1. What will be the output of the code below? Explain your answer.
def extendList(val, list=): list.append(val) return list list1 = extendList(10) list2 = extendList(123,) list3 = extendList('a') print "list1 = %s" % list1 print "list2 = %s" % list2 print "list3 = %s" % list3
How would you modify the definition of
extendList to produce the presumably desired behavior?
The output of the above code will be:
list1 = [10, 'a'] list2 =  list3 = [10, 'a']
Many will mistakenly expect
list1 to be equal to
list3 to be equal to
['a'], thinking that the
list argument will be set to its default value of
 each time
extendList is called.
However, what actually happens is that the new default list is created only once when the function is defined, and that same list is then used subsequently whenever
extendList is invoked without a
list argument being specified. This is because expressions in default arguments are calculated when the function is defined, not when it’s called.
list3 are therefore operating on the same default list, whereas
list2 is operating on a separate list that it created (by passing its own empty list as the value for the
The definition of the
extendList function could be modified as follows, though, to always begin a new list when no
list argument is specified, which is more likely to have been the desired behavior:
def extendList(val, list=None): if list is None: list =  list.append(val) return list
With this revised implementation, the output would be:
list1 =  list2 =  list3 = ['a']
2. What will be the output of the code below? Explain your answer.
def multipliers(): return [lambda x : i * x for i in range(4)] print [m(2) for m in multipliers()]
How would you modify the definition of
multipliers to produce the presumably desired behavior?
The output of the above code will be
[6, 6, 6, 6] (not
[0, 2, 4, 6]).
The reason for this is that Python’s closures are late binding. This means that the values of variables used in closures are looked up at the time the inner function is called. So as a result, when any of the functions returned by
multipliers() are called, the value of
i is looked up in the surrounding scope at that time. By then, regardless of which of the returned functions is called, the
for loop has completed and
i is left with its final value of 3. Therefore, every returned function multiplies the value it is passed by 3, so since a value of 2 is passed in the above code, they all return a value of 6 (i.e., 3 x 2).
(Incidentally, as pointed out in The Hitchhiker’s Guide to Python, there is a somewhat widespread misconception that this has something to do with lambdas, which is not the case. Functions created with a
lambda expression are in no way special and the same behavior is exhibited by functions created using an ordinary
Below are a few examples of ways to circumvent this issue.
One solution would be use a Python generator as follows:
def multipliers(): for i in range(4): yield lambda x : i * x
Another solution is to create a closure that binds immediately to its arguments by using a default argument. For example:
def multipliers(): return [lambda x, i=i : i * x for i in range(4)]
Or alternatively, you can use the
from functools import partial from operator import mul def multipliers(): return [partial(mul, i) for i in range(4)]
3. What will be the output of the code below? Explain your answer.
class Parent(object): x = 1 class Child1(Parent): pass class Child2(Parent): pass print Parent.x, Child1.x, Child2.x Child1.x = 2 print Parent.x, Child1.x, Child2.x Parent.x = 3 print Parent.x, Child1.x, Child2.x
The output of the above code will be:
1 1 1 1 2 1 3 2 3
What confuses or surprises many about this is that the last line of output is
3 2 3 rather than
3 2 1. Why does changing the value of
Parent.x also change the value of
Child2.x, but at the same time not change the value of
The key to the answer is that, in Python, class variables are internally handled as dictionaries. If a variable name is not found in the dictionary of the current class, the class hierarchy (i.e., its parent classes) are searched until the referenced variable name is found (if the referenced variable name is not found in the class itself or anywhere in its hierarchy, an
x = 1 in the
Parent class makes the class variable
x (with a value of 1) referenceable in that class and any of its children. That’s why the first
1 1 1.
Subsequently, if any of its child classes overrides that value (for example, when we execute the statement
Child1.x = 2), then the value is changed in that child only. That’s why the second
1 2 1.
Finally, if the value is then changed in the Parent (for example, when we execute the statement
Parent.x = 3), that change is reflected also by any children that have not yet overridden the value (which in this case would be
Child2). That’s why the third
3 2 3.
4. What will be the output of the code below in Python 2? Explain your answer.
def div1(x,y): print "%s/%s = %s" % (x, y, x/y) def div2(x,y): print "%s//%s = %s" % (x, y, x//y) div1(5,2) div1(5.,2) div2(5,2) div2(5.,2.)
Also, how would the answer differ in Python 3 (assuming, of course, that the above
In Python 2, the output of the above code will be:
5/2 = 2 5.0/2 = 2.5 5//2 = 2 5.0//2.0 = 2.0
By default, Python 2 automatically performs integer arithmetic if both operands are integers. As a result,
Note that you can override this behavior in Python 2 by adding the following import:
from __future__ import division
Also note that the “double-slash” (
//) operator will always perform integer division, regardless of the operand types. That’s why
2.0 even in Python 2.
Python 3, however, does not have this behavior; i.e., it does not perform integer arithmetic if both operands are integers. Therefore, in Python 3, the output will be as follows:
5/2 = 2.5 5.0/2 = 2.5 5//2 = 2 5.0//2.0 = 2.0
5. What will be the output of the code below?
list = ['a', 'b', 'c', 'd', 'e'] print list[10:]
The above code will output
, and will not result in an
As one would expect, attempting to access a member of a list using an index that exceeds the number of members (e.g., attempting to access
list in the list above) results in an
IndexError. However, attempting to access a slice of a list at a starting index that exceeds the number of members in the list will not result in an
IndexError and will simply return an empty list.
What makes this a particularly nasty gotcha is that it can lead to bugs that are really hard to track down since no error is raised at runtime.
6. Consider the following code snippet:
1. list = [ [ ] ] * 5 2. list # output? 3. list.append(10) 4. list # output? 5. list.append(20) 6. list # output? 7. list.append(30) 8. list # output?
What will be the ouput of lines 2, 4, 6, and 8? Explain your answer.
The output will be as follows:
[, , , , ] [, , , , ] [[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]] [[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], 30]
The first line of output is presumably intuitive and easy to understand; i.e.,
list = [ [ ] ] * 5 simply creates a list of 5 lists.
However, the key thing to understand here is that the statement
list = [ [ ] ] * 5 does NOT create a list containing 5 distinct lists; rather, it creates a a list of 5 references to the same list. With this understanding, we can better understand the rest of the output.
list.append(10) appends 10 to the first list. But since all 5 lists refer to the same list, the output is:
[, , , , ].
list.append(20) appends 20 to the second list. But again, since all 5 lists refer to the same list, the output is now:
[[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]].
list.append(30) is appending an entirely new element to the “outer” list, which therefore yields the output:
[[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], 30].
7. Given a list of N numbers, use a single list comprehension to produce a new list that only contains those values that are:
(a) even numbers, and
(b) from elements in the original list that had even indices
For example, if
list contains a value that is even, that value should be included in the new list, since it is also at an even index (i.e., 2) in the original list. However, if
list contains an even number, that number should not be included in the new list since it is at an odd index (i.e., 3) in the original list.
A simple solution to this problem would be as follows
[x for x in list[::2] if x%2 == 0]
For example, given the following list:
# 0 1 2 3 4 5 6 7 8 list = [ 1 , 3 , 5 , 8 , 10 , 13 , 18 , 36 , 78 ]
the list comprehension
[x for x in list[::2] if x%2 == 0] will evaluate to:
[10, 18, 78]
The expression works by first taking the numbers that are at the even indices, and then filtering out all the odd numbers.
8. Given the following subclass of dictionary:
class DefaultDict(dict): def __missing__(self, key): return 
Will the code below work? Why or why not?
d = DefaultDict() d['florp'] = 127
Yes, it will work. With this implementation of the
DefaultDict class, whenever a key is missing, the instance of the dictionary will automatically be instantiated with a list.