How to covnert int64 data into ?day, month and year?
Question:
I have a date feature in the format 20001130 and another 2000-11-30 without any space. How can i write the optimized code that works for both to split the date into day month and year efficiently
Answers:
You can use pandas.to_datetime:
import pandas as pd
pd.to_datetime([20001130, 20001129], format='%Y%m%d')
or with a dataframe.
df = pd.DataFrame({'time': [20001129, 20001130]})
df.time = pd.to_datetime(df.time, format='%Y%m%d')
EDIT
The two date formats should be in one column. In this case, convert all to strings and let pandas.to_datetime interpret the values, as it supports different formats in one column.
df = pd.DataFrame({'time': [20001129, '2000-11-30']})
df.time = pd.to_datetime(df.time.astype(str))
time
0
2000-11-29
1
2000-11-30
I have a date feature in the format 20001130 and another 2000-11-30 without any space. How can i write the optimized code that works for both to split the date into day month and year efficiently
You can use pandas.to_datetime:
import pandas as pd
pd.to_datetime([20001130, 20001129], format='%Y%m%d')
or with a dataframe.
df = pd.DataFrame({'time': [20001129, 20001130]})
df.time = pd.to_datetime(df.time, format='%Y%m%d')
EDIT
The two date formats should be in one column. In this case, convert all to strings and let pandas.to_datetime interpret the values, as it supports different formats in one column.
df = pd.DataFrame({'time': [20001129, '2000-11-30']})
df.time = pd.to_datetime(df.time.astype(str))
| time | |
|---|---|
| 0 | 2000-11-29 |
| 1 | 2000-11-30 |