G11_C6_TeacherCopy
Introduction to lists
| Activity Flow | Slide No. | Topic | Time |
|---|---|---|---|
| TA | 4-7 | WarmUp Quiz | 3 min |
| 8-9 | Revision | 2 min | |
| 10-21 | Introduction to Lists | 10 min | |
| 22-30 | Creating and Iterating through Lists | 10 min | |
| SA | 31-33 | Brick Creation | 5 min |
| TA | 34 | Precap | 2 min |
| Wrap - Up | 35-38 | Quiz | 3 min |
| SA | 39-44 | Additional Activity | 10 min |
Class Structure
| Slide No. | Topic |
|---|---|
| 15 | TA1 Solution |
| 18 | TA2 Solution |
| 21 | TA3 Solution |
| 27 | TA4 Solution |
| 29 | TA5 Solution |
| 31 | TA6 Solution |
| 34 | SA1 solution |
| 41 | SAA1 solution |
| 43 | SAA2 Solution |
Preparation and Reference
Prerequisites
FOR TEACHER
FOR STUDENTS
-
Computer with an Internet connection.
-
The latest browser installed.
-
Spyder IDE.
-
Projector to present the screen.
1. Computer with an Internet connection.
2. The latest browser installed.
3.Spyder IDE.
(WARM-UP QUIZ)
Which loop we would preferably use if we want to perform a task a fixed number of times?
Q1.
for loop
-
A "for" loop is a better choice when the task has to be performed a predefined number of times.
-
A "while" loop is more useful when the decision on whether to perform a task depends on some condition that needs to be dynamically changed.
What will be the output of the following code?
Q2.
print 0 infinitely
for i in range(10):
prod=10*i
while prod<=10:
print(prod)check again
condition never becomes false!!
for i in range 10:
for loop starts: i=0
prod=10*0=0
prod<=10? : True
prod which is "0" is printed
carryOn = True
while carryOn:
for event in pygame.event.get():
if event.type == pygame.QUIT:
carryOn = False
pygame.quit()Revision
Learning Programming
B
Variables that hold multiple values
Single value
Multiple values
List
Learning Programming
B
roll_no=[12,56,90]
Box Brackets
Comma
Lists in real life
roll_no=[12,56,90]
print(roll_no)
TA1:Solution
list.append(item)
dot operator: helps the append function understand which list it has to work on.
Adding items to a list
List to which we want to append an item
Here: roll_no
item to be appended to list
append function:
appends one item to list
roll_no=[12,56,90]
print(roll_no)
roll_no.append(67)
print(roll_no)
roll_no.append(39)
print(roll_no)TA2 :Solution
list.remove(item)
Removing items from a list
dot operator: helps the remove function understand which list it has to work on.
List to which we want to remove an item
Here: roll_no
item to be removed
from list
remove
function:
removes one item to list
roll_no=[12,56,90]
print(roll_no)
roll_no.append(67)
print(roll_no)
roll_no.append(39)
print(roll_no)
roll_no.remove(90)
print(roll_no)TA3: Solution
GREAT!
Manual list creation
Create a list of numbers 1 to 100.
numbers=[1,2,3,4,5,6,...]numbers=[]
for i in range(1,101):
numbers.append(i)1. Create an "empty" list
2. Use a "for" loop to iterate through the desired range.
3. Append the desired element to the created list.
Deploying loops to create lists
Recreating the red bricks
[
]
[
]
1. Create an "empty" list
2. Use a "for" loop to create a rectangle object
and add to the "empty" list.
+
X 6 times
bricksR=[]
for i in range(6):
brick=pygame.Rect(10 + i* 100,60,80,30)
red_bricks.append(brick)TA4: Solution
List comprehension
Too many steps!!
[
]
for
i
in
range(6)
What item to append?
How many times to append?
[
]
[
]
1. Create an "empty" list
2. Use a "for" loop to create
rectangle object and it add to the "empty" list.
+
X6 times
bricksR=[pygame.Rect(10 + i* 100,60,80,30) for i in range(6)]TA5: Solution
Drawing the bricks
[
]
for
i
in
list:
for i in bricksR:
pygame.draw.rect(screen,RED,i)TA6: Solution
Write to code to see the orange bricks on the screen.
SA1: Orange bricks
Hints:
bricksO=[pygame.Rect(10 + i* 100,100,80,30) for i in range(6)]for i in bricksO:
pygame.draw.rect(screen,ORANGE,i)ORANGE = [255,100,0]SA1: Solution
Removing a brick on collision
[
]
for
i
in
list:
Remove brick
What will be the output:
B
2,3,4,5,6,7
Error
No output
Q.1
A
Error
B
C
Since 10 is not present it will throw us a value error.
num=[2,3,4,6,7]
num.remove(10)
print(num)
What will be the output:
B
2,3,4,6,7,10
Error
10,2,3,4,6,7
Q.2
A
Error
B
C
The syntax is :
list.append(item)
num=[2,3,4,6,7]
append(num,10)
print(num)Write a code to see the yellow bricks on the screen
SAA1: Yellow bricks("For" loop)
Hints:
bricksY=[]
for i in range(7):
brick=pygame.Rect(10 + i* 100,140,80,30)
bricksY.append(brick)for i in bricksY:
pygame.draw.rect(screen,YELLOW,i)YELLOW = [255,255,0]SAA1: Solution
Code to be able to see the yellow bricks on the screen
SAA2: Yellow bricks(List Comprehension)
Hints:
bricksY=[pygame.Rect(10 + i* 100,140,80,30) for i in range(7)]for i in bricksY:
pygame.draw.rect(screen,YELLOW,i)YELLOW = [255,255,0]SAA2: Solution
| Activity | Activity Name | Link |
|---|---|---|
| TEACHER ACTIVITY 1,2,3 | List creation | |
| TEACHER ACTIVITY 4 | Red brick creation | |
| TEACHER ACTIVITY 5 | List comprehension | |
| TEACHER ACTIVITY 6 | List iteration | |
| TEACHER ACTIVITY 1 SOLUTION | Solution of TA1 | |
| TEACHER ACTIVITY 2 SOLUTION | Solution of TA2 | |
| TEACHER ACTIVITY 3 SOLUTION | Solution of TA3 | |
| TEACHER ACTIVITY 4 SOLUTION | Solution of TA4 | |
| TEACHER ACTIVITY 5 SOLUTION | Solution of TA5 | |
| TEACHER ACTIVITY 6 SOLUTION | Solution of TA6 | |
| STUDENT ACTIVITY 1 | Orange Brick Creation | |
| TEACHER REFERENCE: STUDENT ACTIVITY 1 SOLUTION | Solution of SA1 | |
| STUDENT ADDITIONAL ACTIVITY 1 | Yellow brick creation | |
| STUDENT ADDITIONAL ACTIVITY 2 | Yellow brick creation (List Comprehension) |
|
| TEACHER REFERENCE: STUDENT ADDITIONAL ACTIVITY 1 SOLUTION | Solution of SAA1 | |
| TEACHER REFERENCE: STUDENT ADDITIONAL ACTIVITY 2 SOLUTION | Solution of SAA2 |
G11 C6_TeacherCopy
By Sanjukta Bhattacharya
