What is the output of the following code snippet: (You can see the picture too) if 'bar' in {'foo': 1, 'bar': 2, 'baz': 3}: print(1) print(2) if 'a' in 'qux': print(3) print(4)
1 2 3 4
1 2 4
It doesn’t generate any output.
4
The following if/elif/else statement will raise a KeyError exception: d = {'a': 0, 'b': 1, 'c': 0}
if d['a'] > 0: print('ok') elif d['b'] > 0: print('ok') elif d['c'] > 0: print('ok') elif d['d'] > 0: print('ok') else: print('not ok')
True
False
Shown below is a diagram of a simple dictionary:
Which of the following is not a valid way to define this dictionary in Python:
d = {'foo': 100, 'bar': 200, 'baz': 300}
d = { ('foo', 100), ('bar', 200), ('baz', 300) }
d = {} d['foo'] = 100 d['bar'] = 200 d['baz'] = 300
d = dict(foo=100, bar=200, baz=300)
d = dict([ ('foo', 100), ('bar', 200), ('baz', 300) ])
List a is defined as follows: a = [1, 2, 3, 4, 5] Select all of the following statements that remove the middle element 3 from a so that it equals [1, 2, 4, 5]:
a[2] = []
del a[2]
a.add(3)
a[0:3] = []
a[2:2] = []
Which of the following would separate a string input_string on the first 2 occurences of the letter “e”?
input_string.split('e', 2)
'e'.split(input_string, maxsplit=2)
'e'.split(input_string, 2)
What is the output of the following code snippet: d = {'foo': 1, 'bar': 2, 'baz': 3} while d: print(d.popitem()) print('Done.')
Done.
foo bar baz
The snippet doesn’t generate any output.
('baz', 3) ('bar', 2) ('foo', 1) Done.
The function sqrt() from the math module computes the square root of a number.
Will the highlighted line of code raise an exception? x = -100 from math import sqrt x > 0 and sqrt(x)
Which of the following mathematical operators can be used to concatenate strings:
+
*
/
-
Suppose you have the following tuple definition: t = ('foo', 'bar', 'baz') Which of the following statements replaces the second element ('bar') with the string 'qux':
It’s a trick question—tuples can’t be modified.
t(1) = 'qux'
t[1:1] = 'qux'
t[1] = 'qux'
What is value of this expression: 'a' + 'x' if '123'.isdigit() else 'y' + 'b'
'ab'
'axb'
'ax'
'axyb'
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