# Section #5 Solutions

May 10th, 2020

Written by Brahm Capoor, Juliette Woodrow, Peter Maldonado, Kara Eng, Tori Qiu and Parth Sarin

## Lists Review: Compatibility Calculations

        
def in_common(l1, l2):
num_in_common = 0
total_elem = len(l1) + len(l2)
for i in range(len(l1)):
if l1[i] in l2:
num_in_common += 1

return num_in_common / total_elem

def calc_score(netflix_history1, netflix_history2, fav_books1, fav_books2):
netflix_score = in_common(netflix_history1, netflix_history2)
book_score = in_common(fav_books1, fav_books2)
return netflix_score + book_score

def new_friend(name_list, compatability_scores):
best_score = -1
best_friend = None

for i in range(len(compatability_scores)):
compatability_score = compatability_scores[i]
if compatability_score > best_score: # we could make this >= as well
best_score = compatability_score
best_friend = name_list[i]




## Nested Lists & Grids

### Enumerating a list

        
def enumerate(lst):
enum_lst = []
for i in range(len(lst)):
enum_lst.append([i, lst[i]])
return enum_lst



### Matrix Math

        
def matrix_constant_multiply(c, m):
prod_matrix = []

for row in m:
new_row = []
prod_matrix.append(new_row)
for elem in row:
new_row.append(elem * c)

return prod_matrix

sum_matrix = []

for row_idx in range(len(m1)):
new_sum_row = []
sum_matrix.append(new_sum_row)

row = m1[row]
for col_idx in range(len(row)):
sum_elem = m1[row_idx][col_idx] + m2[row_idx][col_idx]
new_sum_row.append(sum_elem)

return sum_matrix



### Times Table

        
def make_times_table(m, n):
times_table = []

for i in range(m):
row_num = i + 1
new_row = []
times_table.append(new_row)
for j in range(n):
col_num = j + 1
new_row.append(row_num * col_num)

return times_table



## Strings

### String Slicing

1. s[1:6]
2. s[:2] or s[0:2]
3. s[6:9]
4. s[6:] or s[6:10]
5. s or s[6:7]
6. s[:] or s[0:10] (or just s)

### String Construction

        
def only_one_first_char(s):

if s == "":
return ""

first_char = s
output = first_char
for i in range(1, len(s)):
if(s[i] != first_char):
output += s[i]

return output

def make_gerund(s):

if len(s) >= 3 and s[len(s)-3:] == 'ing':
s = s[0:len(s)-3] + 'ly'
else:
s = s + 'ing'

return s

def put_in_middle(outer, inner):
middle = len(outer) // 2
return outer[0:middle] + inner + outer[middle:]


Word Puzzles

def is_tridrome(word):
"""
Returns whether or not word is a tridrome, i.e., the first three letters
are the same as the last three letters.

Arguments:
word -- The word to check

>>> is_tridrome('ENTERTAINMENT')
True
>>> is_tridrome('UNDERGROUND')
True
>>> is_tridrome('DEFENESTRATION')
False
>>> is_tridrome('PYTHON')
False
>>> is_tridrome('')
False
"""

"""
We need to check that
1. the word is at least six letters, and
2. the first three letters of the word are the same as the last three
letters.

To check the first condition, we check if
len(word) >= 6

For the second condition, we can extract the first three letters of the
word with word[:3] and the last three with word[-3:]. If negative indexing
isn't comfortable, the last three letters of the word can also be sliced
with word[len(word)-3:]!

In Python, we can combine these two conditions using the and keyword,
which short-circuits (i.e., if the first condition is false, it doesn't
check the second condition).

This leads to the rather nice one-line solution:
"""
return len(word) >= 6 and word[:3] == word[-3:]

ALPHABET = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'

def is_peaceful(word):
"""
Returns whether a word is peaceful, i.e., whether its letters appear in
sorted order.

Arguments:
word -- The word to check
>>> is_peaceful('ABORT')
True
>>> is_peaceful('FIRST')
True
>>> is_peaceful('')
True
>>> is_peaceful('PYTHON')
False
>>> is_peaceful('CHOCOLATE')
False
"""

"""
Python provides a really nice sorted function that can sort collections.
It returns a list, and you can turn a list into a string by joining together
its elements with a string:
''.join(lst)

We can, therefore, get a nice one-line solution with:
"""
# return word == ''.join(sorted(word))

"""
Or, you might prefer the iterative solution, using string comparisons to
check if a character is >= the next character (Python lets you compare
strings alphabetically!):
"""
# for i in range(len(word) - 1): # don't check the last letter
#     if word[i] >= word[i+1]:
#         return False
# return True

"""
And finally, we can use a creative application of the .find function to
obtain a character's position in the alphabet:
"""
for i in range(len(word) - 1): # don't check the last letter
"""
We search for the character in ALPHABET.
"""
curr_letter_index = ALPHABET.find(word[i])
next_letter_index = ALPHABET.find(word[i+1])
if curr_letter_index >= next_letter_index:
return False
return True

def is_stacatto(word):
"""
Returns whether a word is a stacatto word, i.e., whether the letters in
even positions are vowels.

Arguments:
word -- The word to check

>>> is_stacatto('AUTOMATIC')
True
>>> is_stacatto('POPULATE')
True
>>> is_stacatto('')
True
>>> is_stacatto('PYTHON')
False
>>> is_stacatto('SPAGHETTI')
False
"""
VOWELS = 'AEIOUY'

for i in range(len(word)):
if i % 2 == 1:
even_letter = word[i]
if not even_letter in VOWELS:
return False # we've found an even letter that isn't a vowel,
# so we can return immediately.

return True

def count_tridromes(filename):
"""
Return the number of tridromes in the file
"""
count = 0
with open(filename, 'r') as f:
for line in f:
word = line.strip().upper()
if is_tridrome(word):
count += 1
return count

def count_peaceful(filename):
"""
Return the number of peaceful words in the file
"""
count = 0
with open(filename, 'r') as f:
for line in f:
word = line.strip().upper()
if is_peaceful(word):
count += 1
return count

def count_stacatto(filename):
"""
Return the number of stacatto words in the file
"""
count = 0
with open(filename, 'r') as f:
for line in f:
word = line.strip().upper()
if is_stacatto(word):
count += 1
return count