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python interview questions :201

    How about dictionary comprehension?

    Similar to its sibling the list comprehension, a dictionary comprehension is nothing more than a shorthand way of creating a new dictionary collection.
    >>> D1 = dict((k,k**3) for k in L)
    >>> D1
    {0: 0, 1: 1, 2: 8, 3: 27, 4: 64, 5: 125}
    
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    • Or for Python 3:
    >>> D2 = {k:k**3 for k in L}
    >>> D2
    {1: 1, 2: 8, 3: 27}
    
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    python interview questions :202

    Sum of all elements in a list ([1,2,3,...,100] with one line of code ?

    >>> Ans = sum(range(101))
    >>> Ans
    
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    5050
    

    python interview questions :203

    How about summing only odd elements?

    >>> sum(range(1,101,2))
    
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     odd

    Learn python - python tutorial - odd - python examples - python programs

    python interview questions :204

    How to truncate division operation?

    • The truncating division operator (also known as floor division) truncates the result to an integer and works with both integers and floating-point numbers.
    • As of this writing, the true division operator (/) also truncates the result to an integer if the operands are integers. Therefore, 7/4 is 1, not 1.75.
    • However, this behavior is scheduled to change in a future version of Python, so you will need to be careful.
    • The modulo operator returns the remainder of the division x // y. For example, 7 % 4 is 3.
    • For floating-point numbers, the modulo operator returns the floating-point remainder of x // y, which is x – (x // y) * y. For complex numbers, the modulo (%) and truncating division operators (//) are invalid.
    • Division of two number like this:
    >>> 5.5/2 
    
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    Output:

    2.75
    
    >>> 5.5//2
    
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    Output:

    2.0

    python interview questions :205

    How to Print a list of file in a directory?

    • If you want to print a list of files in a directory including the sub-directories. You may want to do it recursively.
    import os
    def file_list(dir):
        basedir = dir
        subdir_list = []
        for item in os.listdir(dir):
            fullpath = os.path.join(basedir,item)
            if os.path.isdir(fullpath):
                subdir_list.append(fullpath)
            else:
                print fullpath
    
        for d in subdir_list:
            file_list(d)
    
    file_list('/dir')
    
    
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    Output:

    ./d3/f1
    ./d3/d4/d7/f2
    ./d3/d4/d7/f3
    ./d3/d5/d9/d10/f4
    ./d3/d5/d9/d10/f5
    [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
    
    

    python interview questions :206

    How to Make a prime number list from (1,100)?

    import math
    def isPrime(n):
        if n <= 1:
            return False
        if n == 2:
            return True
        if n % 2 == 0:
            return False
        for t in range(3, int(math.sqrt(n)+1),2):
            if n % t == 0:
                return False
        return True
    print [n for n in range(100) if isPrime(n)]
    
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    Output:

    [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83]
    

    python interview questions :207

    How to reverse a string using recursive function?

    • In the following recursive code, at every call, the first string will go to the end and stored in stack. So, when there is only one character left as an input string, the code starts rewind and retrieve the character one by one.
    • The net effect is it gets the character in reverse. It becomes obvious when we see the printed output below:
    def reverse(input):
        print input
        if len(input) <= 1:
    	    return input
        return reverse(input[1:]) + input[0]
    s = 'reverse'	
    print(reverse(s))
    
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    Output:

    reverse
    everse
    verse
    erse
    rse
    se
    e
    esrever
    

    python interview questions :208

    How to Merge the overlapped range?

    • You want to merge all overlapping ranges and return a list of distinct ranges. For example:
    [(1, 5), (2, 4), (3, 6)] --->  [(1, 6)]
    [(1, 3), (2, 4), (5, 6)] --->  [(1, 4), (5, 6)]
    
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    • Here is the code:
    L = [(4,9), (20, 22), (1, 3), (24, 32), (23, 31), (40,50), (12, 15), (8,13)]
    # L = [(1, 5), (2, 4), (3, 6)]
    # L = [(1, 3), (2, 4), (5, 6)]
    L = sorted(L)
    print L
    
    Lnew = []
    st = L[0][0]
    end = L[0][1]
    
    for item in L[1:]:
        if end >= item[0]:
            if end < item[1]:
                end = item[1]
        else:
            Lnew.append((st,end))
    st = item[0]
            end = item[1]
        
    Lnew.append((st,end))
    
    print Lnew
    
    
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    Output:

    [(1, 3), (4, 9), (8, 13), (12, 15), (20, 22), (23, 31), (24, 32), (40, 50)]
    [(1, 3), (4, 15), (20, 22), (23, 32), (40, 50)]
    

    python interview questions :209

    How to Initialize dictionary with list - I ?

    • Sometimes If you want to construct dictionary whose values are lists.
    • In the following example, we make a dictionary like {'Country': [cities,...], }:
    cities = {'San Francisco': 'US', 'London':'UK',
            'Manchester':'UK', 'Paris':'France',
            'Los Angeles':'US', 'Seoul':'Korea'}
    # => {'US':['San Francisco', 'Los Angeles'], 'UK':[,], ...}
    
    from collections import defaultdict
    # using collections.defaultdict()
    d1 = defaultdict(list) # initialize dict with list
    for k,v in cities.items():
        d1[v].append(k)
    print d1
    
    # using dict.setdefault(key, default=None)
    d2 = {}
    for k,v in cities.items():
           d2.setdefault(v,[]).append(k)
    print d2
    
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    Output:

    defaultdict(<type 'list'>, {'Korea': ['Seoul'], 'US': ['Los Angeles', 'San Francisco'], 'UK': ['Manchester', 'London'], 'France': ['Paris']}) {'Korea': ['Seoul'], 'US': ['Los Angeles', 'San Francisco'], 'UK': ['Manchester', 'London'], 'France': ['Paris']}
    

    python interview questions :210

    How to Initialize dictionary with list - II?

    • A little bit simpler problem. If youhave a list of numbers:
    L = [1,2,4,8,16,32,64,128,256,512,1024,32768,65536,4294967296]
    
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    L = [1,2,4,8,16,32,64,128,256,512,1024,32768,65536,4294967296]
    
    from collections import defaultdict
    d = defaultdict(list)
    for i in L:
        d[len(str(i))].append(i)
    print d
    print {k:v for k,v in d.items()}
    
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    Output:

    defaultdict(<type 'list'>, {1: [1, 2, 4, 8], 2: [16, 32, 64], 3: [128, 256, 512], 4: [1024], 5: [32768, 65536], 10: [4294967296]})
    {1: [1, 2, 4, 8], 2: [16, 32, 64], 3: [128, 256, 512], 4: [1024], 5: [32768, 65536], 10: [4294967296]}
    

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