CompSci 101 Fall 2017 Test 1

Problem 1 Sec 01

<type 'str'> f             
<type 'int'> 4
<type 'str'> ng
<type 'str'> data
<type 'int'> 4
<type 'str'> nts 5 c       
<type 'str'> end
<type 'str'> adogcat
<type 'str'> b-u-s
<type 'bool'> False        
<type 'list'> ['for', 'list']   
<type 'float'> 7.2
<type 'int'> 3

Problem 2 Sec 01
Part A

def calculate(price, days, percent):
    daysPercent = (days * percent)/100.00
    calcPrice = price - (price * daysPercent)
    if calcPrice < 0.0:
        return 0.0
    return calcPrice

ALT SOLN
def calculate(price, days, percent):
    calcPrice = price - (price * days * percent)/100.0
    if calcPrice < 0.0:
        return 0.0
    return calcPrice

ALT SOLN
def calculate(price, days, percent):
    calcPrice = price * (1 -  (days * percent/100.0))
    if calcPrice < 0.0:
        return 0.0
    return calcPrice

Part B  (8 pts)

def yardage(stfield, styard, efield, eyard):
    if stfield == efield: # same
        if styard > eyard:
            return styard - eyard
        else:
            return eyard - styard
    else: 
        return 50 -styard + 50 - eyard

Problem 3 Sec 01
PART A - Mystery (6 pts total)

Q1. (2 pts)

['strike', 'street', 'fish', 'tigers', 'eels']

Q2. (1 pt)

['fish']

Q3. (1 pt)

['fish', 'tigers']

Q4. (2 pts)

This function creates a new list of the words by processing each word in
the list and then deciding where to add it to the new list. For a word, it
gets the position of the two letters let1 and let2 in that word. If let2 is
in a higher numbered position than let1, then the word is appended to the
right side of the new list. Otherwise the word is appended to the
left side of the new list.

PART B - Debugging (6 pts total)

Q1. 2 pts

After the for loop, if exactly one vowel was found, the code searches for
the position of the "last" character in the word and returns that
position. That is not necessarily the position of the vowel. 

Q2. 1 pt
Any word with one vowel where the vowel is not the last letter, such as:
bark, stork, up

Q3. 1 pt
Vowel needs to be the last letter. the, to, pro, be

Q4. 2 pts
Whenever a vowel is counted, you also need to save the vowel in a
variable. Then after the loop, if there is just one vowel, you can use that
variable with find to find the location of that one vowel and return that
location. 

def onlyOneVowel(word):
    count = 0
    vowel = ""            #create a variable to save the vowel
    for ch in word:
        if isVowel(ch):
            count += 1
            vowel = ch    # save the vowel
    if count == 1:
        return word.find(vowel)   # look for the vowel. 
    else:
        return -1

ALT SOLN:
def onlyOneVowel(word):
    count = 0
    indexSave = -1         # save the index
    for index in range(len(word)):   #loop over indices
        ch = word[index]            # extract the character to look at
        if isVowel(ch):
            count += 1
            indexSave = index       # save the index
    if count == 1:
        return indexSave            # return the index, don't need to find it
    else:
        return -1

Problem 4 Sec 01

def reformat (name, num):
    alist = name.split()
    last = " ".join(alist[-1*num:])
    first = alist[0]
    middle = ""
    if len(alist) > num+1:  # yes middle names
        middle = alist[1:-1*num]
    lets = ""
    for word in middle:
        lets += word[0]
    if len(lets) > 0:
        return   last + ", " + first + " " + lets
    else: 
        return last + ", " + first

Problem 5 Sec 01, Problem A (6 pts)  

def getParts(line):
    pos = line.find(" (")
    name = line[:pos]
    pos2 = line.find(")")
    sizestr = line[pos+2:pos2]
    size = float(sizestr)
    rest = line[pos2+1:].split()
    cals = int(rest[0])
    type = " ".join(rest[1:])
    return [name, size, cals, type]

Problem 5 Sec 01, Problem B  (4 pts) 

def fileToList(filename):
    answer = []
    f = open(filename)

    # code below is answer
    for line in f:
        line = line.strip()
        answer.append(getParts(line))

    # code below was given
    return answer
 
Problem 5 Sec 01, Problem C  (8 pts) 

def snacksOfType(snklist, type, amount):
    ans = []
    for alist in snklist:
        if alist[-1] == type:
            if alist[2] <= amount:
                ans.append(alist[0])

    return ans

Problem 5 Sec 01, Problem D (10 pts) 

def mostCalories(snklist, type):
    maxitem = ""
    maxcal = 0
    for alist in snklist:
        if alist[3] == type:
            if alist[2] > maxcal:
                maxcal = alist[2]
                maxitem = alist[0]
    return maxitem

ALT SOLN

def mostCalories(snklist, type):
    typelist = []
    for alist in snklist:
        if alist[3] == type:
            typelist.append(alist)                  
    maxitem = ""
    maxcal = 0
    for alist in typelist:
        if alist[2] > maxcal:
             maxcal = alist[2]
             maxitem = alist[0]
    return maxitem

ALT SOLN

def mostCalories(snklist, type):
    typelist = []
    for alist in snklist:
        if alist[3] == type:
            typelist.append(alist)                  
    maxcallist = []
    for alist in typelist:
         maxcallist.append(alist[2])
    maxcal = max(maxcallist)
    maxitem = ""
    for alist in typelist:
        if alist[2] == maxcal:   # found it
             maxitem = alist[0]
    return maxitem