Exercises¶
Answer the questions or complete the tasks outlined in bold below, use the specific method described if applicable.
** What is 7 to the power of 4?**
7**4
** Split this string:**
s = "Hi there Sam!"
**into a list. **
s = "Hi there Sam!"
s.split()
** Given the variables:**
planet = "Earth"
diameter = 12742
** Use .format() to print the following string: **
The diameter of Earth is 12742 kilometers.
planet = "Earth"
diameter = 12742
print(f"The diameter of {planet} is {diameter} kilometers.")
** Given this nested list, use indexing to grab the word “hello” **
lst = [1,2,[3,4],[5,[100,200,['hello']],23,11],1,7]
lst[3][1][2][0]
** Given this nested dictionary grab the word “hello”. Be prepared, this will be annoying/tricky **
d = {'k1':[1,2,3,{'tricky':['oh','man','inception',{'target':[1,2,3,'hello']}]}]}
d['k1'][3]['tricky'][3]['target'][3]
** What is the main difference between a tuple and a list? **
# Tuple is immutable
** Create a function that grabs the email website domain from a string in the form: **
user@domain.com
So for example, passing “user@domain.com” would return: domain.com
def domainGet(email):
return email.split('@')[1]
domainGet('user@domain.com')
** Create a basic function that returns True if the word ‘dog’ is contained in the input string. Don’t worry about edge cases like a punctuation being attached to the word dog, but do account for capitalization. **
On first pass forgot to include lower
def findDog(st):
return 'dog' in st.lower().split()
findDog('Is there a dog here?')
** Create a function that counts the number of times the word “dog” occurs in a string. Again ignore edge cases. **
def countDog(string):
i = 0
y = string.split()
dogs = list(filter(lambda a:a=='dog', y))
return len(dogs)
ANSWER FROM INSTRUCTOR
def countDog(st): count = 0 for word in st.lower().split(): if word == ‘dog’: count += 1 return count
countDog('This dog runs faster than the other dog dude!')
** Use lambda expressions and the filter() function to filter out words from a list that don’t start with the letter ‘s’. For example:**
seq = ['soup','dog','salad','cat','great']
should be filtered down to:
['soup','salad']
seq = ['soup','dog','salad','cat','great']
list(filter(lambda x: x[0]=='s', seq))
Final Problem¶
**You are driving a little too fast, and a police officer stops you. Write a function to return one of 3 possible results: “No ticket”, “Small ticket”, or “Big Ticket”. If your speed is 60 or less, the result is “No Ticket”. If speed is between 61 and 80 inclusive, the result is “Small Ticket”. If speed is 81 or more, the result is “Big Ticket”. Unless it is your birthday (encoded as a boolean value in the parameters of the function) – on your birthday, your speed can be 5 higher in all cases. **
only difference in my answer and the instructor is no need for “is True” for speed is not needed - also, used return ‘Big Ticket’, etc as opposed to print
def caught_speeding(speed, is_birthday):
if is_birthday is True:
cspeed = speed-5
else:
cspeed = speed
if cspeed < 61:
print("No Ticket")
elif 60 < cspeed < 81:
print("Small Ticket")
elif cspeed > 80:
print("Big Ticket")
else:
print("Oops")
caught_speeding(81,True)
caught_speeding(81,False)