Amortized analysis: Insertion in a list

[quazirfan]

I am trying to calculate the time is takes to append one element in a list. According to amortized analysis, it should be constant regardless how long the list is.

So, here is a little piece of code I've written to test it out with the corresponding output,

def calTime(n):
    from time import time
    data = []
    start = time()
    for r in range(n):
        data.append(None)
    end = time()
    return ((end - start)/n)

for r in [10**x for x in range(15)]:
    print("Size ",r, " time it took ", calTime(r), "sec")

Output:

Input is being redirected from C:\Projects\2698\input.txt
Size  1  time it took  0.0  sec
Size  10  time it took  0.0  sec
Size  100  time it took  0.0  sec  
Size  1000  time it took  0.0  sec  
Size  10000  time it took  1.0018348693847657e-07  sec
Size  100000  time it took  1.099705696105957e-07  sec
Size  1000000  time it took  8.603405952453613e-08  sec
Size  10000000  time it took  8.960003852844239e-08  sec

I have two questions,

1. Is this the correct way to test if amortized cost of insertion in a dynamic array is O(1) - I am assuming it is correct as average insertion time for all values or n are very close to zero.
   
2. I understand that the average time it takes to insert an elements in a list when n is small being rounded to zero. How do I print that small value?
   

[deanhystad]

It is not a rounding problem. It is a timer problem. Your timer mas a smallest minimum time it can measure. Periods less than that may appear to be zero time.