Assessment: Accumulating Lists and Strings >> Python Basics
seqmut-1-9: Which of these is the accumulator variable?
byzo = 'hello world!'
c = 0
for x in byzo:
z = x + "!"
print(z)
c = c + 1
✔️ Yes, this is the accumulator variable. By the end of the program, it will have a full count of how many items are in byzo.
Multiple Choice (assess_question5_1_1_1)
Score: 1.0 / 1
seqmut-1-10: Which of these is the sequence?
cawdra = ['candy', 'daisy', 'pear', 'peach', 'gem', 'crown']
t = 0
for elem in cawdra:
t = t + len(elem)
✔️ Yes, this is the sequence that we iterate over.
Multiple Choice (assess_question5_1_1_2)
Score: 1.0 / 1
seqmut-1-11: Which of these is the iterator (loop) variable?
lst = [5, 10, 3, 8, 94, 2, 4, 9]
num = 0
for item in lst:
num += item
✔️ Yes, this is the iterator variable. It changes each time but is not the whole sequence itself.
Multiple Choice (assess_question5_1_1_3)
Score: 1.0 / 1
seqmut-1-12: What is the iterator (loop) variable in the following?
rest = ["sleep", 'dormir', 'dormire', "slaap", 'sen', 'yuxu', 'yanam']
let = ''
for phrase in rest:
let += phrase[0]
The iterator variable is
Good work!
Fill in the Blank (assess_question5_1_1_4)
Score: 1.0 / 1
Currently there is a string called str1. Write code to create a list called chars which should contain the characters from str1. Each character in str1 should be its own element in the list chars.
str1 = "I love python"
# HINT: what's the accumulator? That should go here.
chars = []
for i in str1:
chars.append(i)
ActiveCode (assess_week5_01)
Expand Differences
| Result | Actual Value | Expected Value | Notes |
|---|---|---|---|
| Pass | [‘I’,… ‘n’] | [‘I’,… ‘n’] | Testing that chars is assigned the correct value. |
You passed: 100.0% of the tests
