Write a brief description of all the following Object Types and Data Structures we've learned about:
Numbers:
Strings:
Lists:
Tuples:
Dictionaries:
In [2]:
print 10*100/10+5.75-5.5
Explain what the cell below will produce and why. Can you change it so the answer is correct?
In [4]:
2.0/3
Out[4]:
Answer these 3 questions without typing code. Then type code to check your answer.
What is the value of the expression 4 * (6 + 5)
What is the value of the expression 4 * 6 + 5
What is the value of the expression 4 + 6 * 5
In [7]:
print 4*(6+5)
print 4*6+5
print 4+6*5
print 3+1.5+4
What is the type of the result of the expression 3 + 1.5 + 4?
What would you use to find a number’s square root, as well as its square?
In [9]:
print 2**(0.5)
Given the string 'hello' give an index command that returns 'e'. Use the code below:
In [10]:
s = 'hello'
# Print out 'e' using indexing
print s[1]
# Code here
Reverse the string 'hello' using indexing:
In [17]:
s ='hello'
# Reverse the string using indexing
print s[::-1]
print s[:3:-1]
# Code here
Given the string hello, give two methods of producing the letter 'o' using indexing.
In [18]:
s ='hello'
# Print out the
print s[4]
print s[-1]
# Code here
Build this list [0,0,0] two separate ways.
In [24]:
a = list([0,0,0])
print a
a = list([0,0])
print a
a.append(0)
print a
Reassign 'hello' in this nested list to say 'goodbye' item in this list:
In [26]:
l = [1,2,[3,4,'hello']]
l[2][2] = 'goodbye'
print l
Sort the list below:
In [48]:
l = [3,4,5,5,6,1]
print l
l.sort()
print l
l = [3,4,5,5,6,1]
print sorted(l)
print l
Using keys and indexing, grab the 'hello' from the following dictionaries:
In [31]:
d = {'simple_key':'hello'}
# Grab 'hello'
print d['simple_key']
In [32]:
d = {'k1':{'k2':'hello'}}
# Grab 'hello'
print d['k1']['k2']
In [39]:
# Getting a little tricker
d = {'k1':[{'nest_key':['this is deep',['hello']]}]}
# Grab hello
print d['k1'][0]['nest_key'][1][0]
In [43]:
# This will be hard and annoying!
d = {'k1':[1,2,{'k2':['this is tricky',{'tough':[1,2,['hello']]}]}]}
print d['k1'][2]['k2'][1]['tough'][2][0]
Can you sort a dictionary? Why or why not?
Not sure about this.
What is the major difference between tuples and lists?
How do you create a tuple?
What is unique about a set?
Use a set to find the unique values of the list below:
In [49]:
l = [1,2,2,33,4,4,11,22,3,3,2]
set(l)
Out[49]:
For the following quiz questions, we will get a preview of comparison operators:
| Operator | Description | Example |
|---|---|---|
| == | If the values of two operands are equal, then the condition becomes true. | (a == b) is not true. |
| != | If values of two operands are not equal, then condition becomes true. | |
| <> | If values of two operands are not equal, then condition becomes true. | (a <> b) is true. This is similar to != operator. |
| > | If the value of left operand is greater than the value of right operand, then condition becomes true. | (a > b) is not true. |
| < | If the value of left operand is less than the value of right operand, then condition becomes true. | (a < b) is true. |
| >= | If the value of left operand is greater than or equal to the value of right operand, then condition becomes true. | (a >= b) is not true. |
| <= | If the value of left operand is less than or equal to the value of right operand, then condition becomes true. | (a <= b) is true. |
What will be the resulting Boolean of the following pieces of code (answer fist then check by typing it in!)
In [50]:
# Answer before running cell
2 > 3
Out[50]:
In [51]:
# Answer before running cell
3 <= 2
Out[51]:
In [52]:
# Answer before running cell
3 == 2.0
Out[52]:
In [53]:
# Answer before running cell
3.0 == 3
Out[53]:
In [54]:
# Answer before running cell
4**0.5 != 2
Out[54]:
Final Question: What is the boolean output of the cell block below?
In [55]:
# two nested lists
l_one = [1,2,[3,4]]
l_two = [1,2,{'k1':4}]
#True or False?
l_one[2][0] >= l_two[2]['k1']
Out[55]: