FAQ: Functional Programming - Review of filter(), map(), and reduce()

This community-built FAQ covers the “Review of filter(), map(), and reduce()” exercise from the lesson “Functional Programming”.

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This exercise can be found in the following Codecademy content:

[Beta] Learn Advanced Python 3

FAQs on the exercise Review of filter(), map(), and reduce()

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Hello!

I have found some mistakes in the lesson code snippets.

nums = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
 
# filter_values is not a higher-order function
def filter_values(predicate, lst):
 
  # Mutable list required because this example is imperative, not declarative
  ret = []
  for i in lst:
    if predicate(i):
      ret.append(i)
  return ret
 
filtered_numbers = filter_values(lambda x: x % 2 == 0, numbers) 
 
print(k) 
 
# This will output the tuple: (2, 4, 6, 8, 10)

“nums” became “numbers” in the end;
“filetered_numbers” became “k” in the end;
“This will output the tuple” will not output the tuple, according to “return ret” statement (ret is a list).

nums = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
 
filtered_numbers = filter(lambda x: x % 2 == 0, numbers) 
 
print(tuple(filtered_numbers))
 
# This will output the tuple: (2, 4, 6, 8, 10)

“nums” became "numbers.

nums = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
 
def mapper(function, lst):
  ret = []
  for i in lst:
    ret.append(function(i))
  return ret
 
mapped_numbers  = mapper(lambda x: x*x, numbers)
 
print(tuple(k))
 
# This will output: (1, 4, 9, 16, 25, 36, 49, 64, 81, 100)

“nums” became “numbers” in the end;
“mapped_numbers” became “k” in the end.

nums = (2, 6, 7, 9, 1, 4, 8)
 
sum = 0
 
for i in t:
  sum += i
 
print(sum) # Output: 37

“nums” became “t”

from functools import reduce
 
nums = (2, 6, 7, 9, 1, 4, 8)
 
k = reduce(lambda x, y: x + y, t) # k is a number
 
print(k) # Output: 37

“nums” became “t”.

Hope this information will help somehow.

1 Like