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在DataFrame Pandas中添加带有日期之间的天数的列

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问题：在DataFrame Pandas中添加带有日期之间的天数的列
回答 0
回答 1
回答 2
回答 3

问题：在DataFrame Pandas中添加带有日期之间的天数的列

我想从“ B”中的日期中减去“ A”中的日期，并添加一个具有差异的新列。

df
          A        B
one 2014-01-01  2014-02-28 
two 2014-02-03  2014-03-01

我尝试了以下操作，但是在尝试将其包含在for循环中时遇到错误…

import datetime
date1=df['A'][0]
date2=df['B'][0]
mdate1 = datetime.datetime.strptime(date1, "%Y-%m-%d").date()
rdate1 = datetime.datetime.strptime(date2, "%Y-%m-%d").date()
delta =  (mdate1 - rdate1).days
print delta

我该怎么办？

点击查看英文原文

I want to subtract dates in ‘A’ from dates in ‘B’ and add a new column with the difference.

df
          A        B
one 2014-01-01  2014-02-28 
two 2014-02-03  2014-03-01

I’ve tried the following, but get an error when I try to include this in a for loop…

import datetime
date1=df['A'][0]
date2=df['B'][0]
mdate1 = datetime.datetime.strptime(date1, "%Y-%m-%d").date()
rdate1 = datetime.datetime.strptime(date2, "%Y-%m-%d").date()
delta =  (mdate1 - rdate1).days
print delta

What should I do?

回答 0

假设这些是datetime列（如果不适用to_datetime），则可以减去它们：

df['A'] = pd.to_datetime(df['A'])
df['B'] = pd.to_datetime(df['B'])

In [11]: df.dtypes  # if already datetime64 you don't need to use to_datetime
Out[11]:
A    datetime64[ns]
B    datetime64[ns]
dtype: object

In [12]: df['A'] - df['B']
Out[12]:
one   -58 days
two   -26 days
dtype: timedelta64[ns]

In [13]: df['C'] = df['A'] - df['B']

In [14]: df
Out[14]:
             A          B        C
one 2014-01-01 2014-02-28 -58 days
two 2014-02-03 2014-03-01 -26 days

注意：请确保您使用的是新熊猫（例如0.13.1），这可能在较旧的版本中不起作用。

点击查看英文原文

Assuming these were datetime columns (if they’re not apply to_datetime) you can just subtract them:

df['A'] = pd.to_datetime(df['A'])
df['B'] = pd.to_datetime(df['B'])

In [11]: df.dtypes  # if already datetime64 you don't need to use to_datetime
Out[11]:
A    datetime64[ns]
B    datetime64[ns]
dtype: object

In [12]: df['A'] - df['B']
Out[12]:
one   -58 days
two   -26 days
dtype: timedelta64[ns]

In [13]: df['C'] = df['A'] - df['B']

In [14]: df
Out[14]:
             A          B        C
one 2014-01-01 2014-02-28 -58 days
two 2014-02-03 2014-03-01 -26 days

Note: ensure you’re using a new of pandas (e.g. 0.13.1), this may not work in older versions.

回答 1

要删除“天”文本元素，您还可以使用系列的dt（）访问器：https : //pandas.pydata.org/pandas-docs/stable/generation/pandas.Series.dt.html

所以， 

df[['A','B']] = df[['A','B']].apply(pd.to_datetime) #if conversion required
df['C'] = (df['B'] - df['A']).dt.days

返回：

             A          B   C
one 2014-01-01 2014-02-28  58
two 2014-02-03 2014-03-01  26
点击查看英文原文

To remove the ‘days’ text element, you can also make use of the dt() accessor for series: https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.dt.html

So, 

df[['A','B']] = df[['A','B']].apply(pd.to_datetime) #if conversion required
df['C'] = (df['B'] - df['A']).dt.days

which returns:

             A          B   C
one 2014-01-01 2014-02-28  58
two 2014-02-03 2014-03-01  26

回答 2

列表理解是最Python（最快捷）的方式的最佳选择：

[int(i.days) for i in (df.B - df.A)]
  1. 我将返回timedelta（例如“ -58天”）
  2. i.days将以长整数值（例如-58L）返回此值
  3. int（i.days）将为您提供-58。

如果您的列不是日期时间格式。较短的语法为：df.A = pd.to_datetime(df.A)

点击查看英文原文

A list comprehension is your best bet for the most Pythonic (and fastest) way to do this:

[int(i.days) for i in (df.B - df.A)]
  1. i will return the timedelta(e.g. ‘-58 days’)
  2. i.days will return this value as a long integer value(e.g. -58L)
  3. int(i.days) will give you the -58 you seek.

If your columns aren’t in datetime format. The shorter syntax would be: df.A = pd.to_datetime(df.A)

回答 3

这个怎么样： 

times['days_since'] = max(list(df.index.values))  
times['days_since'] = times['days_since'] - times['months']  
times
点击查看英文原文

How about this: 

times['days_since'] = max(list(df.index.values))  
times['days_since'] = times['days_since'] - times['months']  
times