Python 如何获取datetime.date对象的工作日
在df中,如下所示:Python 如何获取datetime.date对象的工作日,python,pandas,numpy,datetime,for-loop,Python,Pandas,Numpy,Datetime,For Loop,在df中,如下所示: 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
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
...
在生成绘图的for循环中,我想在绘图标题内打印date
的工作日date
是一个datetime.date对象。但是我在格式化日期时出错了。我试过这样的方法,基于:
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'
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'
其中返回了警告:
/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
尽管已经打印了警告数据框df_date
。但是如何使其从循环中同时返回日期和工作日(例如“2020-04-02,星期四”
)
我应该用这样的东西吗:
weekday = df_date.loc[date, df_date['Weekday']]
要在循环中获取相应的date
工作日?请使用DatetimeIndex的weekday属性。请参见链接页面底部的示例
范例
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
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
情节呢
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,这将为每个独特的日期生成一个绘图,我假设这就是您所追求的
对于上面的有限数据示例,这将生成绘图
Hi Hiho,在文档中声明“此方法在具有datetime值(使用dt访问器)或DatetimeIndex的两个系列上都可用。”但是plt.title(date,date.dayofweek())
返回了AttributeError:'datetime.date'对象没有属性'dayofweek'
@nilsinelabore添加了一个使用(其中一些)的示例你的数据来自上面。希望这显示了如何使用它。嗨,嗨,谢谢你的答案编辑。我试过你的密码。我认为它在系列中作为dataframe的一部分工作得很好,但我仍在努力将其插入Matplotlib标题中。我用df[“weekday”]=df[“Date”].dt.weekday
创建了一个新列,并将标题部分更改为plt.title(Date,df[“weekday]”)
,它返回TypeError:pop()接受2个位置参数,但给出了3个位置参数。我想我不应该在标题上这么做?你知道什么是标题论点吗?如果你认为有必要,我可以上传完整的代码。非常感谢…@nilsinelabore,因为您将许多参数传递给plt.title,它只需要一个字符串值(即标题),因此您必须在标题中获取所需的特定工作日。我已经添加了一个情节的例子,我认为你在追求什么。嗨,你好,谢谢你的答复。实际上,我想在标题中同时包含日期和工作日,例如“2020-04-02,星期四”。你知道我如何调整我的代码来做到这一点吗?
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()
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
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()