Related
Can someone help me understand how process_time() works?
My code is
from time import process_time
t = process_time()
def fibonacci_of(n):
if n in cache: # Base case
return cache[n]
# Compute and cache the Fibonacci number
cache[n] = fibonacci_of(n - 1) + fibonacci_of(n - 2) # Recursive case
return cache[n]
cache = {0: 0, 1: 1}
fib = [fibonacci_of(n) for n in range(1500)]
print(fib[-1])
print(process_time() - t)
And last print is always 0.0.
My expected result is something like 0.764891862869
Docs at https://docs.python.org/3/library/time.html#time.process_time don't help newbie me :(
I tried some other functions and reading docs. But without success.
I'd assume this is OS dependent. Linux lets me get down to ~5 microseconds using process_time, but other operating systems might well not deal well with differences this small and return zero instead.
It's for this reason that Python exposes other timers that are designed to be more accurate over shorter time scales. Specifically, perf_counter is specified as using:
the highest available resolution to measure a short duration
Using this lets me measure down to ~80 nanoseconds, whether I'm using perf_counter or perf_counter_ns.
As documentation suggests:
time.process_time() → float
Return the value (in fractional seconds) of the sum of the system and user
CPU time of the current process. It does not include time
elapsed during sleep. It is process-wide by definition. The reference
point of the returned value is undefined, so that only the difference
between the results of two calls is valid.
Use process_time_ns() to avoid the precision loss caused by the float type.
This last sentence is the most important and differentiates between the very precise function: process_time_ns() and one that is less precise but more appropriate for the long-running processes - time.process_time().
It turns out that when you measure a couple nanoseconds (nano means 10**-9) and try to express it in seconds dividing by 10**9, you often go out of the precision of float (64 bits) and end up rounding up to zero. The float limitations are described in the Python documentation.
To get to know more you can also read a general introduction to precision in floating point arithmetic (ambitious) and its peril (caveats).
Introduction
Today I found a weird behaviour in python while running some experiments with exponentiation and I was wondering if someone here knows what's happening. In my experiments, I was trying to check what is faster in python int**int or float**float. To check that I run some small snippets, and I found a really weird behaviour.
Weird results
My first approach was just to write some for loops and prints to check which one is faster. The snipper I used is this one
import time
# Run powers outside a method
ti = time.time()
for i in range(EXPERIMENTS):
x = 2**2
tf = time.time()
print(f"int**int took {tf-ti:.5f} seconds")
ti = time.time()
for i in range(EXPERIMENTS):
x = 2.**2.
tf = time.time()
print(f"float**float took {tf-ti:.5f} seconds")
After running it I got
int**int took 0.03004
float**float took 0.03070 seconds
Cool, it seems that data types do not affect the execution time. However, since I try to be a clean coder I refactored the repeated logic in a function power_time
import time
# Run powers in a method
def power_time(base, exponent):
ti = time.time()
for i in range(EXPERIMENTS):
x = base ** exponent
tf = time.time()
return tf-ti
print(f"int**int took {power_time(2, 2):.5f} seconds")
print(f"float**float took {power_time(2., 2.):5f} seconds")
And what a surprise of mine when I got these results
int**int took 0.20140 seconds
float**float took 0.05051 seconds
The refactor didn't affect a lot the float case, but it multiplied by ~7 the time required for the int case.
Conclusions and questions
Apparently, running something in a method can slow down your process depending on your data types, and that's really weird to me.
Also, if I run the same experiments but change ** by * or + the weird results disappear, and all the approaches give more or less the same results
Does someone know why is this happening? Am I missing something?
Apparently, running something in a method can slow down your process depending on your data types, and that's really weird to me.
It would be really weird if it was not like this! You can write your class that has it's own ** operator (through implementing the __pow__(self, other) method), and you could, for example, sleep 1s in there. Why should that take as long as taking a float to the power of another?
So, yeah, Python is a dynamically typed language. So, the operations done on data depend on the type of that data, and things can generally take different times.
In your first example, the difference never arises, because a) most probably the values get cached, because right after parsing it's clear that 2**2 is a constant and does not need to get evaluated every loop. Even if that's not the case, b) the time it costs to run a loop in python is hundreds of times that it takes to actually execute the math here – again, dynamically typed, dynamically named.
base**exponent is a whole different story. None about this is constant. So, there's actually going to be a calculation every iteration.
Now, the ** operator (__rpow__ in the Python data model) for Python's built-in float type is specified to do the float exponent (which is something implemented in highly optimized C and assembler), as exponentiation can elegantly be done on floating point numbers. Look for nb_power in cpython's floatobject.c. So, for the float case, the actual calculation is "free" for all that matters, again, because your loop is limited by how much effort it is to resolve all the names, types and functions to call in your loop. Not by doing the actual math, which is trivial.
The ** operator on Python's built-in int type is not as neatly optimized. It's a lot more complicated – it needs to do checks like "if the exponent is negative, return a float," and it does not do elementary math that your computer can do with a simple instruction, it handles arbitrary-length integers (remember, a python integer has as many bytes as it needs. You can save numbers that are larger than 64 bit in a Python integer!), which comes with allocation and deallocations. (I encourage you to read long_pow in CPython's longobject.c; it has 200 lines.)
All in all, integer exponentiation is expensive in python, because of python's type system.
I think I may have implemented this incorrectly because the results do not make sense. I have a Go program that counts to 1000000000:
package main
import (
"fmt"
)
func main() {
for i := 0; i < 1000000000; i++ {}
fmt.Println("Done")
}
It finishes in less than a second. On the other hand I have a Python script:
x = 0
while x < 1000000000:
x+=1
print 'Done'
It finishes in a few minutes.
Why is the Go version so much faster? Are they both counting up to 1000000000 or am I missing something?
One billion is not a very big number. Any reasonably modern machine should be able to do this in a few seconds at most, if it's able to do the work with native types. I verified this by writing an equivalent C program, reading the assembly to make sure that it actually was doing addition, and timing it (it completes in about 1.8 seconds on my machine).
Python, however, doesn't have a concept of natively typed variables (or meaningful type annotations at all), so it has to do hundreds of times as much work in this case. In short, the answer to your headline question is "yes". Go really can be that much faster than Python, even without any kind of compiler trickery like optimizing away a side-effect-free loop.
pypy actually does an impressive job of speeding up this loop
def main():
x = 0
while x < 1000000000:
x+=1
if __name__ == "__main__":
s=time.time()
main()
print time.time() - s
$ python count.py
44.221405983
$ pypy count.py
1.03511095047
~97% speedup!
Clarification for 3 people who didn't "get it". The Python language itself isn't slow. The CPython implementation is a relatively straight forward way of running the code. Pypy is another implementation of the language that does many tricky (especiallt the JIT) things that can make enormous differences. Directly answering the question in the title - Go isn't "that much" faster than Python, Go is that much faster than CPython.
Having said that, the code samples aren't really doing the same thing. Python needs to instantiate 1000000000 of its int objects. Go is just incrementing one memory location.
This scenario will highly favor decent natively-compiled statically-typed languages. Natively compiled statically-typed languages are capable of emitting a very trivial loop of say, 4-6 CPU opcodes that utilizes simple check-condition for termination. This loop has effectively zero branch prediction misses and can be effectively thought of as performing an increment every CPU cycle (this isn't entirely true, but..)
Python implementations have to do significantly more work, primarily due to the dynamic typing. Python must make several different calls (internal and external) just to add two ints together. In Python it must call __add__ (it is effectively i = i.__add__(1), but this syntax will only work in Python 3.x), which in turn has to check the type of the value passed (to make sure it is an int), then it adds the integer values (extracting them from both of the objects), and then the new integer value is wrapped up again in a new object. Finally it re-assigns the new object to the local variable. That's significantly more work than a single opcode to increment, and doesn't even address the loop itself - by comparison, the Go/native version is likely only incrementing a register by side-effect.
Java will fair much better in a trivial benchmark like this and will likely be fairly close to Go; the JIT and static-typing of the counter variable can ensure this (it uses a special integer add JVM instruction). Once again, Python has no such advantage. Now, there are some implementations like PyPy/RPython, which run a static-typing phase and should fare much better than CPython here ..
You've got two things at work here. The first of which is that Go is compiled to machine code and run directly on the CPU while Python is compiled to bytecode run against a (particularly slow) VM.
The second, and more significant, thing impacting performance is that the semantics of the two programs are actually significantly different. The Go version makes a "box" called "x" that holds a number and increments that by 1 on each pass through the program. The Python version actually has to create a new "box" (int object) on each cycle (and, eventually, has to throw them away). We can demonstrate this by modifying your programs slightly:
package main
import (
"fmt"
)
func main() {
for i := 0; i < 10; i++ {
fmt.Printf("%d %p\n", i, &i)
}
}
...and:
x = 0;
while x < 10:
x += 1
print x, id(x)
This is because Go, due to it's C roots, takes a variable name to refer to a place, where Python takes variable names to refer to things. Since an integer is considered a unique, immutable entity in python, we must constantly make new ones. Python should be slower than Go but you've picked a worst-case scenario - in the Benchmarks Game, we see go being, on average, about 25x times faster (100x in the worst case).
You've probably read that, if your Python programs are too slow, you can speed them up by moving things into C. Fortunately, in this case, somebody's already done this for you. If you rewrite your empty loop to use xrange() like so:
for x in xrange(1000000000):
pass
print "Done."
...you'll see it run about twice as fast. If you find loop counters to actually be a major bottleneck in your program, it might be time to investigate a new way of solving the problem.
#troq
I'm a little late to the party but I'd say the answer is yes and no. As #gnibbler pointed out, CPython is slower in the simple implementation but pypy is jit compiled for much faster code when you need it.
If you're doing numeric processing with CPython most will do it with numpy resulting in fast operations on arrays and matrices. Recently I've been doing a lot with numba which allows you to add a simple wrapper to your code. For this one I just added #njit to a function incALot() which runs your code above.
On my machine CPython takes 61 seconds, but with the numba wrapper it takes 7.2 microseconds which will be similar to C and maybe faster than Go. Thats an 8 million times speedup.
So, in Python, if things with numbers seem a bit slow, there are tools to address it - and you still get Python's programmer productivity and the REPL.
def incALot(y):
x = 0
while x < y:
x += 1
#njit('i8(i8)')
def nbIncALot(y):
x = 0
while x < y:
x += 1
return x
size = 1000000000
start = time.time()
incALot(size)
t1 = time.time() - start
start = time.time()
x = nbIncALot(size)
t2 = time.time() - start
print('CPython3 takes %.3fs, Numba takes %.9fs' %(t1, t2))
print('Speedup is: %.1f' % (t1/t2))
print('Just Checking:', x)
CPython3 takes 58.958s, Numba takes 0.000007153s
Speedup is: 8242982.2
Just Checking: 1000000000
Problem is Python is interpreted, GO isn't so there's no real way to bench test speeds. Interpreted languages usually (not always have a vm component) that's where the problem lies, any test you run is being run in interpreted bounds not actual runtime bounds. Go is slightly slower than C in terms of speed and that is mostly due to it using garbage collection instead of manual memory management. That said GO compared to Python is fast because its a compiled language, the only thing lacking in GO is bug testing I stand corrected if I'm wrong.
It is possible that the compiler realized that you didn't use the "i" variable after the loop, so it optimized the final code by removing the loop.
Even if you used it afterwards, the compiler is probably smart enough to substitute the loop with
i = 1000000000;
Hope this helps =)
I'm not familiar with go, but I'd guess that go version ignores the loop since the body of the loop does nothing. On the other hand, in the python version, you are incrementing x in the body of the loop so it's probably actually executing the loop.
Given log(a) and log(b), I want to compute log(a+b) (in a numerically stable way).
I wrote a little function for this:
def log_add(logA,logB):
if logA == log(0):
return logB
if logA<logB:
return log_add(logB,logA)
return log( 1 + math.exp(logB-logA) ) + logA
I wrote a program where this is by far the most time-consuming piece of code. Obviously I could try to optimize it (eliminate the recursive call, for instance).
Do you know of a standard math or numpy function for computing log(a+b) from log(a) and log(b)?
If not, do you know of a simple way to make a single C++ hook for this function? It's not a complicated function (it uses floats), and as I said, it's taking up the majority of my runtime.
Thanks in advance, numerical methods ninja!
Note: Best answer until now is to simply use numpy.logaddexp(logA,logB).
Why exactly do you compare with log(0)? This is equal to -numpy.inf, in this case you come to log(1 + math.exp(-inf-logB) ) + logB Which reduces itself to logB. This call always will give an warning message which is extremely slow.
I could come up with this one-liner. However you'll need to really measure to see if this is actually faster. It does only use one 'complex' calculation function instead of the two that you use, and no recursion is happening, the if is still there but hidden (and maybe optimized) in fabs/maximum.
def log_add(logA,logB):
return numpy.logaddexp(0,-numpy.fabs(logB-logA)) + numpy.maximum(logA,logB)
edit:
I did a quick timeit() with following results :
Your original version took about 120s
My version took about 30s
I removed the compare with log(0) from your version and it came down to 20s
I edited my code to keep the logaddexp but also worked with your recursive if and it went down to 18s.
Updated code, you could also switch the recursive call with an inline updated formula but this made little difference in my timing tests:
def log_add2(logA, logB):
if logA < logB:
return log_add2(logB, logA)
return numpy.logaddexp(0,logB-logA)+logA
Edit 2:
As pv noted in comments, you could actually just do numpy.logaddexp(logA, logB) which comes down to calculating log(exp(logA)+exp(logB)) which is of course equal to log(A+B). I timed it (on the same machine as above) and it went further down to about 10s. So we've come down to about 1/12, not bad ;).
def log_add(logA, logB):
return math.log(math.exp(logA) + math.exp(logB))
is too slow? or inaccurate?
Here are two programs that naively calculate the number of prime numbers <= n.
One is in Python and the other is in Java.
public class prime{
public static void main(String args[]){
int n = Integer.parseInt(args[0]);
int nps = 0;
boolean isp;
for(int i = 1; i <= n; i++){
isp = true;
for(int k = 2; k < i; k++){
if( (i*1.0 / k) == (i/k) ) isp = false;
}
if(isp){nps++;}
}
System.out.println(nps);
}
}
`#!/usr/bin/python`
import sys
n = int(sys.argv[1])
nps = 0
for i in range(1,n+1):
isp = True
for k in range(2,i):
if( (i*1.0 / k) == (i/k) ): isp = False
if isp == True: nps = nps + 1
print nps
Running them on n=10000 I get the following timings.
shell:~$ time python prime.py 10000 && time java prime 10000
1230
real 0m49.833s
user 0m49.815s
sys 0m0.012s
1230
real 0m1.491s
user 0m1.468s
sys 0m0.016s
Am I using for loops in python in an incorrect manner here or is python actually just this much slower?
I'm not looking for an answer that is specifically crafted for calculating primes but rather I am wondering if python code is typically utilized in a smarter fashion.
The Java code was compiled with
javac 1.6.0_20
Run with java version "1.6.0_18"
OpenJDK Runtime Environment (IcedTea6 1.8.1) (6b18-1.8.1-0ubuntu1~9.10.1)
OpenJDK Client VM (build 16.0-b13, mixed mode, sharing)
Python is:
Python 2.6.4 (r264:75706, Dec 7 2009, 18:45:15)
As has been pointed out, straight Python really isn't made for this sort of thing. That the prime checking algorithm is naive is also not the point. However, with two simple things I was able to greatly reduce the time in Python while using the original algorithm.
First, put everything inside of a function, call it main() or something. This decreased the time on my machine in Python from 20.6 seconds to 14.54 seconds. Doing things globally is slower than doing them in a function.
Second, use Psyco, a JIT compiler. This requires adding two lines to the top of the file (and of course having psyco installed):
import psyco
psyco.full()
This brought the final time to 2.77 seconds.
One last note. I decided for kicks to use Cython on this and got the time down to 0.8533. However, knowing how to make the few changes to make it fast Cython code isn't something that I recommend for the casual user.
Yes, Python is slow, about a hundred times slower than C. You can use xrange instead of range for a small speedup, but other than that it's fine.
Ultimately what you're doing wrong is that you do this in plain Python, instead of using optimized libraries such as Numpy or Psyco.
Java comes with a jit compiler that makes a big difference where you're just crunching numbers.
You can make your Python about twice as fast by replacing that complicated test with
if i % k == 0: isp = False
You can also make it about eight times faster (for n=10000) than that by adding a break after that isp = False.
Also, do yourself a favor and skip the even numbers (adding one to nps to start to include 2).
Finally, you only need k to go up to sqrt(i).
Of course, if you make the same changes in the Java, it's still about 10x faster than the optimized Python.
Boy, when you said it was a naive implementation, you sure weren't joking!
But yes, a one to two order of magnitude difference in performance is not unexpected when comparing JIT-compiled, optimized machine code with an interpreted language. An alternative Python implementation such as Jython, which runs on the Java VM, may well be faster for this task; you could give it a whirl. Cython, which allows you to add static typing to Python and get C-like performance in some cases, may be worth investigating as well.
Even when considering the standard Python interpreter, CPython, though, the question is: is Python fast enough for the task at hand? Will the time you save writing the code in a dynamic language like Python make up for the extra time spent running it? If you had to write a given program in Java, would it seem like too much work to be worth the trouble?
Consider, for example, that a Python program running on a modern computer will be about as fast as a Java program running on a 10-year-old computer. The computer you had ten years ago was fast enough for many things, wasn't it?
Python does have a number of features that make it great for numerical work. These include an integer type that supports an unlimited number of digits, a decimal type with unlimited precision, and an optional library called NumPy specifically for calculations. Speed of execution, however, is not generally one of its major claims to fame. Where it excels is in getting the computer to do what you want with minimal cognitive friction.
If you're looking to do it fast, Python probably isn't the way forward, but you could speed it up a bit. First, you're using quite a slow way to test for divisibility. Modulo is quicker. You can also stop the inner loop (with k) as soon as it detects a match. I'd do something like this:
nps = 0
for i in range(1, n+1):
if all(i % k for k in range(2, i)): # i.e. if divisible by none of them
nps += 1
That brings it down from 25 s to 1.5 s for me. Using xrange brings it down to 0.9 s.
You could speed it up further by keeping a list of primes you've already found, and only testing those, rather than every number up to i (if i isn't divisible by 2, it won't be divisible by 4, 6, 8...).
Why don't you post something about the memory usage - and not just the speed? Trying to get a simple servlet on tomcat is wasting 3GB on my server.
What you did with the examples up there is not very good. You need to use numpy. Replace for/range with while loops, thus avoiding the list creation.
At last, python is quite suitable for number crunching, at least by people that do it the right way, and know what Sieve of Eratosthenes is, or mod operation is.
There are lots of things you can do to this algorithm to speed it up, but most of them would also speed up the Java version as well. Some of those will speed up the Python more than the Java, so they're worth testing.
Here's just a couple of changes that speed it up from 11.4 to 2.8 seconds on my system:
nps = 0
for i in range(1,n+1):
isp = True
for k in range(2,i):
isp = isp and (i % k != 0)
if isp: nps = nps + 1
print nps
Python is a language which, ironically, is well-suited for developing algorithms. Even a modified algorithm like this:
# See Thomas K for use of all(), many posters for sqrt optimization
nps = 0
for i in xrange(1, n+1):
if all(i % k for k in xrange(2, 1 + int(i ** 0.5))):
nps += 1
runs in significantly under one second. Code like this:
def eras(n):
last = n + 1
sieve = [0,0] + range(2, last)
sqn = int(round(n ** 0.5))
it = (i for i in xrange(2, sqn + 1) if sieve[i])
for i in it:
sieve[i*i:last:i] = [0] * (n//i - i + 1)
return filter(None, sieve)
is faster still. Or try out these.
The thing is, python is usually fast enough for designing your solution. If it is not fast enough for production, use numpy or Jython to goose more performance out of it. Or move it to a compiled language, taking your algorithm observations learned in python with you.
Yes, Python is one of the slowest practical languages you'll encounter. While loops are marginally faster than for i in xrange(), but ultimately Python will always be much, much slower than anything else.
Python has its place: Prototyping theory and ideas, or in any situation where the ability to produce code fast is more important than the code's performance.
Python is a scripting language. Not a programming language.