如何获取 datetime.date 对象的工作日

Question

在如下的df中：

        id      timestamp               temperature 
27581   27822   2020-01-02 07:53:05.173 19.5    
27582   27823   2020-01-02 07:53:05.273 20.0    
27647   27888   2020-01-02 10:01:46.380 20.5    
27648   27889   2020-01-02 10:01:46.480 21.0    
27649   27890   2020-01-02 10:01:48.463 21.5    
27650   27891   2020-01-02 10:01:48.563 22.0    
27711   27952   2020-01-02 10:32:19.897 21.5    
27712   27953   2020-01-02 10:32:19.997 21.0
27861   28102   2020-01-02 11:34:41.940 21.5    
...

在生成情节的 for 循环中，我想在情节标题内打印 date 的工作日。 date 是一个 datetime.date 对象。但是我在格式化日期时遇到了一些错误。我根据this answer:

尝试了类似的方法

df['Date'] = df['timestamp'].dt.date     
df.set_index(df['timestamp'], inplace=True)


for date in df['Date'].unique():   
  df_date = df[df['Date'] == date]

  ...

  plt.title(date, date.strftime('%B'))    # I want to print both the date and the corresponding weekday.

日期以2020-01-01 格式显示日期，这很好，但工作日部分返回错误：

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-56-6cc9c07f6879> in <module>()

---> 86   plt.title(date, date.strftime('%B'))
     87 
     88   number.append(np.count_nonzero(df2['events'][minLim:maxLim]))

2 frames
/usr/local/lib/python3.6/dist-packages/matplotlib/text.py in update(self, kwargs)
    174         # Update bbox last, as it depends on font properties.
    175         sentinel = object()  # bbox can be None, so use another sentinel.
--> 176         bbox = kwargs.pop("bbox", sentinel)
    177         super().update(kwargs)
    178         if bbox is not sentinel:

AttributeError: 'str' object has no attribute 'pop'

然后我尝试了这个，基于this answer：

df['Date'] = df['timestamp'].dt.date     
df.set_index(df['timestamp'], inplace=True)


for date in df['Date'].unique():   
  df_date = df[df['Date'] == date]

  ...

  year, month, day = (int(x) for x in date.split('-'))    
  answer = datetime.date(year, month, day).weekday()
  plt.title(date, answer)

返回的

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-57-03b7529a410a> in <module>()
     82   number = []
     83   ax.autoscale() 
---> 84   year, month, day = (int(x) for x in date.split('-'))
     85   answer = datetime.date(year, month, day).weekday()
     86   plt.title(date, answer)

AttributeError: 'datetime.date' object has no attribute 'split'

---

更新：

我尝试使用以下方法在日期框架中为每个唯一的“日期”创建“工作日”列：

for date in df['Date'].unique():   
  df_date = df[df['Date'] == date]

  df_date['Date'] = pd.to_datetime(df_date['Date'], errors='coerce')
  df_date['Weekday'] = df_date['Date'].dt.dayofweek  #Add 'Weekday' column.

  print(df_date)

返回警告：

/usr/local/lib/python3.6/dist-packages/ipykernel_launcher.py:7: SettingWithCopyWarning:

A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy

尽管已打印警告数据框df_date。但是我应该如何让它从循环中返回both date 和weekday（例如"2020-04-02, Thursday"）？

我应该使用这样的东西吗：

weekday = df_date.loc[date, df_date['Weekday']]

要在循环中获取date 对应的工作日？

Answer 1

使用 DatetimeIndex 的 weekday 属性 link。请参阅链接页面底部的示例。

例子

import pandas as pd
import matplotlib.pyplot as plt

data = [
    ["2020-01-02 10:01:48.563", "22.0"],
    ["2020-01-02 10:32:19.897", "21.5"],
    ["2020-01-02 10:32:19.997", "21.0"],
    ["2020-01-02 11:34:41.940", "21.5"],
]

df = pd.DataFrame(data)
df.columns = ["date", "temp"]

df["date"] = pd.to_datetime(df["date"])
df["weekday"] = df["date"].dt.weekday
df["day_name"] = df["date"].dt.day_name()

print(df)

for day_name in df["day_name"].unique():
    plt.figure()
    plt.plot(df["date"], df["temp"])
    plt.title(day_name)
plt.show()

给了

                     date  temp  weekday  day_name
0 2020-01-02 10:01:48.563  22.0        3  Thursday
1 2020-01-02 10:32:19.897  21.5        3  Thursday
2 2020-01-02 10:32:19.997  21.0        3  Thursday
3 2020-01-02 11:34:41.940  21.5        3  Thursday

和剧情

---

更新 04/04 以回应评论

import pandas as pd
import matplotlib.pyplot as plt

data = [
["2020-01-02 10:01:48.563", "22.0"],
["2020-01-02 10:32:19.897", "21.5"],
["2020-01-02 10:32:19.997", "21.0"],
["2020-01-02 11:34:41.940", "21.5"],
]

df = pd.DataFrame(data)
df.columns = ["timestamp", "temp"]
df["timestamp"] = pd.to_datetime(df["timestamp"])

df['Date'] = df['timestamp'].dt.date
df.set_index(df['timestamp'], inplace=True)

df['Weekday'] = df.index.day_name() 

for date in df['Date'].unique():
  df_date = df[df['Date'] == date]

  plt.figure()
  plt.plot(df_date["timestamp"], df["temp"])
  plt.title("{}, {}".format(date, df_date["Weekday"].iloc[0]))
  plt.show()

请注意，这将为每个唯一日期生成一个图，我假设这就是您所追求的。

对于上面的有限数据示例，这会产生图