我试图找出一系列日期和日期之间的差异。例如,该系列是 从 5 月 1 日到 6 月 1 日这是
date = pd.DataFrame()
In [0]: date['test'] = pd.date_range("2021-05-01", "2021-06-01", freq = "D")
Out[0]: date
test
0 2021-05-01 00:00:00
1 2021-05-02 00:00:00
2 2021-05-03 00:00:00
3 2021-05-04 00:00:00
4 2021-05-05 00:00:00
5 2021-05-06 00:00:00
6 2021-05-07 00:00:00
7 2021-05-08 00:00:00
8 2021-05-09 00:00:00
9 2021-05-10 00:00:00
In[1]
date['test'] = date['test'].dt.date
Out[1]:
test
0 2021-05-01
1 2021-05-02
2 2021-05-03
3 2021-05-04
4 2021-05-05
5 2021-05-06
6 2021-05-07
7 2021-05-08
8 2021-05-09
9 2021-05-10
In[2]:date['base'] = dt.strptime("2021-05-01",'%Y-%m-%d')
Out[2]:
0 2021-05-01 00:00:00
1 2021-05-01 00:00:00
2 2021-05-01 00:00:00
3 2021-05-01 00:00:00
4 2021-05-01 00:00:00
5 2021-05-01 00:00:00
6 2021-05-01 00:00:00
7 2021-05-01 00:00:00
8 2021-05-01 00:00:00
9 2021-05-01 00:00:00
In[3]:date['base'] = date['base'].dt.date
Out[3]:
base
0 2021-05-01
1 2021-05-01
2 2021-05-01
3 2021-05-01
4 2021-05-01
5 2021-05-01
6 2021-05-01
7 2021-05-01
8 2021-05-01
9 2021-05-01
In[4]:date['test']-date['base']
Out[4]:
diff
0 0 days 00:00:00.000000000
1 1 days 00:00:00.000000000
2 2 days 00:00:00.000000000
3 3 days 00:00:00.000000000
4 4 days 00:00:00.000000000
5 5 days 00:00:00.000000000
6 6 days 00:00:00.000000000
7 7 days 00:00:00.000000000
8 8 days 00:00:00.000000000
9 9 days 00:00:00.000000000
10 10 days 00:00:00.000000000
我唯一能得到的就是这个。我不想要数字 1-10 以外的任何东西,因为我需要它们进行进一步的数值计算,但我无法摆脱它们。另外,我如何构建一个只输出日期而不是后面的 hms 的时间序列?我不想为所有这些手动 .dt.date 并且它有时会搞砸
答案 0 :(得分:1)
您无需为此创建列 base
,只需执行以下操作:
>>> (date['test'] - pd.to_datetime("2021-05-01", format='%Y-%m-%d')).dt.days
0 0
1 1
2 2
3 3
4 4
...
27 27
28 28
29 29
30 30
31 31
Name: test, dtype: int64
答案 1 :(得分:0)
您可以先将时间戳转换为 epoch seconds(它们实际上在内部存储为某个数字,并且可能是纪元秒的一个因素)
使用 pandas datetime to unix timestamp seconds
import pandas as pd
# start df with date column
df = pd.DataFrame({"date": pd.date_range("2021-05-01", "2021-06-01", freq = "D")})
# create a column for datetimes
df["ts"] = (df["date"] - pd.Timestamp("1970-01-01")) // pd.Timedelta("1s")
>>> df
date ts
0 2021-05-01 1619827200
1 2021-05-02 1619913600
2 2021-05-03 1620000000
3 2021-05-04 1620086400
...
31 2021-06-01 1622505600
这将允许您在转换回来之前进行整数数学运算
>>> df["days"] = (df["ts"] - min(df["ts"])) // (60*60*24) # 1 day in seconds
>>> df
date ts days
0 2021-05-01 1619827200 0
1 2021-05-02 1619913600 1
2 2021-05-03 1620000000 2
3 2021-05-04 1620086400 3
...
31 2021-06-01 1622505600 31
答案 2 :(得分:0)
或者,对于简单的基于日的系列,您可以使用索引作为日偏移量(因为 DataFrame 就是这样生成的)!
>>> import pandas as pd
>>> df = pd.DataFrame({"date": pd.date_range("2021-05-01", "2021-06-01", freq = "D")})
>>> df["days"] = df.index
>>> df
date days
0 2021-05-01 0
1 2021-05-02 1
2 2021-05-03 2
3 2021-05-04 3
...
31 2021-06-01 31