there is a very detailed table on python wiki which answers your question.
However, in your particular example you should use enumerate to get an index of an iterable within a loop. like so:
for i, item in enumerate(some_seq):
bar(item, i)
there is a very detailed table on python wiki which answers your question.
However, in your particular example you should use enumerate to get an index of an iterable within a loop. like so:
for i, item in enumerate(some_seq):
bar(item, i)
The answer is "undefined". The Python language doesn't define the underlying implementation. Here are some links to a mailing list thread you might be interested in.
It is true that Python's lists have been implemented as contiguous vectors in the C implementations of Python so far.
I'm not saying that the O() behaviours of these things should be kept a secret or anything. But you need to interpret them in the context of how Python works generally.
Also, the more Pythonic way of writing your loop would be this:
def foo(some_list):
for item in some_list:
bar(item)
What is the time complexity of the “in” operation
algorithm - Time Complexity of List creation in Python - Stack Overflow
Complexity of *in* operator in Python - Stack Overflow
Why is the time complexity of Python's list.append() method O(1)? - Stack Overflow
Videos
I’m not the biggest python user. But I was looking at a friends code yesterday and they had something like:
For x in (list of 40000)
For y in (list of 2.7 million)
If x = y
Append something This was obviously super slow so they changed it to something like:
For x in (list of 2.7 million)
If y in (list of 40000)
Append something
This moved much faster. I get the point of one for loop being faster than two, but what is that “in” exists function doing that makes it so much faster. I always thought that to check if something exists is O(n) which shouldn’t be faster. Also this was for ML purposes so they were likely using numpy stuff.
The complexity of in depends entirely on what L is. e in L will become L.__contains__(e).
See this time complexity document for the complexity of several built-in types.
Here is the summary for in:
- list - Average: O(n)
- set/dict - Average: O(1), Worst: O(n)
The O(n) worst case for sets and dicts is very uncommon, but it can happen if __hash__ is implemented poorly. This only happens if everything in your set has the same hash value.
It depends entirely on the type of the container. Hashing containers (dict, set) use the hash and are essentially O(1). Typical sequences (list, tuple) are implemented as you guess and are O(n). Trees would be average O(log n). And so on. Each of these types would have an appropriate __contains__ method with its big-O characteristics.
It's amortized O(1), not O(1).
Let's say the list reserved size is 8 elements and it doubles in size when space runs out. You want to push 50 elements.
The first 8 elements push in O(1). The nineth triggers reallocation and 8 copies, followed by an O(1) push. The next 7 push in O(1). The seventeenth triggers reallocation and 16 copies, followed by an O(1) push. The next 15 push in O(1). The thirty-third triggers reallocation and 32 copies, followed by an O(1) push. The next 31 push in O(1). This continues as the size of list is doubled again at pushing the 65th, 129th, 257th element, etc..
So all of the pushes have O(1) complexity, we had 64 copies at O(1), and 3 reallocations at O(n), with n = 8, 16, and 32. Note that this is a geometric series and asymptotically equals O(n) with n = the final size of the list. That means the whole operation of pushing n objects onto the list is O(n). If we amortize that per element, it's O(n)/n = O(1).
If you look at the footnote in the document you linked, you can see that they include a caveat:
These operations rely on the "Amortized" part of "Amortized Worst Case". Individual actions may take surprisingly long, depending on the history of the container.
Using amortized analysis, even if we have to occasionally perform expensive operations, we can get a lower bound on the 'average' cost of operations when you consider them as a sequence, instead of individually.
So, any individual operation could be very expensive - O(n) or O(n^2) or something even bigger - but since we know these operations are rare, we guarantee that a sequence of O(n) operations can be done in O(n) time.