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Python: How can I remove multiple occurrences of "\\" from a string in Python? [closed]

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string = "C:\\folder\\important\\week1.xlsx"
I need to extract the file name alone, "week1.xlsx" from this string.
But for some reason, it doesn't work.

You can use basename:
import os
str(os.path.basename("C:\\folder\\important\\week1.xlsx"))
=> 'week1.xlsx'

Try:
filename = string[string.rfind('\\')+1:]
Here's more info on rfind().

How do I perform the below datetime operation? [closed]

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I have a date with format in 'yyyy-dd-mm'. How do I convert to 'yyyy-mm-dd' format, been bit of a struggle with strftime and strptime!
Assuming the date is '1919-01-12', I want to see it in the format '1919-12-01'.

from datetime import datetime
date = '1919-01-12'
formatted_date = datetime.strptime(date, '%Y-%d-%m').strftime('%Y-%m-%d')
datetime.strptime(date, '%Y-%d-%m') creates a datetime object from a string, .strftime('%Y-%m-%d') formats the datetime as a string accoring to the specified pattern

try this code
import datetime
print (datetime.datetime.strptime("1919-01-12", "%Y-%d-%m").strftime("%Y-%m-%d"))

Why do we have to use two times datetime function for and object in python [closed]

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import datetime
now = datetime.datetime.now()
print("Current date and time:")
print(now.strftime("%T-%m-%d %H:%M:%S"))
the now object created is why we have to use two times the datetime function

You are not using "two times the datetime function". In datetime.datetime.now(), the function is now. The first datetime is the package, and the second datetime is sub-package. The objects are grouped hierarchically for sake of better organization, for example alongside datetime.datetime there is also datetime.date (compare datetime.datetime.now() with datetime.date.today()).
If you do not want to use both levels each time, then instead of:
import datetime
you could do:
from datetime import datetime

Extracting time data from odd Dd HH:MM:SS.mm format [closed]

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I am trying to import experimental time data which is not really a time, date, or datetime according to the python formalism. It is a time-elapsed format as Dd HH:MM:SS.m where the first time point would be 0d 00:00:00.00 and a point 26 hours later would be 1d 02:00:00.00.
Is there a way to use the datetime module to extract the appropriate time information without hard-coding a string search for the day and adding a multiple of 24 to the hour counter?

To convert your 'odd' format to datetime.timedelta object try:
from datetime import timedelta
import re
input = '1d 02:00:00.00'
def to_timedelta(input):
reg = re.search('([0-9])+d ([0-9]){2}:([0-9]){2}:([0-9]){2}\.([0-9]){2}', input)
ints = tuple(int(t) for t in reg.groups())
return timedelta(days=ints[0],
hours=ints[1],
minutes=ints[2],
seconds=ints[3],
milliseconds=ints[4])
td = to_timedelta(input)
print(td)

how to convert timecode into seconds:fractions_of_a_second? [closed]

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Given timecode in this format
00:00:00,0
What is the best way to convert this into seconds.fractions_of_a_second?

Here's a link for some python code to handle SMPTE timecodes http://code.google.com/p/pytimecode/
Hope it helps...

The datetime module is your friend in this case
import datetime
time_string = "17:48:12,98"
t = datetime.datetime.strptime(time_string, "%H:%M:%S,%f")
seconds = 60 * t.minute * t.hour
print (seconds, t.microsecond)