基于时间的点之间的平均值

Question

我正在尝试使用 Python 来获取时间，以及在点之间移动的对象之间的平均速度。

数据看起来有点像这样，

location    initialtime id  speed distance
    1   2020-09-18T12:03:14.485952Z car_uno 72 9km
    2   2020-09-18T12:10:14.485952Z car_uno 83 8km
    3   2020-09-18T11:59:14.484781Z car_duo 70 9km
    7   2020-09-18T12:00:14.484653Z car_trio    85 8km
    8   2020-09-18T12:12:14.484653Z car_trio    70 7.5km

我目前使用的功能基本上是这样的，

Speeds.index = pd.to_datetime(Speeds.index)
..etc

现在，如果我通常这样做，我只会获取 id 的唯一值，

for x in speeds.id.unique():
    Speeds[speeds.id=="x"]...

但是这个方法真的不行。

简单地查看是否有多个 id 点随时间推移，然后取给定时间的平均速度的最佳方法是什么？否则，如果没有多个值，则只返回速度本身。

我可以使用更简单的 pandas 过滤器吗？

预期的输出很简单，

area - id - initial time - journey time - average speed.

关键是要获得经过两点的车辆的平均时间和行程时间

Answer 1

要获得平均速度和行程时间，您可以使用groupby() 并传入确定完整行程的列，例如id 或area。

import pandas as pd
from io import StringIO

data = StringIO("""
area    initialtime id  speed
    1   2020-09-18T12:03:14.485952Z car_uno 72
    2   2020-09-18T12:10:14.485952Z car_uno 83
    3   2020-09-18T11:59:14.484781Z car_duo 70
    7   2020-09-18T12:00:14.484653Z car_trio    85
    8   2020-09-18T12:12:14.484653Z car_trio    70
""")

df = pd.read_csv(data, delim_whitespace=True)

df["initialtime"] = pd.to_datetime(df["initialtime"])

# change to ["id", "area"] if need more granular aggregation
group_cols = ["id"]

time = df.groupby(group_cols)["initialtime"].agg([max, min]).eval('max-min').reset_index(name="journey_time")

speed = df.groupby(group_cols)["speed"].mean().reset_index(name="average_speed")

pd.merge(time, speed, on=group_cols)

         id journey_time  average_speed
0   car_duo     00:00:00           70.0
1  car_trio     00:12:00           77.5
2   car_uno     00:07:00           77.5

Answer 2

我尝试使用一个非常直观的解决方案。我假设数据已经加载到df。

df['initialtime'] = pd.to_datetime(df['initialtime'])

result = []

for car in df['id'].unique():
    _df = df[df['id'] == car].sort_values('initialtime', ascending=True)

    #  Where the car is leaving "from" and where it's heading "to"
    _df['From'] = _df['location']
    _df['To'] = _df['location'].shift(-1, fill_value=_df['location'].iloc[0])

   #  Auxiliary columns
    _df['end_time'] = _df['initialtime'].shift(-1, fill_value=_df['initialtime'].iloc[0])
    _df['end_speed'] = _df['speed'].shift(-1, fill_value=_df['speed'].iloc[0])
   
    #  Desired columns
    _df['journey_time'] = _df['end_time'] - _df['initialtime']
    _df['avg_speed'] = (_df['speed'] + _df['end_speed']) / 2

    _df = _df[_df['journey_time'] >= pd.Timedelta(0)]
    _df.drop(['location', 'distance', 'speed', 'end_time', 'end_speed'], 
             axis=1, inplace=True)

    result.append(_df)

final_df = pd.concat(result).reset_index(drop=True)

最终的DataFrame如下：

                       initialtime        id  From  To    journey_time  avg_speed
0 2020-09-18 12:03:14.485952+00:00   car_uno     1   2 0 days 00:07:00       77.5
1 2020-09-18 11:59:14.484781+00:00   car_duo     3   3 0 days 00:00:00       70.0
2 2020-09-18 12:00:14.484653+00:00  car_trio     7   8 0 days 00:12:00       77.5

Answer 3

这是另一种方法。我的结果与其他帖子不同，所以我可能误解了要求。简而言之，我将每个平均速度计算为总距离除以总时间（对于每辆车）。

from io import StringIO
import pandas as pd

# speed in km / hour; distance in km
data = '''location    initial-time id  speed distance
1   2020-09-18T12:03:14.485952Z car_uno 72 9
2   2020-09-18T12:10:14.485952Z car_uno 83 8
3   2020-09-18T11:59:14.484781Z car_duo 70 9
7   2020-09-18T12:00:14.484653Z car_trio    85 8
8   2020-09-18T12:12:14.484653Z car_trio    70 7.5
'''

现在创建数据框并执行计算

# create data frame
df = pd.read_csv(StringIO(data), delim_whitespace=True)
df['elapsed-time'] = df['distance'] / df['speed'] # in hours

# utility function
def hours_to_hms(elapsed):
    ''' Convert `elapsed` (in hours) to hh:mm:ss (round to nearest sec)'''
    h, m = divmod(elapsed, 1)
    m *= 60
    _, s = divmod(m, 1)
    s *= 60
    hms = '{:02d}:{:02d}:{:02d}'.format(int(h), int(m), int(round(s, 0)))
    return hms

# perform calculations
start_time = df.groupby('id')['initial-time'].min()
journey_hrs = df.groupby('id')['elapsed-time'].sum().rename('elapsed-hrs')
hms = journey_hrs.apply(lambda x: hours_to_hms(x)).rename('hh:mm:ss')
ave_speed = ((df.groupby('id')['distance'].sum() 
             / df.groupby('id')['elapsed-time'].sum())
             .rename('ave speed (km/hr)')
             .round(2))

# assemble results
result = pd.concat([start_time, journey_hrs, hms, ave_speed], axis=1)

print(result)
                         initial-time  elapsed-hrs  hh:mm:ss  \
id                                                             
car_duo   2020-09-18T11:59:14.484781Z     0.128571  00:07:43   
car_trio  2020-09-18T12:00:14.484653Z     0.201261  00:12:05   
car_uno   2020-09-18T12:03:14.485952Z     0.221386  00:13:17   

          ave speed (km/hr)  
id                           
car_duo               70.00  
car_trio              77.01  
car_uno               76.79  

Answer 4

您应该提供更好的数据集（即具有相同时间点）以便我们更好地了解输入，并提供预期输出示例以便我们了解平均速度的计算。

因此，如果 df 是包含您的输入数据的数据框，我只是猜测您可能正在寻找 df.groupby('initialtime')['speed'].mean()。