Python interview questions

How are arguments passed by value or by reference?

Answer:

Everything in Python is an object and all variables hold references to the objects. The references values are according to the functions; as a result you cannot change the value of the references. However, you can change the objects if it is mutable.

What is the difference between list and tuple?

Answer:

The difference between list and tuple is that list is mutable while tuple is not. Tuple can be hashed for e.g as a key for dictionaries.

What are Python decorators?

Answer:

A Python decorator is a specific change that we make in Python syntax to alter functions easily.

What are the tools that help to find bugs or perform static analysis?

Answer:

PyChecker is a static analysis tool that detects the bugs in Python source code and warns about the style and complexity of the bug. Pylint is another tool that verifies whether the module meets the coding standard.

How memory is managed in Python?

Answer:

• Python memory is managed by Python private heap space. All Python objects and data structures are located in a private heap. The programmer does not have an access to this private heap and interpreter takes care of this Python private heap.
• The allocation of Python heap space for Python objects is done by Python memory manager.  The core API gives access to some tools for the programmer to code.
• Python also have an inbuilt garbage collector, which recycle all the unused memory and frees the memory and makes it available to the heap space.

How Python is interpreted?

Answer:

Python language is an interpreted language. Python program runs directly from the source code. It converts the source code that is written by the programmer into an intermediate language, which is again translated into machine language that has to be executed.

What is pickling and unpickling?

Answer:

Pickle module accepts any Python object and converts it into a string representation and dumps it into a file by using dump function, this process is called pickling.  While the process of retrieving original Python objects from the stored string representation is called unpickling.

What is PEP 8?

Answer:

PEP 8 is a coding convention, a set of recommendation, about how to write your Python code more readable.

What is Python? What are the benefits of using Python?

Answer:

Python is a programming language with objects, modules, threads, exceptions and automatic memory management. The benefits of pythons are that it is simple and easy, portable, extensible, build-in data structure and it is an open source.

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

Answer:

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.

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[2] 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[3] 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.

Answer:

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.

Consider the following code snippet:

1. list = [ [ ] ] * 5
2. list  # output?
3. list[0].append(10)
4. list  # output?
5. list[1].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.

Answer:

The output will be as follows:

[[], [], [], [], []]
[[10], [10], [10], [10], [10]]
[[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]]
[[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], 30]

Here’s why:

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[0].append(10) appends 10 to the first list. But since all 5 lists refer to the same list, the output is: [[10], [10], [10], [10], [10]].

Similarly, list[1].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]].

In contrast, 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].

What will be the output of the code below?

list = ['a', 'b', 'c', 'd', 'e']
print list[10:]

Answer:

The above code will output [], and will not result in an IndexError.

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[10] 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.

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 print statements were converted to Python 3 syntax)?

Answer:

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, 5/2 yields 2, while 5./2 yields 2.5.

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 5.0//2.0 yields 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

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

Answer:

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 Child1.x?

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 AttributeError occurs).

Therefore, setting 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 print statement outputs 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 print statement outputs 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 print statement outputs 3 2 3.

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?

Answer:

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 def.)

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 functools.partial function:

from functools import partial
from operator import mul
def multipliers():
  return [partial(mul, i) for i in range(4)]

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?

Answer:

The output of the above code will be:

list1 = [10, 'a']
list2 = [123]
list3 = [10, 'a']

Many will mistakenly expect list1 to be equal to [10] and 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.

list1 and 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 list parameter).

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 = [10]
list2 = [123]
list3 = ['a']

What Is NumPy And How Is It Better Than A List In Python?

Answer:

NumPy is a Python package for scientific computing which can deal with large data sizes. It includes a powerful N-dimensional array object and a set of advanced functions.

Also, the NumPy arrays are superior to the built-in lists. There are a no. of reasons for this.

• NumPy arrays are more compact than lists.
• Reading and writing items is faster with NumPy.
• Using NumPy is more convenient than to the standard list.
• NumPy arrays are more efficient as they augment the functionality of lists in Python.

How Do You Count The Occurrences Of Each Item Present In The List Without Explicitly Mentioning Them?

Answer:

Unlike sets, lists can have items with same values. In Python, the list has a <count()> function which returns the occurrences of a particular item.

Turn A List Into A Dictionary.

Answer:

In a dictionary, each item represents a key-value pair. So converting a list isn’t as straight forward as it were for other data types.

However, we can achieve the conversion by breaking the list into a set of pairs and then call the <zip()> function to return them as tuples.

Passing the tuples into the <dict()> function would finally turn them into a dictionary.

Example

weekdays = ['sun','mon','tue','wed','thu','fri']
listAsDict = dict(zip(weekdays[0::2], weekdays[1::2]))
print(listAsDict)
#output: {'sun': 'mon', 'thu': 'fri', 'tue': 'wed'}