Where we are: Lecture 3, part 1 of 3

Comparison operators give True or False

Every decision starts with a question. Python answers with True or False (a bool).

= stores, == compares

x = 5
print(x == 5)
x = 10
print(x == 5)

Answer: = stores, == compares

x = 5
print(x == 5)
x = 10
print(x == 5)

True
False

Anatomy of an if statement

marks = 72
if marks >= 50:
    print("You passed!")
print("Result checked")

You passed!
Result checked

When the condition is False, the block is skipped

marks = 35
if marks >= 50:
    print("You passed!")
print("Result checked")

Result checked

if / else: exactly one of two paths

marks = 35
if marks >= 50:
    print("Pass")
else:
    print("Try again")

Try again

elif: more than two paths

score = 72
if score >= 80:
    print("A")
elif score >= 65:
    print("B")
elif score >= 50:
    print("C")
else:
    print("F")

B

Tracing the elif chain

Python checks from the top and stops at the first True. Later conditions are not even checked.

if score >= 80:
    print("A")
elif score >= 65:
    print("B")
elif score >= 50:
    print("C")
else:
    print("F")

Watch it run: the elif chain

Does the order of elif matter?

score = 92
if score >= 50:
    print("C")
elif score >= 65:
    print("B")
elif score >= 80:
    print("A")

Answer: Does the order of elif matter?

score = 92
if score >= 50:
    print("C")
elif score >= 65:
    print("B")
elif score >= 80:
    print("A")

C

and, or, not: combining conditions

and needs both. or needs at least one. not flips True and False.

Combining conditions in an if

age = 20
has_id = True
if age >= 18 and has_id:
    print("Can enter")
else:
    print("Cannot enter")

Can enter

An if inside an if (nesting)

marks = 72
attendance = 60
if marks >= 50:
    if attendance >= 75:
        print("Pass")
    else:
        print("Pass marks, low attendance")
else:
    print("Fail")

Pass marks, low attendance

Decisions about text: the first spam rule

message = "Claim your FREE prize"
text = message.lower()
if "free" in text or "prize" in text:
    print("Suspicious")
else:
    print("Looks normal")

Suspicious

match-case: one value, many fixed options

number = "01711111111"
match number[:3]:
    case "017" | "013":
        print("Grameenphone")
    case "019" | "014":
        print("Banglalink")
    case "018":
        print("Robi")
    case _:
        print("Unknown operator")

Grameenphone

Common mistakes with decisions

Read the last line of an error first. It names the problem.

if marks >= 50
    print("ok")

SyntaxError: expected ':'

if marks = 50:
    print("ok")

SyntaxError: invalid syntax. Maybe you meant '==' or ':=' instead of '='?

if marks >= 50:
print("ok")

IndentationError: expected an indented block after 'if' statement on line 1

if "5" > 3:
    print("ok")

TypeError: '>' not supported between instances of 'str' and 'int'

Solutions 1 and 2

Test every branch: try values that should give each possible answer.

n = 7
if n % 2 == 0:
    print("Even")
else:
    print("Odd")

a = 4
b = 9
c = 2
if a >= b and a >= c:
    print(a)
elif b >= c:
    print(b)
else:
    print(c)

Odd
9

Solution 3: leap year

The brackets group the first two conditions. Test the tricky years: 1900 and 2000.

year = 1900
if (year % 4 == 0 and year % 100 != 0) or year % 400 == 0:
    print("Leap")
else:
    print("Not leap")

Not leap

Solution 4: operator check

The else matters: without it, a wrong-length number would print INV and then also an operator name.

number = "0151234567"
if len(number) != 11:
    print("INV")
else:
    match number[:3]:
        case "017" | "013":
            print("Grameenphone")
        case "019" | "014":
            print("Banglalink")
        case "015":
            print("Teletalk")
        case "016":
            print("Airtel")
        case "018":
            print("Robi")
        case _:
            print("INV")

INV

Today's toolkit

Homework 3.1

Due before Lecture 3.2 · [date]