如何在不使用数据时间的情况下计算时间差？

Question

我试图计算两次之间的时间差，我真的很难使用datetime时间，因为我只有一个时间值，没有日期值，当我写类似t1 = datetime.strptime(item[1], '%H:%M:%S')例如， item[1] = "00:01:34"然后打印t1我得到了类似t1: 1900-01-01 00:01:34的东西。

当我计算timedelta = t2 - t1然后我尝试打印结果时，它不会打印任何内容，因为它进入了一个循环。 我已经尝试过使用time / .time()但它给了我一个TypeError: unsupported operand type(s) for -: 'builtin_function_or_method' and 'builtin_function_or_method' 。 如果我指向timedelta = (t2 - t1).seconds ，它会报告Class 'time' does not define 'sub', so the '-' operator cannot be used on its instances ，如果我运行代码，它甚至会给我TypeError如果我使用.time() 。

为了清楚起见，我的代码是这样的：

listaDivisa = [
['2010-01-05', '12:32:05', 'at the kitchen entrance from the dining room', 'ON']
['2010-01-05', '12:32:05', 'in the kitchen', 'ON']
['2010-01-05', '12:32:08', 'in the living room', 'ON']
['2010-01-05', '12:32:08', 'in the kitchen', 'OFF']
['2010-01-05', '12:32:10', 'at the kitchen entrance from the dining room', 'OFF']
['2010-01-05', '12:32:10', 'in the kitchen', 'ON']
['2010-01-05', '12:32:11', 'in the kitchen', 'OFF']
['2010-01-05', '12:32:11', 'in the living room', 'OFF']
['2010-01-06', '02:32:11', 'in the kitchen', 'ON']
['2010-01-06', '02:32:20', 'in the kitchen', 'OFF']
['2010-01-06', '02:34:23', 'in the living room', 'ON']
['2010-01-06', '02:34:42', 'in the living room', 'OFF']]
# this list contains approximately 3000 of this activities, obviously I put only 
# a few just for example

listaDict = {}

for p in listaDivisa:
    if p[2] not in listaDict.keys():
        listaDict[p[2]] = dict()

for i, item in enumerate(listaDivisa):
    for j in range(i + 1, len(listaDivisa) - 1):
        if item[0] == listaDivisa[j][0]:
            if item[2] == listaDivisa[j][2]:
                if item[3] == "ON" and listaDivisa[j][3] == "OFF":
                    t1 = datetime.strptime(item[1], '%H:%M:%S')
                    t2 = datetime.strptime(listaDivisa[j][1], '%H:%M:%S')
                    timedelta = (t2 - t1).seconds

                    listaDict[item[2]][item[
                        0]] = "active for " + str(
                        timedelta) + " seconds"

for key, value in listaDict.items():
    print(key, ' : ', value)

我试图达到的结果是这样的嵌套字典：

in the kitchen:['2009-10-16': 'active for 2341 seconds', '2009-10-17': 'active for 0 seconds' ..... ]
in the living room:['2009-10-16': 'active for 20 seconds', '2009-10-17': 'active for 3 seconds' ..... ]
at the kitchen entrance from the dining room:['2009-10-16': 'active for 6473 seconds', '2009-10-17': 'active for 374 seconds' ... ]

Answer 1

这是使用pandas解决您的问题的方法。 我相信其他用户会想办法重构下面的一些内容来改进它，但我相信它可以完成工作。 我已经包含评论来解释每个阶段发生的事情。 如果有任何不清楚的地方，或者结果不是你想要的，请告诉我。

import pandas as pd

listaDivisa = [
['2010-01-05', '12:32:05', 'at the kitchen entrance from the dining room', 'ON'],
['2010-01-05', '12:32:05', 'in the kitchen', 'ON'],
['2010-01-05', '12:32:08', 'in the living room', 'ON'],
['2010-01-05', '12:32:08', 'in the kitchen', 'OFF'],
['2010-01-05', '12:32:10', 'at the kitchen entrance from the dining room', 'OFF'],
['2010-01-05', '12:32:10', 'in the kitchen', 'ON'],
['2010-01-05', '12:32:11', 'in the kitchen', 'OFF'],
['2010-01-05', '12:32:11', 'in the living room', 'OFF'],
['2010-01-06', '02:32:11', 'in the kitchen', 'ON'],
['2010-01-06', '02:32:20', 'in the kitchen', 'OFF'],
['2010-01-06', '02:34:23', 'in the living room', 'ON'],
['2010-01-06', '02:34:42', 'in the living room', 'OFF']]

# 1) list of lists to df
df = pd.DataFrame(listaDivisa, columns=['Date','Time','Activity','Status'])

# 2) add datetime col
df['Datetime'] = pd.to_datetime(df.Date + ' ' + df.Time)

# 3) groupby date & activity, calc diff between datetime vals (= timedeltas) 
# and convert those timedeltas to seconds
df['Timedelta'] = df.groupby(['Date','Activity'])['Datetime'].transform('diff').dt.total_seconds()

# 4) create new df just on df.Status == 'OFF', these will be the correct timedeltas
new_df = df.loc[df.Status=='OFF',['Date','Activity','Timedelta']]

# =============================================================================
# new_df at this stage:
# 
#           Date                                      Activity  Timedelta
# 3   2010-01-05                                in the kitchen        3.0
# 4   2010-01-05  at the kitchen entrance from the dining room        5.0
# 6   2010-01-05                                in the kitchen        1.0
# 7   2010-01-05                            in the living room        3.0
# 9   2010-01-06                                in the kitchen        9.0
# 11  2010-01-06                            in the living room       19.0
# =============================================================================

# 5) now groupby to get sum per activity & date and reset the index 
# e.g. on '2010-01-05' we have 2 timedeltas for 'in the kitchen': 3 sec + 1 sec = 4

new_df = new_df.groupby(['Activity','Date'])['Timedelta'].sum().reset_index()
  
# =============================================================================
# e.g. new_df now looks like:
#                                        Activity        Date  Timedelta
# 0  at the kitchen entrance from the dining room  2010-01-05        5.0
# 1                                in the kitchen  2010-01-05        4.0
# 2                                in the kitchen  2010-01-06        9.0
# 3                            in the living room  2010-01-05        3.0
# 4                            in the living room  2010-01-06       19.0
# =============================================================================

# 6) turn df.Timedelta into strings
new_df['Timedelta'] = new_df['Timedelta'].apply(lambda x: f'active for {int(x)} seconds')

# 7) set df.Activity as index
new_df.set_index('Activity', drop=True, inplace=True)

# 8) define a dictionary and iter over rows to populate it
a_dict = {}

for row, data in new_df.iterrows():
    # if row already exists in dict, we want to *add* a key:value pair
    if row in a_dict:
        a_dict[row].update({ data[0]: data[1]})
    # else, we want to create a new key, and set its init value to key:value pair
    else:
        a_dict[row] = { data[0]: data[1] } 
    
a_dict

{'at the kitchen entrance from the dining room': {'2010-01-05': 'active for 5 seconds'},
 'in the kitchen': {'2010-01-05': 'active for 4 seconds',
  '2010-01-06': 'active for 9 seconds'},
 'in the living room': {'2010-01-05': 'active for 3 seconds',
  '2010-01-06': 'active for 19 seconds'}}

因此，现在您可以执行以下操作：

a_dict['in the kitchen']['2010-01-05']
'active for 4 seconds'

# As mentioned: we have 2 entries for 'in the kitchen' on '2010-01-05'
# summed to 4 seconds'