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First solution in Clear category for Median by Rafal.U
def checkio(data):
x = len(data)
data.sort() #sortowanie listy
if x%2 == 0:
return (data[int(x/2)-1] + data[int(x/2)])/2 # wyliczenie mediany dla parzystej liczby elementów w liście
else:
return data[int(x/2)] # mediana dla nieparzystej liczby elementów listy, czyli środkowy lelement listy
#replace this for solution
return data[0]
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert checkio([1, 2, 3, 4, 5]) == 3, "Sorted list"
assert checkio([3, 1, 2, 5, 3]) == 3, "Not sorted list"
assert checkio([1, 300, 2, 200, 1]) == 2, "It's not an average"
assert checkio([3, 6, 20, 99, 10, 15]) == 12.5, "Even length"
print("Start the long test")
assert checkio(list(range(1000000))) == 499999.5, "Long."
print("The local tests are done.")
Dec. 11, 2016