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011 Python data types : Answers to exercises

Exercise 1

  • If the ASCII code for e is 101 and the code for n is 110, what is the code for a?
# ANSWER
msg = '''
e -> 101
n -> 110

since e is the 5th letter of the alphabet and n is the 14th
it follows that a is 101 - 4 == 110 - 13 = 97

so

a -> 97

Confirm this e.g. on http://www.asciitable.com/
'''
print(msg)

e -> 101
n -> 110

since e is the 5th letter of the alphabet and n is the 14th
it follows that a is 101 - 4 == 110 - 13 = 97

so

a -> 97

Confirm this e.g. on http://www.asciitable.com/

Exercise 2

  • in a code cell below, create a variable called name and set it to your name
  • print the string name, and its length
  • comment on why the length is the value you find
# ANSWER

# in a code cell below, create a variable called `name` and set it to your name
name = 'Charlie Walker'

# print the string name, and its length
print(name,len(name))

# comment on why the length is the value you find
msg = '''
    The string 'Charlie' has 7 characters
    the string 'Walker' has 6 characters
    plus we have a space iun the middle
    so, the length we would expect is 7 + 6 + 1 = 14

    This is what is returned by len(name)
'''
print(msg)

Charlie Walker 14

    The string 'Charlie' has 7 characters
    the string 'Walker' has 6 characters
    plus we have a space iun the middle
    so, the length we would expect is 7 + 6 + 1 = 14

    This is what is returned by len(name)

Answer

We could find this by examining ASCII code tables and see that a (lower case a) has the code 97.

Alternatively, we could search for help on this topic, and find that the python function ord() converts from string to ASCII code: 

# ANSWER
print("the ASCII code for 'a' is",ord('a'))

the ASCII code for 'a' is 97

Alternatively, we might notice that n is the 14th letter of the alphabet, and ethe 5th, so the code seems to be 97 + N where N is the order the letter appears in the alphabet. We can confirm this with the 15th letter o which we see from above has the code 111.

Exercise 3

  • insert a new cell below here
  • set a variable called message to contain the string hello world
  • print the value and data type of the variable message
# ANSWER

# set a variable called message to contain the string hello world
message = 'hello world'

# print the value and data type of the variable message
print(message,type(message))

hello world <class 'str'>

Exercise 4

Since the energy level expressed in \(J\) is quite small, we might more conveniently express it in units of eV. Given that:

\[ 1\ Electron\ volt\ (eV) = 1.602176565 \times 10^{-19} J \]
  • Insert a new cell below here
  • calculate the energy associated with a blue photon at 450 nm, in eV
  • confirm your answer using a web calculator
# Answer
# Copy mostly from above:

# values of c and h
c = 2.99792458e8
h = 6.62607015e-34

print(h * c)
# wavelength in nm: BLUE
# calculate the energy associated with a blue photon at 450 nm, in eV
l_nm = 450.0

# wavelength in m
l_m = l_nm * 1e-9

# Planck-Einstein in J
E_J = h * c / l_m

# conversion formula given above
E_eV = E_J / 1.602176565e-19
print('Photon of wavelength', l_nm, 'nm')
print('has an energy of', E_eV, 'eV')
# which compares with 2.75520 eV given in the web calculator

1.9864458571489286e-25
Photon of wavelength 450.0 nm
has an energy of 2.7552045282834468 eV

Exercise 5

  • insert a new cell below here
  • using integer arithmetic, print the result of:
  • 2 to the power of 8
  • 1024 divided by 2
  • set a variable called x to the result of 7 (floor) divided by 3.
  • print the value of x, and confirm its data type is int
#
# answer
# using integer arithmetic, print the result of:

# 2 to the power of 8
print(2**8)

# 1024 divided by 2 integer division (floor)
print(1024 // 2)

# set a variable called x to the result of 7 divided by 3
x = 7 // 3
print('Integer: 7 divided by 3 is', x, type(x))

# We contrast this with the use of /
# which results in a variable of type float
x = 7 / 3
print('7 divided by 3 is', x, type(x))

256
512
Integer: 7 divided by 3 is 2 <class 'int'>
7 divided by 3 is 2.3333333333333335 <class 'float'>

Exercise 6

  • Insert a new cell below here
  • Set a variable called is_class_today to the value True
  • print the variable name, its value, and its data type
# ANSWER
# Insert a new cell below here
# Set a variable called `is_class_today` to the value `True`

is_class_today = True

# print the variable name, its value, and its data type

print('is_class_today',is_class_today,type(is_class_today))

is_class_today True <class 'bool'>

Exercise 7

  • Insert a new cell below here
  • write a statement to set a variable x to True and print the value of x and not x
  • what does not not x give? Make sure you understand why 
# ANSWER

# write a statement to set a variable x to True 
x = True

# and print the value of x and not x
print('x is',x)
print('not x is',not x)

# what does not not x give? 
print('not not x is',not not x)
msg = '''
answer
------
not not x is the same as just x

A double negative cancels out, in effect
'''
print(msg)

x is True
not x is False
not not x is True

answer
------
not not x is the same as just x

A double negative cancels out, in effect

# ANSWER
# do the testing here e.g.
print (True or False)

True
blank A and B A or B
blank blank blank

Exercise 9

  • Copy the 3 variable truth table from above onto paper
  • fill out a column with A and B
  • fill out a column with ((A and B) or C)
  • Try some other compound statements

If you are unsure, or to check your answers, test the response using code, below.

# ANSWER
# do the testing here e.g.
print ((True and False) or True)

True
blank A and B ((A and B) or C)
blank blank blank 
# ANSWER

# write a statement to set a variable x to True a
x = True

# print the value of x 
print('x is',x)

# print the value of not x 
print('not x is',not x)

# what does not not x give?
print('not not x is',not not x)
# not not cancels out (double negative)

x is True
not x is False
not not x is True

# ANSWER

# Set a variable called `is_class_today` to the value `True`
is_class_today = True

# print the variable name, its value, and its data type
print('is_class_today',is_class_today,type(is_class_today))

is_class_today True <class 'bool'>

Exercise 10

  • insert a new cell below here
  • copy the code in the cell above, set start_number to 0, and run
    • What are the boolean representations of 0 and 1?
  • What would happen if you set start_number to the string 'zero', and why?
# ANSWER

# copy the code in the cell above, and set start_number to 0
start_number = 0

print("starting with", start_number)
int_number = int(start_number)
print('int_number', int_number, type(int_number))
# now convert to float
float_number = float(int_number)
print('float_number', float_number, type(float_number))
# now convert to str
str_number = str(int_number)
print('str_number', str_number, type(str_number))
# now convert to bool
bool_number = bool(int_number)
print('bool_number', bool_number, type(bool_number))

# What is the boolean representation of 0?
msg = '''

What would happen if you set start_number to the string 'zero', and why?

Answer
------
  1 -> True
  0 -> False

  If we set start_number to 'zero' then int('zero') will fail
  because it cannot convert a word representation of this sort to an integer
  only a character representation such as '0' or '1'
'''
print(msg)

starting with 0
int_number 0 <class 'int'>
float_number 0.0 <class 'float'>
str_number 0 <class 'str'>
bool_number False <class 'bool'>


What would happen if you set start_number to the string 'zero', and why?

Answer
------
  1 -> True
  0 -> False

  If we set start_number to 'zero' then int('zero') will fail
  because it cannot convert a word representation of this sort to an integer
  only a character representation such as '0' or '1'
Last update: October 8, 2020