英文:
Python Network Hours
问题
以下是代码部分的翻译：
def business_hours(start, end):start = pd.Series(start)end = pd.Series(end)mask = (pd.notnull(start) & pd.notnull(end)) & (start.dt.hour >= 9) & (end.dt.hour <= 17) & (start.dt.weekday < 5) & (end.dt.weekday < 5)result = np.empty(len(start), dtype=float)result.fill(np.nan)result[mask] = np.where((start[mask].dt.hour >= 9) & (end[mask].dt.hour <= 17), (end[mask] - start[mask]).astype('timedelta64[h]').astype(float), 0)return result
希望这能帮助你解决问题。
英文:
Really struggling with this one so any help would be much appreciated.
GOAL - workout the hours between two datetime columns excluding weekends and only taking the hours between the working times of 9 & 17.
Now I have reused a function that I use for network days but the output is wrong and I can't seem to figure out how to get it working.
As an example I have In my data a start date and end date that are as follows
> Start_Date = 2017-07-11 19:33:00
> End_Date = 2017/07/12 12:01:00
and the output I'm after is
>3.02
However the function I do have is returning 16!
Function below -
start = pd.Series(start)end = pd.Series(end)mask = (pd.notnull(start) & pd.notnull(end)) & (start.dt.hour >= 9) & (end.dt.hour <= 17) & (start.dt.weekday < 5) & (end.dt.weekday < 5)result = np.empty(len(start), dtype=float)result.fill(np.nan)result[mask] = np.where((start[mask].dt.hour >= 9) & (end[mask].dt.hour <= 17), (end[mask] - start[mask]).astype('timedelta64[h]').astype(float), 0)return result ```</details># 答案1**得分**: 0看起来你需要的是 `businesstimedelta`：```pythonimport datetimeimport businesstimedeltastart = datetime.datetime.strptime("2017-07-11 19:33:00", "%Y-%m-%d %H:%M:%S")end = datetime.datetime.strptime("2017-07-12 12:01:00", "%Y-%m-%d %H:%M:%S")# 定义一个工作日规则workday = businesstimedelta.WorkDayRule(start_time=datetime.time(9),end_time=datetime.time(17),working_days=[0, 1, 2, 3, 4])businesshours = businesstimedelta.Rules([workday])# 计算差异diff = businesshours.difference(start, end)print(diff)
输出:
<BusinessTimeDelta 3 hours 60 seconds>
https://pypi.org/project/businesstimedelta/
英文:
It looks like what you need is businesstimedelta
import datetimeimport businesstimedeltastart = datetime.datetime.strptime("2017-07-11 19:33:00", "%Y-%m-%d %H:%M:%S")end = datetime.datetime.strptime("2017-07-12 12:01:00", "%Y-%m-%d %H:%M:%S")# Define a working day ruleworkday = businesstimedelta.WorkDayRule(start_time=datetime.time(9),end_time=datetime.time(17),working_days=[0, 1, 2, 3, 4])businesshours = businesstimedelta.Rules([workday])# Calculate the differencediff = businesshours.difference(start, end)print(diff)
Output:
<BusinessTimeDelta 3 hours 60 seconds>
答案2
得分: 0
以下是代码的翻译部分：
# 导入必要的库import datetimeimport businesstimedelta# 定义工作日规则workday = businesstimedelta.WorkDayRule(start_time=datetime.time(9),end_time=datetime.time(17),working_days=[0, 1, 2, 3, 4])# 结合两者businesshrs = businesstimedelta.Rules([workday])# 定义一个函数，将时间差转换为分钟def business_Mins(df, start, end):try:# 创建一个布尔掩码，用于选择非空的时间戳mask = pd.notnull(df[start]) & pd.notnull(df[end])result = np.empty(len(df), dtype=object)# 对每一行应用工作日时间差，并将小时转换为分钟result[mask] = df.loc[mask].apply(lambda x: businesshrs.difference(x[start],x[end]).hours, axis=1)result[~mask] = np.nanreturn result * 60except KeyError as e:print(f"错误：在数据框中找不到一个或多个列 - {e}")return None# 调用函数，将结果存储在名为'Contact_SLA'的列中df['Contact_SLA'] = business_Mins(df, 'Date and Time of Instruction', 'Date and Time of Attempted Contact')
请注意，这是给定代码的翻译，包括导入库、定义工作日规则、定义函数以及在数据框中应用函数的部分。
英文:
So, I really struggled finding out how to apply the above into a function but finally after much banging of the head came up with the below. Sharing for the next person in my situation. I wanted to convert to minutes so if not required just remove the *60 at the return
import datetimeimport businesstimedelta# Define a working dayworkday = businesstimedelta.WorkDayRule(start_time=datetime.time(9),end_time=datetime.time(17),working_days=[0, 1, 2, 3, 4])# Combine the twobusinesshrs = businesstimedelta.Rules([workday])def business_Mins(df, start, end):try:mask = pd.notnull(df[start]) & pd.notnull(df[end])result = np.empty(len(df), dtype=object)result[mask] = df.loc[mask].apply(lambda x: businesshrs.difference(x[start],x[end]).hours, axis=1)result[~mask] = np.nanreturn result * 60except KeyError as e:print(f"Error: One or more columns not found in the dataframe - {e}")return Nonedf['Contact_SLA'] = business_Mins(df, 'Date and Time of Instruction', 'Date and Time of Attempted Contact')
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