Python Interview Questions and Answers

How do you remove duplicates from a list?
If you don't mind reordering the list, sort it and then scan from the end of the list, deleting duplicates as you go:

if List:
List.sort()
last = List[-1]
for i in range(len(List)-2, -1, -1):
if last==List[i]: del List[i]
else: last=List[i]

If all elements of the list may be used as dictionary keys (i.e. they are all hash able) this is often faster

d = {}
for x in List: d[x]=x
List = d.values()

How do you make an array in Python?
Use a list:
["this", 1, "is", "an", "array"]

Lists are equivalent to C or Pascal arrays in their time complexity; the primary difference is that a Python list can contain objects of many different types.

The array module also provides methods for creating arrays of fixed types with compact representations, but they are slower to index than lists. Also note that the Numeric extensions and others define array-like structures with various characteristics as well.

To get Lisp-style linked lists, you can emulate cons cells using tuples:

lisp_list = ("like", ("this", ("example", None) ) )

If mutability is desired, you could use lists instead of tuples. Here the analogue of lisp car is lisp_list[0] and the analogue of cdr is lisp_list[1]. Only do this if you're sure you really need to, because it's usually a lot slower than using Python lists.

How do I create a multidimensional list?
You probably tried to make a multidimensional array like this:

A = [[None] * 2] * 3

This looks correct if you print it:

>>> A
[[None, None], [None, None], [None, None]]

But when you assign a value, it shows up in multiple places:

>>> A[0][0] = 5
>>> A
[[5, None], [5, None], [5, None]]

The reason is that replicating a list with * doesn't create copies, it only creates references to the existing objects. The *3 creates a list containing 3 references to the same list of length two. Changes to one row will show in all rows, which is almost certainly not what you want.

The suggested approach is to create a list of the desired length first and then fill in each element with a newly created list:

A = [None]*3
for i in range(3):
A[i] = [None] * 2

This generates a list containing 3 different lists of length two. You can also use a list comprehension:

w,h = 2,3
A = [ [None]*w for i in range(h) ]

Or, you can use an extension that provides a matrix datatype; Numeric Python is the best known.

How do I apply a method to a sequence of objects?
Use a list comprehension:

result = [obj.method() for obj in List]

More generically, you can try the following function:

def method_map(objects, method, arguments):
"""method_map([a,b], "meth", (1,2)) gives [a.meth(1,2), b.meth(1,2)]"""
nobjects = len(objects)
methods = map(getattr, objects, [method]*nobjects)
return map(apply, methods, [arguments]*nobjects)

I want to do a complicated sort: can you do a Schwartzman Transform in Python?
Yes, it's quite simple with list comprehensions.

The technique, attributed to Randal Schwartz of the Perl community, sorts the elements of a list by a metric which maps each element to its "sort value". To sort a list of strings by their uppercase values:

tmp1 = [ (x.upper(), x) for x in L ] # Schwartzman transform
tmp1.sort()
Usorted = [ x[1] for x in tmp1 ]

To sort by the integer value of a subfield extending from positions 10-15 in each string:

tmp2 = [ (int(s[10:15]), s) for s in L ] # Schwartzman transform
tmp2.sort()
Isorted = [ x[1] for x in tmp2 ]

Note that Isorted may also be computed by

def intfield(s):
return int(s[10:15])

def Icmp(s1, s2):
return cmp(intfield(s1), intfield(s2))

Isorted = L[:]
Isorted.sort(Icmp)

but since this method calls intfield() many times for each element of L, it is slower than the Schwartzman Transform.

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