Python - 取列中第一个日期的时间差

Question

给定日期列，我想创建另一个列差异来计算距第一个日期的天数。

date                    diff
2011-01-01 00:00:10      0
2011-01-01 00:00:11      0.000011 days
2011-02-01 00:00:11      30.000011 days 
2013-02-01 00:00:11      395.000011 days
2014-02-01 00:00:11      760.000011 days

日期为日期时间。到目前为止我尝试了什么：

df = df.sort_values(['date'], ascending=True)
df.set_index('date', inplace = True)
first = df.index[0]
df['diff'] = (first - df.index.shift()).fillna(0)

Answer 1

您可以使用这种方法而无需设置新索引

原始数据帧

df
                 date        diff
0 2011-01-01 00:00:10    0.000000
1 2011-01-01 00:00:11    0.000011
2 2011-02-01 00:00:11   30.000011
3 2013-02-01 00:00:11  395.000011
4 2014-02-01 00:00:11  760.000011

可能的答案

df['diff_new'] = df['date'] - df.loc[0,'date']

                 date        diff           diff_new
0 2011-01-01 00:00:10    0.000000    0 days 00:00:00
1 2011-01-01 00:00:11    0.000011    0 days 00:00:01
2 2011-02-01 00:00:11   30.000011   31 days 00:00:01
3 2013-02-01 00:00:11  395.000011  762 days 00:00:01
4 2014-02-01 00:00:11  760.000011 1127 days 00:00:01

顺便说一句，我得到了您在第三行的原始数据中显示的不同日期差异。您可以手动与this online tool to calculate date differences in days进行比较。

Answer 2

这是我将天数作为浮点数值的方法：

dates = pd.to_datetime(df.date) # make sure we are working with dates and not strings
df["diff"] = (dates - dates[0]).apply(lambda x: x.total_seconds() / 86400))

生成的df：

                  date         diff
0  2011-01-01 00:00:10     0.000000
1  2011-01-01 00:00:11     0.000012
2  2011-02-01 00:00:11    31.000012
3  2013-02-01 00:00:11   762.000012
4  2014-02-01 00:00:11  1127.000012

Answer 3

这就是你尝试的......

>>> df
                  date
0  2011-01-01 00:00:10
1  2011-01-01 00:00:11
2  2011-02-01 00:00:11
3  2013-02-01 00:00:11
4  2014-02-01 00:00:11

首先将它们转换为时间戳，这样数据才能正确地框起来，一旦转换，只需区分DataFrame：

>>> df2 = df.apply(lambda x: [pd.Timestamp(ts) for ts in x])
>>> df['diff']  = (df2 - df2.shift()).fillna(0)
>>> df
                  date              diff
0  2011-01-01 00:00:10   0 days 00:00:00
1  2011-01-01 00:00:11   0 days 00:00:01
2  2011-02-01 00:00:11  31 days 00:00:00
3  2013-02-01 00:00:11 731 days 00:00:00
4  2014-02-01 00:00:11 365 days 00:00:00

Answer 4

你可以试试

df['diff'] = df.date - df.date.min()

df
                 date               diff
0 2011-01-01 00:00:10    0 days 00:00:00
1 2011-01-01 00:00:11    0 days 00:00:01
2 2011-02-01 00:00:11   31 days 00:00:01
3 2013-02-01 00:00:11  762 days 00:00:01
4 2014-02-01 00:00:11 1127 days 00:00:01