在python中计算每五秒的平均值答案

【问题标题】：compute the average of values for every five seconds in python在python中计算每五秒的平均值
【发布时间】：2021-12-16 02:32:43
【问题描述】：

我有一个像下面这样的数据集，它的时间列是基于毫秒的。

pid_col ,timestamp_col ,value_col
31,2019-03-29 07:14:56.999999756,0.0
31,2019-03-29 07:14:57.250000,0.614595
31,2019-03-29 07:14:57.500000,0.678615
31,2019-03-29 07:14:57.750000,0.687578
31,2019-03-29 07:14:58.000000244,0.559804
31,2019-03-29 07:14:58.250000,0.522672
31,2019-03-29 07:14:58.499999512,0.51627
31,2019-03-29 07:14:58.750000,0.51627
31,2019-03-29 07:14:59.000000244,0.517551
31,2019-03-29 07:14:59.250000,0.51627
31,2019-03-29 07:14:59.500000244,0.509868
31,2019-03-29 07:14:59.750000488,0.513709
31,2019-03-29 07:15:00,0.513709
31,2019-03-29 07:15:00.249999512,0.518831
31,2019-03-29 07:15:00.500000,0.531635

我如何计算每 5 秒的平均值？对于这个数据集，我应该每 5 秒精确计算一次值的平均值...我的意思是前 5 秒的值应该在 7:14:56 之间计算直到 7:15:01 等每 5 秒一次。这是我的代码：

col_list = ["timestamp", "pid","value"]
df = read_csv("data.csv", usecols=col_list)
df['timestamp'] = to_datetime(df['timestamp'], unit='ms')
df = df.groupby(['pid', Grouper(freq='5S', key='timestamp')], as_index=False) \
      .agg({'timestamp': 'first', 'value': 'mean'})

感谢您的帮助

【问题讨论】：

标签： python pandas numpy datetime time-series

【解决方案1】：

有一个很好的库叫做datetime，它能够在日期之间进行操作。例如：

from datetime import datetime, timedelta

# datetime(year, month, day, hour, minute, second, microsecond)
time0 = datetime(2019, 3, 29, 7, 14, 57, 500000)
print(time0)

fiveseconds = timedelta(seconds=5)
print(fiveseconds)

time1 = time0 + fiveseconds
print(time1)

给出输出

2019-03-29 07:14:57.500000
0:00:05
2019-03-29 07:15:02.500000

然后你可以比较一下：

from datetime import datetime, timedelta

time0 = datetime(2019, 3, 29, 7, 14, 57, 500000)

fourseconds = timedelta(seconds=4)
fiveseconds = timedelta(seconds=5)
sixseconds = timedelta(seconds=6)

time1 = time0 + fiveseconds
print(time1 < (time0 + fourseconds))  # False
print(time1 < (time0 + sixseconds))  # True

所以，对于你的问题：

from datetime import datetime, timedelta
from numpy import floor


def convert(timestr):
    """
    It receives a string, like ""2019-03-29 07:14:57.250000"
    And returns a datetime instance
    """
    date = timestr.split(" ")
    year, month, day = date[0].split("-")
    year = int(year)
    month = int(month)
    day = int(day)
    hour, minute, second = date[1].split(":")
    hour = int(hour)
    minute = int(minute)
    intsecond = int(second.split(".")[0])
    if "." in second:
        microsecond = int(floor(1e+6 * float("0." + second.split(".")[1])))
    else:
        microsecond = 0
    return datetime(year, month, day, hour, minute, intsecond, microsecond)


listtimes = ["2019-03-29 07:14:56.999999756",
             "2019-03-29 07:14:57.250000",
             "2019-03-29 07:14:57.500000",
             "2019-03-29 07:14:57.750000",
             "2019-03-29 07:14:58.000000244",
             "2019-03-29 07:14:58.250000",
             "2019-03-29 07:14:58.499999512",
             "2019-03-29 07:14:58.750000",
             "2019-03-29 07:14:59.000000244",
             "2019-03-29 07:14:59.250000",
             "2019-03-29 07:14:59.500000244",
             "2019-03-29 07:14:59.750000488",
             "2019-03-29 07:15:00",
             "2019-03-29 07:15:00.249999512",
             "2019-03-29 07:15:00.500000"]

listvalues = [0.0,
              0.614595,
              0.678615,
              0.687578,
              0.559804,
              0.522672,
              0.51627,
              0.51627,
              0.517551,
              0.51627,
              0.509868,
              0.513709,
              0.513709,
              0.518831,
              0.531635]

dt = timedelta(seconds=5)

averagevalues = []
time0 = convert(listtimes[0])
time1 = time0 + dt
counter = 0
mysum = 0
for i, v in enumerate(listvalues):
    if convert(listtimes[i]) >= time1:
        averagevalues.append(mysum / counter)
        counter = 0
        mysum = 0
        time1 += dt

    counter += 1
    mysum += v

if counter != 0:
    averagevalues.append(mysum / counter)
print(averagevalues)

给出结果

[0.5144918]

因此，如果您有更大的值列表和更大的时间，列表averagevalues 将分组每个5 seconds 的平均值。在这个例子中，所有时间都在2019-03-29 07:14:56 和"2019-03-29 07:15:01 之间，所以我们在averagevalues 中只有一个值

【讨论】：