Question

我的数据框如下所示

Unit #  Start Date  End Date

7417      2/6/2017  3/5/2017

我想将它拆分数周输出应该是

Unit #  Start Date  End Date

7417 2/6/2017 2/12/2017
7417 2/13/2017 2/19/2017
7417 2/20/2017 2/26/2017
7417 2/27/2017 3/05/2017

有人可以帮我吗？

Answer 1

你没有写任何代码，所以我想给你一个完整的解决方案。

这是在两个日期之间迭代几周的方法：

import datetime

date_format = "%m/%d/%Y"
d1 = datetime.datetime.strptime("2/6/2017", date_format).date()
d2 = datetime.datetime.strptime("3/5/2017", date_format).date()
d = d1
step = datetime.timedelta(days=7)

while d < d2:
    print(d.strftime(date_format))
    d += step

# 02/06/2017
# 02/13/2017
# 02/20/2017
# 02/27/2017

Answer 2

假设您有一个包含多个Unit #的数据框，您可以使用以下代码来实现您的目标：

import datetime
import pandas as pd

df = pd.DataFrame([[7417, "2/6/2017", "3/5/2017"],[7418, "3/6/2017", "4/7/2017"]],
columns = ["Unit #", "Start Date", "End Date"])

# Convert dtaframe to dates
df['Start Date'] = pd.to_datetime(df['Start Date'])
df['End Date'] = pd.to_datetime(df['End Date'])

df_out = pd.DataFrame()
week = 7

# Iterate over dataframe rows
for index, row in df.iterrows():
    date = row["Start Date"]
    date_end = row["End Date"]
    unit = row["Unit #"]
    # Get the weeks for the row
    while date < date_end:
        date_next = date + datetime.timedelta(week - 1)
        df_out = df_out.append([[unit, date, date_next]])
        date = date_next + datetime.timedelta(1)

# Remove extra index and assign columns as original dataframe
df_out = df_out.reset_index(drop=True)
df_out.columns = df.columns

因此，如果您的输入数据框是：

>>> df
   Unit #          Start Date            End Date
0    7417 2017-02-06 00:00:00 2017-03-05 00:00:00
1    7418 2017-03-06 00:00:00 2017-04-07 00:00:00

输出df_out看起来像：

>>> df_out
   Unit #           Start Date             End Date
0    7417  2017-02-06 00:00:00  2017-02-12 00:00:00
1    7417  2017-02-13 00:00:00  2017-02-19 00:00:00
2    7417  2017-02-20 00:00:00  2017-02-26 00:00:00
3    7417  2017-02-27 00:00:00  2017-03-05 00:00:00
4    7418  2017-03-06 00:00:00  2017-03-12 00:00:00
5    7418  2017-03-13 00:00:00  2017-03-19 00:00:00
6    7418  2017-03-20 00:00:00  2017-03-26 00:00:00
7    7418  2017-03-27 00:00:00  2017-04-02 00:00:00
8    7418  2017-04-03 00:00:00  2017-04-09 00:00:00

将日期范围分为行周

2 个答案: