Python 将日期添加到数据框中的日期
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Add days to dates in dataframe
提问by BigHandsome
I am stymied at the moment. I am sure that I am missing something simple, but how do you move a series of dates forward by x units? In my more specific case I want to add 180 days to a date series within a dataframe.
我现在受阻了。我确定我遗漏了一些简单的东西，但是如何将一系列日期向前移动 x 个单位？在我更具体的情况下，我想向数据框中的日期系列添加 180 天。
Here is what I have so far:
这是我到目前为止所拥有的：
import pandas, numpy, StringIO, datetime
txt = '''ID,DATE
002691c9cec109e64558848f1358ac16,2003-08-13 00:00:00
002691c9cec109e64558848f1358ac16,2003-08-13 00:00:00
0088f218a1f00e0fe1b94919dc68ec33,2006-05-07 00:00:00
0088f218a1f00e0fe1b94919dc68ec33,2006-06-03 00:00:00
00d34668025906d55ae2e529615f530a,2006-03-09 00:00:00
00d34668025906d55ae2e529615f530a,2006-03-09 00:00:00
0101d3286dfbd58642a7527ecbddb92e,2007-10-13 00:00:00
0101d3286dfbd58642a7527ecbddb92e,2007-10-27 00:00:00
0103bd73af66e5a44f7867c0bb2203cc,2001-02-01 00:00:00
0103bd73af66e5a44f7867c0bb2203cc,2008-01-20 00:00:00
'''
df = pandas.read_csv(StringIO.StringIO(txt))
df = df.sort('DATE')
df.DATE = pandas.to_datetime(df.DATE)
df['X_DATE'] = df['DATE'].shift(180, freq=pandas.datetools.Day)
This code generates a type error. For reference I am using:
此代码生成类型错误。作为参考，我正在使用：
Python 2.7.4 Pandas '0.12.0.dev-6e7c4d6' Numpy '1.7.1'
Python 2.7.4 Pandas '0.12.0.dev-6e7c4d6' Numpy '1.7.1'
采纳答案by DSM
If I understand you, you don't actually want shift, you simply want to make a new column next to the existing DATEwhich is 180 days after. In that case, you can use timedelta:
如果我理解你，你实际上并不想要shift，你只是想在DATE180 天后的现有列旁边创建一个新列。在这种情况下，您可以使用timedelta：
>>> from datetime import timedelta
>>> df.head()
ID DATE
8 0103bd73af66e5a44f7867c0bb2203cc 2001-02-01 00:00:00
0 002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00
1 002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00
5 00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00
4 00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00
>>> df["X_DATE"] = df["DATE"] + timedelta(days=180)
>>> df.head()
ID DATE X_DATE
8 0103bd73af66e5a44f7867c0bb2203cc 2001-02-01 00:00:00 2001-07-31 00:00:00
0 002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00 2004-02-09 00:00:00
1 002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00 2004-02-09 00:00:00
5 00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00 2006-09-05 00:00:00
4 00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00 2006-09-05 00:00:00
Does that help any?
这有帮助吗？
回答by dreyco676
For future readers if you want to change different rows by different amounts you will need to use Pandas TimedeltaIndex instead to pass a series of timedeltas.
对于未来的读者，如果您想以不同的数量更改不同的行，您将需要使用 Pandas TimedeltaIndex 来传递一系列 timedeltas。
For example I might want to shift my data to the nearest report period and each record could have started on a different day of the week.
例如，我可能希望将我的数据转移到最近的报告期，并且每条记录可能在一周中的不同日期开始。
import pandas as pd
days_to_shift = pd.TimedeltaIndex(6 - launch_df['launch_dt'].dt.dayofweek)
launch_df['launch_dt'] = launch_df['launch_dt'] + days_to_shift
回答by Zero
You could use pd.DateOffset. Which seems to be faster than timedelta.
你可以使用pd.DateOffset. 这似乎比timedelta.
In [930]: df['x_DATE'] = df['DATE'] + pd.DateOffset(days=180)
In [931]: df
Out[931]:
ID DATE x_DATE
8 0103bd73af66e5a44f7867c0bb2203cc 2001-02-01 2001-07-31
0 002691c9cec109e64558848f1358ac16 2003-08-13 2004-02-09
1 002691c9cec109e64558848f1358ac16 2003-08-13 2004-02-09
4 00d34668025906d55ae2e529615f530a 2006-03-09 2006-09-05
5 00d34668025906d55ae2e529615f530a 2006-03-09 2006-09-05
2 0088f218a1f00e0fe1b94919dc68ec33 2006-05-07 2006-11-03
3 0088f218a1f00e0fe1b94919dc68ec33 2006-06-03 2006-11-30
6 0101d3286dfbd58642a7527ecbddb92e 2007-10-13 2008-04-10
7 0101d3286dfbd58642a7527ecbddb92e 2007-10-27 2008-04-24
9 0103bd73af66e5a44f7867c0bb2203cc 2008-01-20 2008-07-18
Timings
时间安排
Medium
中等的
In [948]: df.shape
Out[948]: (10000, 3)
In [950]: %timeit df['DATE'] + pd.DateOffset(days=180)
1000 loops, best of 3: 1.51 ms per loop
In [949]: %timeit df['DATE'] + timedelta(days=180)
100 loops, best of 3: 2.71 ms per loop
Large
大的
In [952]: df.shape
Out[952]: (100000, 3)
In [953]: %timeit df['DATE'] + pd.DateOffset(days=180)
100 loops, best of 3: 4.16 ms per loop
In [955]: %timeit df['DATE'] + timedelta(days=180)
10 loops, best of 3: 20 ms per loop
