获得Pandas系列的一周的第一天

I have the following df :

我有以下df：

import pandas as pd
from datetime import datetime, timedelta

df = pd.DataFrame([
        ["A", "2018-08-03"],
        ["B", "2018-08-20"]
])
df.columns = ["Item", "Date"]

I want to get the first day of the week for every line of my df. I tried to do this :

我希望得到我的df的每一行的一周的第一天。我试着这样做：

df['Date'] =  pd.to_datetime(df['Date'], format='%Y-%m-%d')
df["Day_of_Week"] = df.Date.dt.weekday

df["First_day_of_the_week"] = df.Date - timedelta(days=df.Day_of_Week)

But I got that error message :

但是我收到了错误消息：

TypeError: unsupported type for timedelta days component: Series

How can I get the first day of the week for a Series ? My expected result is that :

我怎样才能获得系列赛的第一天？我的预期结果是：

"A", "2018-08-03", "2018-07-30"
“A”，“2018-08-03”，“2018-07-30”
"B", "2018-08-20", "2018-08-20"
“B”，“2018-08-20”，“2018-08-20”

4 个解决方案

#1

Unfortunately timedelta doesn't support a vectorized form so I would go for an apply

不幸的是timedelta不支持矢量化形式，所以我会去申请

df["First_day_of_the_week"] = df.apply(lambda x: x['Date'] - timedelta(days=x['Day_of_Week']), axis=1)

EDIT

编辑

timedelta doesn't support vectorized arguments but can be multiplied by a vector :)

timedelta不支持向量化参数，但可以乘以向量:)

df["First_day_of_the_week"] = df.Date - df.Day_of_Week * timedelta(days=1)

#2

Leave out your 'Day of week" calculation and do this.

省略“星期几”计算并执行此操作。

df["First_day_of_the_week"] = df['Date'].apply(lambda x: (x - timedelta(days=x.dayofweek)))
print(df)

giving

给

  Item       Date First_day_of_the_week
0    A 2018-08-03            2018-07-30
1    B 2018-08-20            2018-08-20

#3

A vectorised solution is possible with NumPy:

使用NumPy可以实现矢量化解决方案：

df['First_day'] = df['Date'] - df['Date'].dt.weekday * np.timedelta64(1, 'D')

print(df)

  Item       Date  First_day
0    A 2018-08-03 2018-07-30
1    B 2018-08-20 2018-08-20

#4

You can stay in Pandas and use its DateOffset objects:

您可以留在Pandas并使用其DateOffset对象：

>>> from pandas.tseries.offsets import Week

>>> df.Date.where(df.Date.dt.weekday == 0, df.Date - Week(weekday=0))
0   2018-07-30
1   2018-08-20
Name: Date, dtype: datetime64[ns]

The trick being that you need to not do the subtraction where the weekday is already Monday (weekday == 0). This says, "in cases where weekday is already zero, do nothing; else, return Monday of that week."

诀窍是你不需要在工作日已经是星期一（工作日== 0）的情况下进行减法。这说，“在工作日已经为零的情况下，什么都不做;否则，返回那周的星期一。”

#1