Pandas 按日期-时间列的日级分组聚合

我有一个如下所示的数据帧。这是一个医生预约的数据

  Doctor     Appointment              Show
  A          2020-01-18 12:00:00      Yes
  A          2020-01-18 12:30:00      Yes
  A          2020-01-18 13:00:00      No
  A          2020-01-18 13:30:00      Yes
  B          2020-01-18 12:00:00      Yes
  B          2020-01-18 12:30:00      Yes
  B          2020-01-18 13:00:00      No
  B          2020-01-18 13:30:00      Yes
  B          2020-01-18 16:00:00      No
  B          2020-01-18 16:30:00      Yes
  A          2020-01-19 12:00:00      Yes
  A          2020-01-19 12:30:00      Yes
  A          2020-01-19 13:00:00      No
  A          2020-01-19 13:30:00      Yes
  A          2020-01-19 14:00:00      Yes
  A          2020-01-19 14:30:00      No
  A          2020-01-19 16:00:00      No
  A          2020-01-19 16:30:00      Yes
  B          2020-01-19 12:00:00      Yes
  B          2020-01-19 12:30:00      Yes
  B          2020-01-19 13:00:00      No
  B          2020-01-19 13:30:00      Yes
  B          2020-01-19 14:00:00      No
  B          2020-01-19 14:30:00      Yes
  B          2020-01-19 15:00:00      No
  B          2020-01-18 15:30:00      Yes

从上面的数据框中，我想在pandas中创建一个函数，它将输出以下内容

我在下面试过了

def Doctor_date_summary(doctor, date):
   Number of slots = df.groupby([doctor, date] ).sum()

预期产出：

Doctor_date_summary(Doctor, date)
If Doctor = A, date = 2020-01-19

Number of slots = 8
Number of show up = 5
show up percentage = 62.5

---

如果该日期的“显示”列中的“是”数=5，则您可以首先从以下位置创建“日期”列：

然后可以使用布尔索引：

def Doctor_date_summary(Doctor, date):
    number_of_show_up = np.sum((df['Doctor']==Doctor) & (df['day']==date) & (df['Show']=='Yes'))
    number_of_slots = np.sum((df['Doctor']==Doctor) & (df['day']==date))

    return number_of_show_up, number_of_slots, 100*number_of_show_up/number_of_slots

最后：

number_of_show_up, number_of_slots, percentage = Doctor_date_summary('A', '2020-01-19')

print("Number of slots = {}".format(number_of_slots))
print("Number of show up = {}".format(number_of_show_up))
print("show up percentage = {:.1f}".format(percentage))

Number of slots = 8
Number of show up = 5
show up percentage = 62.5

---

您可以在函数中分别创建每个掩码，然后按

&

按位

和

和

求和

按计数链：

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

def Doctor_date_summary(doctor, date):
    m1 = df['Doctor'] == doctor
    m2 = df['Appointment'].dt.normalize() == date
    m3 = df['Show'] == 'Yes'
    show_up = (m1 & m2 & m3).sum()
    no = (m1 & m2).sum()
    return show_up, no

up, no = Doctor_date_summary('A', '2020-01-19')

最后用于输出的是

f-string

s：

print(f"Number of slots = {up}")
print(f"Number of show up = {no}")
print(f"show up percentage = {up/no*100}")
Number of slots = 5
Number of show up = 8
show up percentage = 62.5

---

一个问题-你是否需要像我的问题中那样计算所有数据，然后按日期和医生查看LCT？或者只需要选择一些值并像另一个问题一样计数？只需要选择一些值并像另一个问题一样计数。不需要全部，只需要选择一些

print(f"Number of slots = {up}")
print(f"Number of show up = {no}")
print(f"show up percentage = {up/no*100}")
Number of slots = 5
Number of show up = 8
show up percentage = 62.5