Test Practice Flashcards
What is the output of the following code?
a , b = 10 , 20
print( (b , a)[a < b] )
10
a < b evaluates to True
True = 1
So, the 1 index of the tuple (b, a) is a, which is set to the value of 10
UTF-8
1 - can handle any Unicode point
2 - is a type of encoding
3 - the 8 in UTF-8 means 8-bit values are used in encoding
4 - ASCII characters are also valid UTF-8 text
5 - stands for Unicode transformation format
What I’d the output of the following code snippet?
i = 0
while i != 0:
i -= 1
else:
i += 1
print(i)
1
what does this print? And why?
g = “good lord”
print(g[::1])
good lord
A blank index at the first colon defaults to 0
A blank index at the second colon defaults to the end of the string
The number after the third colon indicated the step of the slice, so it would:
Start at 0
End at the end of the string
Step through by 1 index at a time
What does the following code snippet print and why?
s = “PYTHON”
print( s[ : : -5] )
NP
since the first two colons are blank, the
1 - string starts slicing at 0
2 - ends slicing at the end of the string
3 - steps at through from the END of the string at a negative 5 steps
So it will print the index at the end of the string first ( N )and then go -5 indexes back from that point and print the P
So NP is the print out
How can you pass arguments to a function?
A keyword argument
A positional argument
A mix of both positional and keyword
What is the output of the following code snippet?
try:
print(“START”)
n = 1 / 0
print(“END”)
except:
print(“FAIL”)
print(“BACK TO NORMAL”)
START
FAIL
BACK TO NORMAL
the code executes until the ZeroDivisionError
Then it moves to the except portion where it performs both Prints
Do keyword arguments precede positional arguments?
No. Keyword arguments must follow positional arguments
Can keyword arguments be passed to a function in any order?
Yes. So long as they follow any positional arguments.
What is the output of the following code?
list_1 = [ ‘a’, ‘b’, ‘c’ ]
list_2 = list_1
del list_1[0]
print ( list_2 )
[‘b’, ‘c’]
list_2 refers to the same place in memory as list_1
So changing data in list_1 will change it in list_2 as well
Look at this code-what is the output? Why?
list_a = [1 , 2 , 3]
list_b = list_a
list_c = list_b[ : ]
list_a.pop(1)
print(list_b)
print(list_c)
[1,3]
[1,2,3]
list b is a memory copy of list_a so any changes to list_a will reflect in list_b
list_c is just a regular copy of list_b though so no changes to list_a will affect it
Which of the following will display a message on the screen?
print()
str()
len()
input()
print()
And
input()
Input allows you to print a message to the console so the user knows to provide some type of input to the program
Insert the correct snippet so that the program produces the expected output.
Expected output:
True
Code:
list = [False, True, “2”, 3, 4, 5]
# insert code here
print (b)
Code options:
A. b = 0 not in list
B. b = list[0]
C. b = 0 in list
D. b = False
C
Is 0 in the list? Yes it is. False evaluates to 0 (and True evaluates to 1) so it is True that 0 is in the list named “list”
Assuming that the tuple is a correctly created tuple, the fact that tuples are immutable means that the following instruction:
my_tuple[1] = my_tuple[1] + my_tuple[0]
A. is illegal
B. may be illegal if the tuple contains strings
C. can be executed if and only if the tuple contains at least two elements
D. is fully correct
A
The code will result in an error
You CAN add two tuple values (or concatenate two tuple str values), but the resulting value must be assigned to a new variable.
Like this:
this_tuple[0] = my_tuple[1] + my_tuple[0]
What is the expected output of the following code?
x = [0, 1, 2]
x. insert(0, 1)
del x[1]
print (sum(x))
4
insert a 1 at index 0 which moves the 0 already AT 0 to index 1
Deleting index 1 just deletes the 0 so now you’re left with x = [1, 1, 2]
Summing list x produces 4 (1 + 1 + 2)
What is the expected output of the following code?
list1 = [1, 3]
list2 = list1
list1[0] = 4
print (list2)
4, 3
list 2 is a clone of list 1 — they point to the same spot in memory. So updating list 1 will also update list 2
What is the expected output of the following code?
data = [‘Peter’, 404, 3.03, “tone”, 4, 17]
print (data[1:3])
404, 3.03
Python slicing by index is not inclusive. So the start point is 1 (and 404) and the end point is 3 (“tone”), but the slice is not inclusive so the values returned are data[1] and data[2]
Take a look at the snippet, and choose the true statements: (Choose two.)
nums = [1, 2, 3]
vals = nums
del vals [1:2]
A. nums is longer than vals
B. nums and vals are of the same length
C. vals is longer than nums
D. nums and vals refer to the same list
B & D
Since both lists refer to the same spot in memory, editing one will edit the other.
So all attributes of both lists will be exactly the same.
In this case they both equal [1,3]
Because we delete the value at index 1 from the list. Since we start at index 1 and stop at index 2 — remember that slicing is not inclusive.
What will be the output?
x = 0
a = 2
b = -5
if a > 0:
if b < 0:
x = x + 5
elif a > 5:
x = x + 4
else:
x = x + 3
else:
x = x + 2
print(x)
5
How many hash signs (#) will be printed after the following snippet?
val = 1
while val < 10:
print(‘#’)
val = val «_space;2
2
