I have timestamps given in the following format in my pandas DataFrame df: 2015-03-09 11:09:05.0.
How can I transform them into this format 2015-03-09T11:09:05.0 (i.e. separated by T)?
df["dt"] = df["dt"].apply(lambda x: ???)
-
I dont see difference...jezrael– jezrael2017-10-16 12:07:30 +00:00Commented Oct 16, 2017 at 12:07
-
@jezrael: Sorry, I updated.Dinosaurius– Dinosaurius2017-10-16 12:08:09 +00:00Commented Oct 16, 2017 at 12:08
Add a comment
|
2 Answers
You were almost there. You are looking for the the isoformat. https://docs.python.org/3.6/library/datetime.html#datetime.date.isoformat
import pandas as pd
df = pd.DataFrame({'dt':pd.to_datetime(['2015-03-09 11:09:05.0'])})
df["dt"] = df["dt"].apply(lambda x: x.isoformat())
df
Returns
dt
0 2015-03-09T11:09:05
You can change the T (default) by inserting a parameter to isoformat(), e.g. df["dt"] = df["dt"].apply(lambda x: x.isoformat(" "))
1 Comment
jezrael
Really nice, I dont know it ;)
Use strftime with custom format:
df = pd.DataFrame({'dt':pd.to_datetime(['2015-03-09 11:09:05.0'])})
print (df)
df["dt"] = df["dt"].dt.strftime('%Y-%m-%dT%H:%M:%S.%f')
print (df)
dt
0 2015-03-09T11:09:05.000000
Or convert to string, split by whitespace and join by T:
df["dt"] = df["dt"].astype(str).str.split().str.join('T')