This is the main part of the assignment that will have you implement many small functions in Python.
We recommend reading over this part and Part 2 before writing any code so that you get into the practice of writing your tests alongside the code you are developing.
For this part you will be using hw1.py
. There is a little bit of starter code there for a function called funky_sum
. There is an example test in hw1_test.py
to test funky_sum
.
For all of the following problems, you should add a function with the expected name to hw1.py
and implement the function as described. For every problem in this homework, you should make no assumptions about the parameters unless otherwise described. For every problem, you may assume we pass parameters of the expected types described for that problem and that those parameters are not None
.
count_divisible_digits
Write a function called count_divisible_digits
that takes two integer numbers n
and m
as an arguments and returns the number of digits in n
that are divisible by m
. If m
is 0, then count_divisible_digits
should return 0. For this problem, any digit in n
that is 0 is divisible by any number. Here are some example calls:
count_divisible_digits(650899, 3) # returns 4 because 0, 6, 9 and 9 are divisible by 3
count_divisible_digits(-204, 5) # returns 1 because 0 is divisible by 5
count_divisible_digits(24, 5) # returns 0
count_divisible_digits(1, 0) # returns 0
To receive full credit on this problem, you should avoid building up a string containing the digits of the number. Instead, you should solve this problem by manipulating the number itself. You can assume m
is not negative.
Hint: You will likely want to use a strategy that involves integer divsiion and the mod operator. A common pattern when working with digits is to use n // 10
and n % 10
to pull a number apart, digit by digit.
is_relatively_prime
Write a function called is_relatively_prime
that takes two integer numbers n
and m
and returns True
if n
and m
are relatively prime to each other, returning False
otherwise. Two numbers are relatively prime if they share no common factors besides 1. 1 is relatively prime with every number. You may assume that n
and m
are at least 1 for this problem. Some example calls are shown below where the comment underneath the call shows the factors for each of the given numbers:
is_relatively_prime(12, 13) # True
# factors of 12 = [1, 2, 3, 4, 6, 12] and factors of 13 = [1, 13]
is_relatively_prime(4, 24) # False, shares 2 and 4 as common factors
# factors of 4 = [1, 2, 4] and factors of 24 = [1, 2, 3, 4, 6, 8, 12, 24]
is_relatively_prime(5, 9) # True
# factors of 5 = [1, 5] and factors of 9 = [1, 3, 9]
is_relatively_prime(8, 9) # True
# factors of 8 = [1, 2, 4, 8] and factors of 9 = [1, 3, 9]
is_relatively_prime(8, 1) # True
# factors of 8 = [1, 2, 4, 8] and factors of 1 = [1]
travel
Write a function travel
which takes a string of North, East, South, West directions, a starting location on a grid x
, and a starting location on a grid y
. Your function should return a string that indicates the new position after following the directions starting from the given x
, y
. The returned string should be in the format '(x_new, y_new)'
.
The directions string will use 'N'
to indicate increasing the y-coordinate, 'E'
to indicate increasing the x-coordinate, 'S'
to indicate decreasing the y-coordinate, and 'W'
to indicate decreasing the x-coordinate. The case of the characters should be ignored. Any characters that are not 'N'
, 'E'
, 'W'
, or 'S'
(ignoring letter-casing) should be ignored.
travel('NW!ewnW', 1, 2) # '(-1, 4)'
# 'N', (1,2) => (1,3)
# 'W', (1,3) => (0,3)
# '!', ignored
# 'e', (0,3) => (1,3)
# 'w', (1,3) => (0,3)
# 'n', (0,3) => (0,4)
# 'W', (0,4) => (-1,4)
swip_swap
Write a function called swip_swap
that takes a string source
and characters c1
and c2
and returns the string source
with all occurrences of c1
and c2
swapped. You may assume that the c1
and c2
are single characters. Here are some example calls:
swip_swap('foobar', 'f', 'o') # 'offbar'
swip_swap('foobar', 'b', 'c') # 'foocar'
swip_swap('foobar', 'z', 'c') # 'foobar'
compress
Write a function compress
which takes a string as an argument and returns a new string such that each character is followed by its count, and any adjacent duplicate characters are removed (replaced by the single character). You may assume the string only contains letters. Here is an example call
compress('cooooooooooooooooolkangaroo') # 'c1o17l1k1a1n1g1a1r1o2'
compress('aaa') # 'a3'
compress('') # ''
longest_line_length
Write a function longest_line_length
that takes a string file_name
and returns the length of the longest line in the file. The length of the line is just the number of characters in it (whitespace or not). If the file is empty, the function should return None
. You may assume the file exists for the given file name.
Suppose we had a file called poem.txt
with the contents:
she sells
sea
shells by
the sea shore
Then the following call would return:
longest_line_length('poem.txt') # 13
In this file, line 2 would have 4 characters because of the new-line character at the end. This is not something you need to think about though since len
will return the proper thing for the strings. Because the last line has 13 characters (there is no new-line), the return of this function is 13.
None
?longest_word
Write a function longest_word
that takes a string file_name
and returns the longest word in the file with which line it appears on. If there are ties for the longest word, it should return the one that appears first in the file. If the file is empty or there are no words in the file, the function should return None
. You may assume that the file_name
describes a file that exists.
Suppose we had a file called poem.txt
with the contents:
she sells
sea
shells by
the sea shore
Then the following call would return:
longest_word('poem.txt') # '3: shells'
Because the longest word is "shells" and appears on line 3.
mode_digit
Write a function mode_digit
that takes an integer number n
and returns the digit that appears most frequently in that number. The given number may be positive or negative, but the most frequent digit returned should always be non-negative. If there is a tie for the most frequent digit, the digit with the greatest value should be returned.
To receive full credit, your solution must not use any nested loops or recursion (if you don't know what that is, you aren't using it), and you must have no more than 4 if/elif/else
branches. You are allowed to use extra storage to solve this problem. Here are some example calls
mode_digit(12121) # 1
mode_digit(0) # 0
mode_digit(-122) # 2
mode_digit(1211232231) # 2