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please help fix the following Scenario: Your team is working with data where dates are caputred as a string, "9/15/2019". You are tasked to create a python function named 'mdy_date' that will accept a string input, convert it to a python datetime object, and return the datetime object. Create the function in the cell below.

please help fix the following Scenario: Your team is working with data where dates are caputred as a string, "9/15/2019". You are tasked to create a python function named 'mdy_date' that will accept a string input, convert it to a python datetime object, and return the datetime object. Create the function in the cell below.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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please help fix the following

Scenario: Your team is working with data where dates are caputred as a string, "9/15/2019". You are tasked to create a python function named 'mdy_date' that will accept a string input, convert it to a python datetime object, and return the datetime object. Create the function in the cell below. 

 

from datatime import datatime def mdy_date(data_string): date_object = datatime.strptime (data_string, '%m/%d/%Y') return(date_object) File "<ipython-input-42-b51ad327b9a1>", line 3 date_object = datatime.strptime (data_string, '%m/%d/%Y') IndentationError: expected an indented block mdy_date("9/15/2019") # Do not change anything in this cell. # Running this cell should return the exact output: # datetime.datetime (2019, 9, 15, 0, 0)
Transcribed Image Text:from datatime import datatime def mdy_date(data_string): date_object = datatime.strptime (data_string, '%m/%d/%Y') return(date_object) File "<ipython-input-42-b51ad327b9a1>", line 3 date_object = datatime.strptime (data_string, '%m/%d/%Y') IndentationError: expected an indented block mdy_date("9/15/2019") # Do not change anything in this cell. # Running this cell should return the exact output: # datetime.datetime (2019, 9, 15, 0, 0)
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