0
Name: val, dtype: float64

资料来源：

In [33]: df
Out[33]:
               ts    val
0    34200.095354    0.0
1    34200.095354    0.0
2    34200.095354    0.0
3    34200.095354    0.0
4    34200.095354    0.0
5    34200.095354    0.0
6    34200.095354    0.0
7    34200.095354    0.0
8    34200.095354    0.0
9    34200.095354    0.0
..            ...    ...
237  34230.365951    0.0
238  34233.152929   -1.0
239  34233.152986 -100.0
240  34233.152996 -100.0
241  34233.153011    0.0
242  34233.154379    0.0
243  34233.154699 -200.0
244  34233.155680    0.0
245  34233.176736  100.0
246  34233.187006  100.0

[247 rows x 2 columns]

df['ts']//10*10

-将时间戳舍入至10秒，因此我们可以在以后将其用于分组：

In [34]: df['ts']//10*10
Out[34]:
0      34200.0
1      34200.0
2      34200.0
3      34200.0
4      34200.0
5      34200.0
6      34200.0
7      34200.0
8      34200.0
9      34200.0
        ...
237    34230.0
238    34230.0
239    34230.0
240    34230.0
241    34230.0
242    34230.0
243    34230.0
244    34230.0
245    34230.0
246    34230.0
Name: ts, Length: 247, dtype: float64

---

您可能需要考虑使用<代码>熊猫> /COD>，它内置了大量的时间序列功能。是的。我将使用

df.set\u索引（'my\u time\u column'，inplace=True）

。然后你可以做类似的事情：

df.groupby（df.index.seconds）.sum（）

你能解释一下你是如何得到这样一个时间戳的吗？它看起来不像是一个标准的UNIX时代（#of（nano-），（micro-），（mili-），秒）@python_newbie，从现在的午夜开始？@python_newbie是的，所以，实际上，它应该是

a.index.second

，不带“s”，但这只适用于

DatetimeIndex

。但是，如果您只有浮点值，请尝试使用

a.groupby（df['a']//10）.sum（）

注意，

//

只是整数除法，因此

123.456//10==12.0

，因此，基本上，相同10秒插槽中的所有时间都将分组在一起。编辑我看到MaxU已经击败了我…@python\u新手，我在答案中添加了一个简短的解释-请检查得到了！非常感谢。