import re 

datePattern = re.compile('\d\d\d\d-(0[1-9]|1[0-2])-(0[1-9]|[12]\d|3[01])$') 

def isValidDate(s): 
    return datePattern.match(s) != None 

Explanation: Since we are using match(…) to compare the compiled pattern in variable datePattern to the string in parameter s given to our function isValidDate(…), we don’t have to worry about additional characters before the start of the date because match(…) will always try to match the pattern to the start of the string. However, we use $ as the last character in our pattern to make sure there are no additional characters following the date. That means the pattern has the form

“…-…-…$”

where the dots have to be replaced with some regular expression notation for the year, month, and day parts. The year part is easy, since we allow for any 4-digit number here. So we can use \d\d\d\d here, or alternatively \d{4,4} (remember that \d stands for the predefined class of all digits).

For the month, we need to distinguish two cases: either it is a 0 followed by one of the digits 1-9 (but not another 0) or a 1 followed by one of the digits 0-2. We therefore write this part as a case distinction (…|…) with the left part 0[1-9] representing the first option and the second part 1[0-2] representing the second option.

For the day, we need to distinguish three cases: (1) a 0 followed by one of the digits 1-9, (2) a 1 or 2 followed by any digit, or (3) a 3 followed by a 0 or a 1. Therefore we use a case-distinction with three options (…|…|…) for this part. The first part 0[1-9] is for option (1), the second part [12]\d for option (2), and the third part 3[01] for the third option.

Part 1:

import operator 

from functools import reduce 

l = [True, False, True] 

r = reduce(operator.and_, l, True) 

print(r)  #  output will be False in this case 

To check whether or not at least one element is True, the call has to be changed to:

r = reduce(operator.or_, l, False) 

Part 2:

import operator 

from functools import reduce 

l = [-4, 2, 1, -6 ] 

r = reduce(operator.and_, map(lambda n: n > 0, l), True) 

print(r)   # will print False in this case 

We use map(…) with a lambda expression for checking whether or not an individual element from the list is >0. Then we apply the reduce(…) version from part 1 to the resulting list of Boolean values we get from map(…) to check whether or not all elements are True.

Part 3:

import operator

l = [True, False, True] 

def myReduce(f, l, i): 
	intermediateResult = i 
	for element in l: 
		intermediateResult = f(intermediateResult, element) 
	return intermediateResult 

r = myReduce(operator.and_, l, True) 
print(r)  #  output will be False in this case

Maybe you were expecting that an implementation of reduce would be more complicated, but it’s actually quite simple. We set up a variable to always contain the intermediate result while working through the elements in the list and initialize it with the initial value provided in the third parameter i. When looping through the elements, we always apply the function given in parameter f to the intermediate result and the element itself and update the intermediate result variable with the result of this operation. At the end, we return the value of this variable as the result of the entire reduce operation.

import pandas as pd 

# create the data frame from a list of tuples 
data = pd.DataFrame( [('Mike',7,10,5.5),
     ('Lisa', 6.5, 9, 8),
     ('George', 4, 3, 7),
     ('Maria', 7, 9.5, 4),
     ('Frank', 5, 5, 5) ] )
     
# set column names
data.columns = ['Name', 'Assignment 1', 'Assignment 2', 'Assignment 3']

# set row names
data.index = range(1,len(data)+1)

# show table 
print(data)
 
# add column with averages
data['Average'] = (data['Assignment 1'] + data['Assignment 2'] + data['Assignment 3']) / 3
 
# part a (all students with a1 score < 7)
print(data[ data['Assignment 1'] < 7])
 
# part b (all students with a1 and a2 score > 6)
print(data[ (data['Assignment 1'] > 6) & (data['Assignment 2'] > 6)])

# part c (at least one assignment < 5) 
print( data[ data[ ['Assignment 1', 'Assignment 2', 'Assignment 3'] ].min(axis = 1) < 5 ] )
 
# part d (name starts with M, only Name and Average columns)
print(data [ data [ 'Name' ].map(lambda x: x.startswith('M')) ] [ ['Name','Average'] ])

# sort by Name 
print(data.sort_values(by = ['Name']))

If any of these steps is unclear to you, please ask for further explanation on the forums.