print sum(1 for x in alist if x[1] == 8)
This code runs fine, but it is so slow. Is there a way better than this. Because, my list is very large and the computation takes a lot of time. Do you know a better and faster way to do it?
You'd have to create indexes or cached counts to speed up such code; trade memory for speed.
Wherever you handle your list (add to it, remove from it, edit entries) you also maintain your indices. For example, if you had a counts dict with ids as keys and their frequency as values, all you had to do is look up the count directly, and ensure that the counts stayed up-to-date as you manipulate alist.
The best way to manage this is by encapsulating your list in a custom type, so that you can control all manipulations of the data structure and maintain the extra information.
Not sure how much faster it would be but
len([x for x in alist if x[1] == 8])
is a little clearer.
I would use numpy. My numpy skills are a little bit rusty, but len(np_array == 8) would give you what you need for a single depth array. I think for you it would be len(np_array[:,1]) but I would have to check (this assumes your problem could use numpy arrays)
Related
I want to perform calculations on a list and assign this to a second list, but I want to do this in the most efficient way possible as I'll be using a lot of data. What is the best way to do this? My current version uses append:
f=time_series_data
output=[]
for i, f in enumerate(time_series_data):
if f > x:
output.append(calculation with f)
etc etc
should I use append or declare the output list as a list of zeros at the beginning?
Appending the values is not slower compared to other ways possible to accomplish this.
The code looks fine and creating a list of zeroes would not help any further. Although it can create problems as you might not know how many values will pass the condition f > x.
Since you wrote etc etc I am not sure how long or what operations you need to do there. If possible try using list comprehension. That would be a little faster.
You can have a look at below article which compared the speed for list creation using 3 methods, viz, list comprehension, append, pre-initialization.
https://levelup.gitconnected.com/faster-lists-in-python-4c4287502f0a
Many of us know that, enumerate is being using in a situation you use the for loop and need to know the index. However, it has its downsides. According to my tests with the timeit module, just using enumerate makes the code 2x slower. Adding this a tuple assignment makes it slower up to 3x. These numbers may come as fast enough for any programmer, but people dealing with algorithms know that every bit of code you can optimize, is a huge advantage. Now to my question,
An example of this usage would be, the need of finding indexes of multiple elements in a list. Say that there is two elements we need to find. The first two solutions that occur to me is like so:
x, y = 0, 0
for ind, val in enumerate(lst):
if x and y:
break
if val == "a":
x = ind
elif val == "b":
y = ind
The solution above iterates the list, assign the values, than break if the two is found.
x = lst.index("a")
y = lst.index("b")
This is an other solution, which I didn't want to use because it appeared really naive. It iterates over the same list twice, to find two elements. The first solution, does this in a single iteration. So by complexity terms, even though we make extra assignments in the first solution, it should be faster than the second one in larger lists. But my assumption failed.
Here is the code I tested the performance: https://codeshare.io/XfvGA
The second solution was 2x to 10x faster than the first one, changing with the position of these two elements. There are several possibilities which this would occur.
There is an optimization in index() method that I am unaware of.
Lower level assignments being made in index() method. Possible use of C++ code.
The conditions and extra assignments in the first solution, makes it slower than expected.
Even these reasons fall short of explaining the speed of iterating the list twice over iterating it once. Though languages have much difference in time while running code, iteration process itself is independant from the programming language, if you need to check a million elements, you still have to check a million elements (Could be exampled by map() being not much faster than using a loop to change values).
So though I need you to examine the cases I presented, in order to clarify what is being asked here, question can be put together like this. We know that Python's for loop is actually a while running in background (possibly in C ?). So this means, the index is being stored as it is incremented somewhere in the memory. If there was a way to access it, this would eliminate the cost of calling and unpacking enumerate. My question is:
Is there such a way exists ?, If not, could be made (why, or why not) ?
The sources I used for more information on the subject:
Python speed
Python objects time complexity
Performance tips for Python
I dont think that the enumerate is the problem, to prove this you can do:
x, y = 0, 0
for val in a:
if x and y:
break
if val == "a":
x = val
elif val == "b":
y = val
This doesnt do the same thing you wanted in the first place (you dont get the index) but if you messure it with timeit, you will find that the diffrence is not so significant, meaning that the enumerate is not the source of the problem ( in my case it was 0.185 to 0.155 when running your example, so it is faster but the second solution got 0.055 at my computer )
The reason that lst.index is faster is that it is implemented in C .
You can see it's source code here:
https://svn.python.org/projects/python/trunk/Objects/listobject.c
the index function is called listindex in this file and is defined like
static PyObject *
listindex(PyListObject *self, PyObject *args)
( i couldnt find a way to add a link directly to the function )
You are trying to be un-Pythonic, which isn't going to end terribly well for you. If you really need to have that iterator count information available, there is a well-known and optimized way to do that: enumerate(). If you need to find an item in a list, there is a well-known and optimized way to do that: lst.index(). As DorElias showed above/below, enumerate is not the problem, it's that you're attempting to reinvent the wheel with the rest of your for loop. enumerate is going to be the best-supported (clearest, fastest, etc.) way to maintain an iteration count in every situation where an iteration count is actually the thing you need.
All,
I am going to compute some feature values using the following python codes. But, because the input sizes are too big, it is very time-consuming. Please help me to optimize the codes.
leaving_volume=len([x for x in pickup_ids if x not in dropoff_ids])
arriving_volume=len([x for x in dropoff_ids if x not in pickup_ids])
transition_volume=len([x for x in dropoff_ids if x in pickup_ids])
union_ids=list(set(pickup_ids + dropoff_ids))
busstop_ids=[x for x in union_ids if self.geoitems[x].fare>0]
busstop_density=np.sum([Util.Geodist(self.geoitems[x].orilat, self.geoitems[x].orilng, self.geoitems[x].destlat, self.geoitems[x].destlng)/(1000*self.geoitems[x].fare) for x in busstop_ids])/len(busstop_ids) if len(busstop_ids) > 0 else 0
busstop_ids=[x for x in union_ids if self.geoitems[x].balance>0]
smartcard_balance=np.sum([self.geoitems[x].balance for x in busstop_ids])/len(busstop_ids) if len(busstop_ids) > 0 else 0
Hi, All,
Here is my revised version. I run this code on my GPS traces data. It is faster.
intersect_ids=set(pickup_ids).intersection( set(dropoff_ids) )
union_ids=list(set(pickup_ids + dropoff_ids))
leaving_ids=set(pickup_ids)-intersect_ids
leaving_volume=len(leaving_ids)
arriving_ids=set(dropoff_ids)-intersect_ids
arriving_volume=len(arriving_ids)
transition_volume=len(intersect_ids)
busstop_density=np.mean([Util.Geodist(self.geoitems[x].orilat, self.geoitems[x].orilng, self.geoitems[x].destlat, self.geoitems[x].destlng)/(1000*self.geoitems[x].fare) for x in union_ids if self.geoitems[x].fare>0])
if not busstop_density > 0:
busstop_density = 0
smartcard_balance=np.mean([self.geoitems[x].balance for x in union_ids if self.geoitems[x].balance>0])
if not smartcard_balance > 0:
smartcard_balance = 0
Many thanks for the help.
Just a few things I noticed, as some Python efficiency trivia:
if x not in dropoff_ids
Checking for membership using the in operator is more efficient on a set than a list. But iterating with for through a list is probably more efficient than on a set. So if you want your first two lines to be as efficient as possible you should have both types of data structure around beforehand.
list(set(pickup_ids + dropoff_ids))
It's more efficient to create your sets before you combine data, rather than creating a long list and constructing a set from it. Luckily you probably already have the set versions around now (see the first comment)!
Above all you need to ask yourself the question:
Is the time I save by constructing extra data structures worth the time it takes to construct them?
Next one:
np.sum([...])
I've been trained by Python to think of constructing a list and then applying a function that theoretically only requires a generator as a code smell. I'm not sure if this applies in numpy, since from what I remember it's not completely straightforward to pull data from a generator and put it in a numpy structure.
It looks like this is just a small fragment of your code. If you're really concerned about efficiency I'd recommend making use of numpy arrays rather than lists, and trying to stick within numpy's built-in data structures and function as much as possible. They are likely more highly optimized for raw data crunching in C than the built-in Python functions.
If you're really, really concerned about efficiency then you should probably be doing this data analysis straight-up in C. Especially if you don't have much more code than what you've presented here it might be pretty easy to translate over.
I can only support what machine yerning wrote in his this post. If you are thinking of switching to numpy so if your variables pickup_ids and dropoff_ids were numpy arrays (which maybe they already are else do:
dropoff_ids = np.array( dropoff_ids, dtype='i' )
pickup_ids = np.array( pickup_ids, dtype='i' )
then you can make use of the functions np.in1d() which will give you a True/False array which you can just sum over to get the total number of True entries.
leaving_volume = (-np.in1d( pickup_ids, dropoff_ids )).sum()
transition_volume= np.in1d( dropoff_ids, pickup_ids).sum()
arriving_volume = (-np.in1d( dropoff_ids, pickup_ids)).sum()
somehow I have the feeling that transition_volume = len(pickup_ids) - arriving_volume but I'm not 100% sure right now.
Another function that could be useful to you is np.unique() if you want to get rid of duplicate entries which in a way will turn your array into a set.
I am new to python and my problem is the following:
I have defined a function func(a,b) that return a value, given two input values.
Now I have my data stored in lists or numpy arrays A,Band would like to use func for every combination. (A and B have over one million entries)
ATM i use this snippet:
for p in A:
for k in B:
value = func(p,k)
This takes really really a lot of time.
So i was thinking that maybe something like this:
C=(map(func,zip(A,B)))
But this method only works pairwise... Any ideas?
Thanks for help
First issue
You need to calculate the output of f for many pairs of values. The "standard" way to speed up this kind of loops (calculations) is to make your function f accept (NumPy) arrays as input, and do the calculation on the whole array at once (ie, no looping as seen from Python). Check any NumPy tutorial to get an introduction.
Second issue
If A and B have over a million entries each, there are one trillion combinations. For 64 bits numbers, that means you'll need 7.3 TiB of space just to store the result of your calculation. Do you have enough hard drive to just store the result?
Third issue
If A and B where much smaller, in your particular case you'd be able to do this:
values = f(*meshgrid(A, B))
meshgrid returns the cartesian product of A and B, so it's simply a way to generate the points that have to be evaluated.
Summary
You need to use NumPy effectively to avoid Python loops. (Or if all else fails or they can't easily be vectorized, write those loops in a compiled language, for instance by using Cython)
Working with terabytes of data is hard. Do you really need that much data?
Any solution that calls a function f 1e12 times in a loop is bound to be slow, specially in CPython (which is the default Python implementation. If you're not really sure and you're using NumPy, you're using it too).
suppose, itertools.product does what you need:
from itertools import product
pro = product(A,B)
C = map(lambda x: func(*x), pro)
so far as it is generator it doesn't require additional memory
One million times one million is one trillion. Calling f one trillion times will take a while.
Unless you have a way of reducing the number of values to compute, you can't do better than the above.
If you use NumPy, you should definitely look the np.vectorize function which is designed for this kind of problems...
how would i search through a list with ~5 mil 128bit (or 256, depending on how you look at it) strings quickly and find the duplicates (in python)? i can turn the strings into numbers, but i don't think that's going to help much. since i haven't learned much information theory, is there anything about this in information theory?
and since these are hashes already, there's no point in hashing them again
If it fits into memeory, use set(). I think it will be faster than sort. O(n log n) for 5 million items is going to cost you.
If it does not fit into memory, say you've lot more than 5 million record, divide and conquer. Break the records at the mid point like 1 x 2^127. Apply any of the above methods. I guess information theory helps by stating that a good hash function will distribute the keys evenly. So the divide by mid point method should work great.
You can also apply divide and conquer even if it fit into memory. Sorting 2 x 2.5 mil records is faster than sorting 5 mil records.
Load them into memory (5M x 64B = 320MB), sort them, and scan through them finding the duplicates.
In Python2.7+ you can use collections.Counter for older Python use collections.deaultdict(int). Either way is O(n).
first make a list with some hashes in it
>>> import hashlib
>>> s=[hashlib.sha1(str(x)).digest() for x in (1,2,3,4,5,1,2)]
>>> s
['5j\x19+y\x13\xb0LTWM\x18\xc2\x8dF\xe69T(\xab', '\xdaK\x927\xba\xcc\xcd\xf1\x9c\x07`\xca\xb7\xae\xc4\xa85\x90\x10\xb0', 'w\xdeh\xda\xec\xd8#\xba\xbb\xb5\x8e\xdb\x1c\x8e\x14\xd7\x10n\x83\xbb', '\x1bdS\x89$s\xa4g\xd0sr\xd4^\xb0Z\xbc 1dz', '\xac4x\xd6\x9a<\x81\xfab\xe6\x0f\\6\x96\x16ZN^j\xc4', '5j\x19+y\x13\xb0LTWM\x18\xc2\x8dF\xe69T(\xab', '\xdaK\x927\xba\xcc\xcd\xf1\x9c\x07`\xca\xb7\xae\xc4\xa85\x90\x10\xb0']
If you are using Python2.7 or later
>>> from collections import Counter
>>> c=Counter(s)
>>> duplicates = [k for k in c if c[k]>1]
>>> print duplicates
['\xdaK\x927\xba\xcc\xcd\xf1\x9c\x07`\xca\xb7\xae\xc4\xa85\x90\x10\xb0', '5j\x19+y\x13\xb0LTWM\x18\xc2\x8dF\xe69T(\xab']
if you are using Python2.6 or earlier
>>> from collections import defaultdict
>>> d=defaultdict(int)
>>> for i in s:
... d[i]+=1
...
>>> duplicates = [k for k in d if d[k]>1]
>>> print duplicates
['\xdaK\x927\xba\xcc\xcd\xf1\x9c\x07`\xca\xb7\xae\xc4\xa85\x90\x10\xb0', '5j\x19+y\x13\xb0LTWM\x18\xc2\x8dF\xe69T(\xab']
Is this array sorted?
I think the fastest solution can be a heap sort or quick sort, and after go through the array, and find the duplicates.
You say you have a list of about 5 million strings, and the list may contain duplicates. You don't say (1) what you want to do with the duplicates (log them, delete all but one occurrence, ...) (2) what you want to do with the non-duplicates (3) whether this list is a stand-alone structure or whether the strings are keys to some other data that you haven't mentioned (4) why you haven't deleted duplicates at input time instead building a list containing duplicates.
As a Data Structures and Algorithms 101 exercise, the answer you have accepted is a nonsense. If you have enough memory, detecting duplicates using a set should be faster than sorting a list and scanning it. Note that deleting M items from a list of size N is O(MN). The code for each of the various alternatives is short and rather obvious; why don't you try writing them, timing them, and reporting back?
If this is a real-world problem that you have, you need to provide much more information if you want a sensible answer.