Lists

def mostCommonElement(L):
    d = {}
    for item in L:
        if item in d:
            d[item] += 1
        else:
            d[item] = 1
    counts = [(k,v) for k, v in d.items()]  # This basically makes a list that looks like
                                            # [(item, numoccurrences), (item2, numoccurrences2), ...]
    counts.sort(lambda x,y: cmp(y[1], x[1]))   # This sorts that list in reverse, based on the second
                                               # element in each little tuple/pair
    return counts[0][0]

def findMedian(L):
    import copy
    listCopy = copy.copy(L)
    listCopy.sort()
    return listCopy[int(len(listCopy)/2)]  # Note that int will just floor() the result if it's
                                           # something.5, if you want this to be different change int() to ceil()

def findMedian(nums):

	"""
	Finds the median of a set of numbers (The one in the middle).
	"""
	
	tmp = len(nums) / 2

	if len(nums) % 2:
		median = nums[tmp]
	else:
		median = (nums[tmp] + nums[tmp - 1]) / 2.0
		
	return median

Traceback (most recent call last):
  File "./bleh.py", line 39, in ?
    main()
  File "./bleh.py", line 33, in main
    common = mostCommonElement(nums)
  File "./bleh.py", line 16, in mostCommonElement
    counts = [(k,v) for k,v in d]
TypeError: unpack non-sequence

def mostCommonElement(nums):

	"""
	Finds the mode of a set of numbers (Most common number).	
	"""
	
	modes = []
	total = 0
	top = 0
	skip = 0
	
	for x in nums:
		for i in nums:
			if x == i:
				total += 1
	
		if total > top:
			top = total
			modes = []
			modes.append(x)
		
		if total == top:
			for j in modes:
				if x == j:
					skip = 1
					break
				else:
					skip = 0
			
			if not skip:
				modes.append(x)
	
		total = 0
		skip = 0

	return modes

def most(sequence):
	index = {}
	for value in sequence:
		index[sequence.count(value)] = value
	items = index.items()
	return items[-1][1]

        if item in d:
            d[item] += 1
        else:
            d[item] = 1

>>> int('a', 16)
10
>>>