Pandas 以秒为单位计算时差，不包括工作日

Question

我有一个包含两列的数据集。一个是Message Received Timestamp，另一个是Message Sent Timestamp。时间的格式看起来像这样2021-06-05T07:00:34Z。我想计算这两列之间的时间差（以秒为单位），并通过排除工作日来确保差小于24 hours。

我知道np.busday_count 方法，但我想以秒为单位计算差异。请在下面查看我当前的代码，它没有考虑工作日：(pd.to_datetime(df["Message Received Timestamp"]) - pd.to_datetime(df["Message Sent Timestamp"])) < pd.Timedelta("1 day")

非常感谢您在计算工作日（以秒为单位）时提供的帮助。非常感谢。

Answer 1

我会做这样的事情......不过不要认为这是理所当然的，在某些特定情况下它可能是错误的。

import numpy as np
import pandas as pd

# test dates
date1 = '2021-03-05T07:00:34Z'
date2 = '2021-10-04T11:41:01Z'

d1 = pd.to_datetime(date1)
d2 = pd.to_datetime(date2)

# business days delta
bus_days_delta = (d2 - d1).days - np.busday_count(date1.partition("T")[0], date2.partition("T")[0])
date2real = d2 - pd.to_timedelta(bus_days_delta, unit='d')

# business day delta in seconds
real_delta = date2real - pd.to_datetime(date1)
print(real_delta)

Answer 2

您可以使用Timedelta.total_seconds() 获取两个日期之间的总时间差（以秒为单位），如下所示：

df["Diff_Total_Seconds"] = (pd.to_datetime(df["Message Received Timestamp"]) - pd.to_datetime(df["Message Sent Timestamp"])).dt.total_seconds() 

然后通过将np.busday_count 乘以 86400（一天中的秒数）来计算工作日的总秒数：

df["Busday_Total_Seconds"] = (np.busday_count(pd.to_datetime(df["Message Sent Timestamp"]).values.astype('datetime64[D]'), pd.to_datetime(df["Message Received Timestamp"]).values.astype('datetime64[D]')) - 1) * 86400

最后，得到不包括工作日的时差，以秒为单位。

df["Diff_Total_Seconds_Excl_Busday"] = df["Diff_Total_Seconds"] - df["Busday_Total_Seconds"] 

测试结果：

  Message Received Timestamp Message Sent Timestamp  Diff_Total_Seconds  Busday_Total_Seconds  Diff_Total_Seconds_Excl_Busday
0       2021-06-07T06:41:01Z   2021-06-04T07:00:34Z            258027.0                     0                        258027.0
1       2021-06-09T06:41:01Z   2021-06-08T07:00:34Z             85227.0                     0                         85227.0

编辑

在与 OP 进一步讨论后，我们提出了一些特殊情况的更新代码。

df["Message Received Timestamp"] = pd.to_datetime(df["Message Received Timestamp"])
df["Message Sent Timestamp"] = pd.to_datetime(df["Message Sent Timestamp"])
df["Diff_Total_Sec"] = (df["Message Received Timestamp"] - df["Message Sent Timestamp"]).dt.total_seconds()
df["Diff_Days"] = (df["Message Received Timestamp"].dt.normalize() - df["Message Sent Timestamp"].dt.normalize()).dt.days
df["NonBDay_Total_Sec"] = (df["Diff_Days"] - np.busday_count(df["Message Sent Timestamp"].values.astype('datetime64[D]'), df["Message Received Timestamp"].values.astype('datetime64[D]'))) * 86400
df["NonBDay_Total_Sec"] = np.where((df["Message Received Timestamp"].dt.dayofweek == 6) & (df["NonBDay_Total_Sec"] < 86400 * 2), 86400 * 2, df["NonBDay_Total_Sec"])
df["NonBDay_Total_Sec"] = np.where((df["Message Received Timestamp"].dt.dayofweek == 5) & (df["NonBDay_Total_Sec"] < 86400), 86400, df["NonBDay_Total_Sec"])
df["Diff_Total_Sec_Excl_NonBDay"] = df["Diff_Total_Sec"] - df["NonBDay_Total_Sec"] 
df["Validity_Result"] = (df["Diff_Total_Sec_Excl_NonBDay"] < 86400) & (df["Message Sent Timestamp"] <= df["Message Received Timestamp"])

测试结果：

   Message Received Timestamp    Message Sent Timestamp  Diff_Total_Sec  Diff_Days  NonBDay_Total_Sec  Diff_Total_Sec_Excl_NonBDay  Validity_Result
0   2021-06-07 06:41:01+00:00 2021-06-04 07:00:34+00:00        258027.0          3             172800                      85227.0             True
1   2021-06-07 06:41:01+00:00 2021-06-07 05:41:01+00:00          3600.0          0                  0                       3600.0             True
2   2021-06-07 06:41:01+00:00 2021-06-07 07:41:01+00:00         -3600.0          0                  0                      -3600.0            False
3   2021-06-07 06:41:01+00:00 2021-06-03 06:41:01+00:00        345600.0          4             172800                     172800.0            False
4   2021-06-07 06:41:01+00:00 2021-06-10 06:41:01+00:00       -259200.0         -3                  0                    -259200.0            False
5   2021-06-07 06:41:01+00:00 2021-06-03 07:00:01+00:00        344460.0          4             172800                     171660.0            False
6   2021-06-07 06:41:01+00:00 2021-06-04 07:00:01+00:00        258060.0          3             172800                      85260.0             True
7   2021-06-07 06:41:01+00:00 2021-06-04 05:00:01+00:00        265260.0          3             172800                      92460.0            False
8   2021-06-06 22:01:03+00:00 2021-06-04 09:38:38+00:00        217345.0          2             172800                      44545.0             True
9   2021-06-05 07:01:36+00:00 2021-06-04 03:49:42+00:00         97914.0          1              86400                      11514.0             True
10  2021-06-07 01:10:09+00:00 2021-06-03 03:32:06+00:00        337083.0          4             172800                     164283.0            False
11  2021-06-06 09:03:34+00:00 2021-06-04 08:36:36+00:00        174418.0          2             172800                       1618.0             True