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Converting string to datetime [hh:mm:ss]

I have two columns Date_x and Date_y. I would like to compare them (i.e Date_x + 1 hour < Date_y)
Format of the strings looks as follows "2020-01-29 11:31:32.754292 UTC"
I have tried converting it using datetime:
from datetime import datetime as dt
df["Date_x"] = [dt.strptime(x, '%Y-%m-%d %H:%M:%S.%f') for x in df['Date_x']]
However, it throws an error regarding the UTC part. I tried removing it with no avail.
Last traceback:
time data '2020-01-29 18:30:28' does not match format '%Y-%m-%d %H:%M:%S.%f'
How would you go about converting the string to hh:mm:ss only?

You could use an if statement:
df["Date_x"] = [dt.strptime(x, '%Y-%m-%d %H:%M:%S.%f') if '.' in x else dt.strptime(x, '%Y-%m-%d %H:%M:%S') for x in df['Date_x']]
But why not just pd.to_datetime:
df["Date_x"] = pd.to_datetime(df["Date_x"], infer_datetime_format=True)

format 01-01-16 7:43 string to datetime

I have the following strings that I'd like to convert to datetime objects:
'01-01-16 7:43'
'01-01-16 3:24'
However, when I try to use strptime it always results in a does not match format error.
Pandas to_datetime function nicely handles the automatic conversion, but I'd like to solve it with the datetime library as well.
format_ = '%m-%d-%Y %H:%M'
my_date = datetime.strptime("01-01-16 4:51", format_)
ValueError: time data '01-01-16 4:51' does not match format '%m-%d-%Y %H:%M'

as i see your date time string '01-01-16 7:43'
its a 2-digit year not 4-digit year
that in order to parse through a 2-digit year, e.g. '16' rather than '2016', a %y is required instead of a %Y.
you can do that like this
from datetime import datetime
datetime_str = '01-01-16 7:43'
datetime_object = datetime.strptime(datetime_str, '%m-%d-%y %H:%M')
print(type(datetime_object))
print(datetime_object)
give you output 2016-01-01 07:43:00

First of all, if you want to match 2016 you should write %Y while for 16 you should write %y.
That means you should write:
format_ = '%m-%d-%y %H:%M'
Check this link for all format codes.

How to extract date from the string format timestamp?

I want to extract the date (eg.2018-07-16) from strings (eg. 2018-07-16 10:17:53.460035).
The strings have two formats: "2018-07-16 10:17:53.460035" and "2018-05-20 14:37:21".
When I use strptime(d, "%Y-%m-%d %H:%M:%S.%f") to convert the strings before extracting the date, it pops this error:
ValueError: time data '2018-05-20 14:37:21' does not match format
%Y-%m-%d %H:%M:%S.%f'
How can I convert both time formats to DateTime type and extract date from it?

Use to_datetime from pandas.
import pandas as pd
a = "2018-07-16 10:17:53.460035"
b = "2018-05-20 14:37:21"
print(pd.to_datetime(a).date())
print(pd.to_datetime(b).date())

You don't need the .%f at the end for the first format, that is what is causing the format error.
t = "2018-05-20 14:37:21"
strptime(t, "%Y-%m-%d %H:%M:%S")
You need to create a second format for the other time string:
t = "2018-07-16 10:17:53.460035"
strptime(t, "%Y-%m-%d %H:%M:%S.%f")
Edit: Here is another example which excepts both
time_stamps = ["2018-05-20 14:37:21", "2018-07-16 10:17:53.460035"]
for stamp in time_stamps:
fmt = "%Y-%m-%d %H:%M:%S"
try:
time = datetime.datetime.strptime(stamp, fmt+".%f")
except ValueError:
time = datetime.datetime.strptime(stamp, fmt)
print(time)

Python: convert datetime format to another

So i have this kind f format:
'%Y-%m-%d %H:%M:%S.%f'
date_time = datetime.now()
Ouput
'2019-04-15 07:52:14.211697'
And i want to change is into this format: '%Y-%m-%d %H:%M:%S,%f'
This is what i have try:
time = datetime.strptime(str(date_time), '%Y-%m-%d %H:%M:%S,%f')
And this is the error:
ValueError: time data '2019-04-15 07:52:14.211697' does not match
format '%Y-%m-%d %H:%M:%S,%f'
Edit
So i have this string:
maches = regex.findall(
'[0-9]{4}-(?:0[1-9]|1[0-2])-(?:0[1-9]|[1-2][0-9]|3[0-1]) (?:2[0-3]|[01][0-9]):[0-5][0-9]:[0-5][0-9],[0-9][0-9][0-9]',
line)
match[0] = '2019-03-13 17:35:35,855'
And i want to convert it to Datetime:
time = datetime.strptime(maches[0], '%Y-%m-%d %H:%M:%S,%f')
And this returned another format:
2019-03-13 17:35:35.855000
Why ?

You want to use strftime:
str_time = datetime.strftime(date_time, '%Y-%m-%d %H:%M:%S,%f')
A quick comparison of strftime and strptime:
strftime() is used to convert a datetime object to a string
strptime() is used to convert a date string to a datetime object
Also here's a good resource in the documentation describing the differences: strftime() and strptime() Behavior
Response to Question Edit
In response to your edit strptime takes a string date and converts it into a datetime object. The format you pass to it just tells it how to parse the string to datetime. It does not dictate it's print format. You are calling print on a datetime object in this case so the output is dictated by the __str__ method on the datetime object.
If you want to print it in a certain way pass it to strftime with the format you want to output it as:
obj_time = datetime.strptime(maches[0], '%Y-%m-%d %H:%M:%S,%f')
print(datetime.strftime(obj_time , '%Y-%m-%d %H:%M:%S,%f'))

Use strftime.
date_time = datetime.now()
print(date_time)
print(datetime.strftime(date_time, '%Y-%m-%d %H:%M:%S,%f'))
2019-04-15 10:26:08.637630
2019-04-15 10:26:08,637630

Use this you get this format '%Y-%m-%d %H:%M:%S,%f'
print '{:%Y-%m-%d %H:%M:%S.%f}'.format(date_time)

replace hours in a datetime type python

I have a datetime type mydate in %Y-%m-%dT%H:%M:%S format.
I want to replace the hours
I did this using mydate.replace() method
Now I want to comapre it with another specific date -> myNEWdate whose format is %Y-%m-%d %H:%M:%S :
newdate = mydate.replace(hour = islot)
print newdate
appointmentDict[mydate]['time_start'] = datetime.strptime(str(newdate),"%Y/%m/%d %H:%M:%S")
The date is printed as 2015-06-26 08:00:00
and I get the error
ValueError: time data '2015-06-26 08:00:00' does not match format '%Y/%m/%d %H:%M:%S'
What should I do to resolve this

You need to set the correct format
datetime.strptime(str(newdate),"%Y-%m-%d %H:%M:%S")
To solve the exception. Although converting from datetime 2 string and backwards doesn't make much sense, as mentioned in the comments.