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how to resolve this "inf" problem with python code

i have a problem with this python code for inverting a Number
like Nb = 358 ---> inv = 853
but in the end i got 'inf' msg from the prog , and its runs normally in C language
def envers(Nb):
inv = 0
cond = True
while cond:
s = Nb % 10
inv = (inv*10)+ s
Nb = Nb/10
if Nb == 0:
cond = False
return inv
data = int(input("give num"))
res = envers(data)
print(res)

This is likely much easier to do via string manipulation, which has a friendly and simple syntax (which are a major reason to choose to use Python)
>>> int(input("enter a number to reverse: ")[::-1])
enter a number to reverse: 1234
4321
How this works
input() returns a string
strings are iterable and [::-1] is used to reverse it
finally convert to an int
Add error checking to taste (for example to to ensure you really received a number)

Here is a simple implementation for a numerical approach:
def envers(Nb):
out = 0
while Nb>0:
Nb, r = divmod(Nb, 10)
out = 10*out + r
return out
envers(1234)
# 4321
envers(358)
# 853
envers(1020)
# 201
Without divmod:
def envers(Nb):
out = 0
while Nb>0:
r = Nb % 10
Nb //= 10
out = 10*out + r
return out

When you set
Nb = Nb / 10
Nb becomes a float (say 1 -> 0.1) and will be able to keep being divided until it reaches a certain limit.
By that point, your inv value will reach python's limits and become 'inf'.
Replacing this line with
Nb = Nb // 10
using Python's builtin integer division will fix the issue.

Numba dictionary: signature in JIT() decorator

My function takes a list of numpy arrays and a dictionary (or a list of dictionaries) as input arguments and returns a list of values. The list of numpy arrays is long, and arrays may be of different shape. Though I can pass numpy arrays separately, for housekeeping purposes I really would like to form a tuple of numpy arrays and pass them as such into my function.
Without dictionary (which is specially formed according to numba >=0.43) the whole setup works fine - see the script below. Because the structure of input and output is of Tuple form, JIT requires signature - it cannot figure out the type of data structure without it. However no matter how I try to declare my dictionary 'd' into the JIT decorator, I cannot manage to get the script working.
Please help with ideas or a solution if one exists.
Many thanks
'''python:
import numpy as np
from numba import njit
from numba import types
from numba.typed import Dict
#njit( 'Tuple( (f8,f8) )(Tuple( (f8[:],f8[:]) ))' )
def somefunction(lst_arr):
arr1, arr2 = lst_arr
summ = 0
prod = 1
for i in arr2:
summ += i
for j in arr1:
prod *= j
result = (summ,prod)
return result
a = np.arange(5)+1.0
b = np.arange(5)+11.0
arg = (a,b)
print(a,b)
print(somefunction(arg))
# ~~ The Dict.empty() constructs a typed dictionary.
d = Dict.empty(
key_type=types.unicode_type,
value_type=types.float64,)
d['k1'] = 1.5
d['k2'] = 0.5
'''
I expect to pass 'd'-dictionary into 'somefunction' and use it inside with dict keys...Form example as follows: result = (summ * d['k1'], prod * d['k2'])
import numpy as np
from numba import njit
from numba import types
from numba.typed import Dict
#njit( 'Tuple( (f8,f8) )(Tuple( (f8[:],f8[:]) ), Dict)' )
def somefunction(lst_arr, mydict):
arr1, arr2 = lst_arr
summ = 0
prod = 1
for i in arr2:
summ += i
for j in arr1:
prod *= j
result = (summ*mydict['k1'],prod*mydict['k2'])
return result
# ~~ Input numpy arrays
a = np.arange(5)+1.0
b = np.arange(5)+11.0
arg = (a,b)
# ~~ Input dictionary for the function
d = Dict.empty(
key_type=types.unicode_type,
value_type=types.float64)
d['k1'] = 1.5
d['k2'] = 0.5
# ~~ Run function and print results
print(somefunction(arg, d))

I am using the version 0.45.1. You can simply pass the dictionary without having to declare the type in the dictionary:
d = Dict.empty(
key_type=types.unicode_type,
value_type=types.float64[:],
)
d['k1'] = np.arange(5) + 1.0
d['k2'] = np.arange(5) + 11.0
# Numba will infer the type on it's own.
#njit
def somefunction2(d):
prod = 1
# I am assuming you want sum of second array and product of second
result = (d['k2'].sum(), d['k1'].prod())
return result
print(somefunction(d))
# Output : (65.0, 120.0)
For reference, you check this example from the official documentation.
Update:
In your case you can simply let jit infer the types on it's own and it should work, the following code works for me:
import numpy as np
from numba import njit
from numba import types
from numba.typed import Dict
from numba.types import DictType
# Let jit infer the types on it's own
#njit
def somefunction(lst_arr, mydict):
arr1, arr2 = lst_arr
summ = 0
prod = 1
for i in arr2:
summ += i
for j in arr1:
prod *= j
result = (summ*mydict['k1'],prod*mydict['k2'])
return result
# ~~ Input numpy arrays
a = np.arange(5)+1.0
b = np.arange(10)+11.0 #<--------------- This is of different shape
arg = (a,b)
# ~~ Input dictionary for the function
d = Dict.empty(
key_type=types.unicode_type,
value_type=types.float64)
d['k1'] = 1.5
d['k2'] = 0.5
# This works now
print(somefunction(arg, d))
You can see the official documentation here:
Unless necessary, it is recommended to let Numba infer argument types by using the signature-less variant of #jit.
I tried various methods, but this is the only one that worked for the problem you specified.

accelerated FFT to be invoked from Python Numba CUDA kernel

I need to calculate the Fourier transform of a 256 element float64 signal. The requirement is as such that I need to invoke these FFTs from inside a cuda.jitted section and it must be completed within 25usec. Alas cuda.jit-compiled functions do not allow to invoke external libraries => I wrote my own. Alas my single-core code is still way too slow (~250usec on a Quadro P4000). Is there a better way?
I created a single core FFT-function that gives correct results, but is alas 10x too slow. I don't understand how to make good use of multiple cores.
---fft.py
from numba import cuda, boolean, void, int32, float32, float64, complex128
import math, sys, cmath
def _transform_radix2(vector, inverse, out):
n = len(vector)
levels = int32(math.log(float32(n))/math.log(float32(2)))
assert 2**levels==n # error: Length is not a power of 2
#uncomment either Numba.Cuda or Numpy memory allocation, (intelligent conditional compileation??)
exptable = cuda.local.array(1024, dtype=complex128)
#exptable = np.zeros(1024, np.complex128)
assert (n // 2) <= len(exptable) # error: FFT length > MAXFFTSIZE
coef = complex128((2j if inverse else -2j) * math.pi / n)
for i in range(n // 2):
exptable[i] = cmath.exp(i * coef)
for i in range(n):
x = i
y = 0
for j in range(levels):
y = (y << 1) | (x & 1)
x >>= 1
out[i] = vector[y]
size = 2
while size <= n:
halfsize = size // 2
tablestep = n // size
for i in range(0, n, size):
k = 0
for j in range(i, i + halfsize):
temp = out[j + halfsize] * exptable[k]
out[j + halfsize] = out[j] - temp
out[j] += temp
k += tablestep
size *= 2
scale=float64(n if inverse else 1)
for i in range(n):
out[i]=out[i]/scale # the inverse requires a scaling
# now create the Numba.cuda version to be called by a GPU
gtransform_radix2 = cuda.jit(device=True)(_transform_radix2)
---test.py
from numba import cuda, void, float64, complex128, boolean
import cupy as cp
import numpy as np
import timeit
import fft
#cuda.jit(void(float64[:],boolean, complex128[:]))
def fftbench(y, inverse, FT):
Y = cuda.local.array(256, dtype=complex128)
for i in range(len(y)):
Y[i]=complex128(y[i])
fft.gtransform_radix2(Y, False, FT)
str='\nbest [%2d/%2d] iterations, min:[%9.3f], max:[%9.3f], mean:[%9.3f], std:[%9.3f] usec'
a=[127.734375 ,130.87890625 ,132.1953125 ,129.62109375 ,118.6015625
,110.2890625 ,106.55078125 ,104.8203125 ,106.1875 ,109.328125
,113.5 ,118.6640625 ,125.71875 ,127.625 ,120.890625
,114.04296875 ,112.0078125 ,112.71484375 ,110.18359375 ,104.8828125
,104.47265625 ,106.65625 ,109.53515625 ,110.73828125 ,111.2421875
,112.28125 ,112.38671875 ,112.7734375 ,112.7421875 ,113.1328125
,113.24609375 ,113.15625 ,113.66015625 ,114.19921875 ,114.5
,114.5546875 ,115.09765625 ,115.2890625 ,115.7265625 ,115.41796875
,115.73828125 ,116. ,116.55078125 ,116.5625 ,116.33984375
,116.63671875 ,117.015625 ,117.25 ,117.41015625 ,117.6640625
,117.859375 ,117.91015625 ,118.38671875 ,118.51171875 ,118.69921875
,118.80859375 ,118.67578125 ,118.78125 ,118.49609375 ,119.0078125
,119.09375 ,119.15234375 ,119.33984375 ,119.31640625 ,119.6640625
,119.890625 ,119.80078125 ,119.69140625 ,119.65625 ,119.83984375
,119.9609375 ,120.15625 ,120.2734375 ,120.47265625 ,120.671875
,120.796875 ,120.4609375 ,121.1171875 ,121.35546875 ,120.94921875
,120.984375 ,121.35546875 ,120.87109375 ,120.8359375 ,121.2265625
,121.2109375 ,120.859375 ,121.17578125 ,121.60546875 ,121.84375
,121.5859375 ,121.6796875 ,121.671875 ,121.78125 ,121.796875
,121.8828125 ,121.9921875 ,121.8984375 ,122.1640625 ,121.9375
,122. ,122.3515625 ,122.359375 ,122.1875 ,122.01171875
,121.91015625 ,122.11328125 ,122.1171875 ,122.6484375 ,122.81640625
,122.33984375 ,122.265625 ,122.78125 ,122.44921875 ,122.34765625
,122.59765625 ,122.63671875 ,122.6796875 ,122.6171875 ,122.34375
,122.359375 ,122.7109375 ,122.83984375 ,122.546875 ,122.25390625
,122.06640625 ,122.578125 ,122.7109375 ,122.83203125 ,122.5390625
,122.2421875 ,122.06640625 ,122.265625 ,122.13671875 ,121.8046875
,121.87890625 ,121.88671875 ,122.2265625 ,121.63671875 ,121.14453125
,120.84375 ,120.390625 ,119.875 ,119.34765625 ,119.0390625
,118.4609375 ,117.828125 ,117.1953125 ,116.9921875 ,116.046875
,115.16015625 ,114.359375 ,113.1875 ,110.390625 ,108.41796875
,111.90234375 ,117.296875 ,127.0234375 ,147.58984375 ,158.625
,129.8515625 ,120.96484375 ,124.90234375 ,130.17578125 ,136.47265625
,143.9296875 ,150.24609375 ,141. ,117.71484375 ,109.80859375
,115.24609375 ,118.44140625 ,120.640625 ,120.9921875 ,111.828125
,101.6953125 ,111.21484375 ,114.91015625 ,115.2265625 ,118.21875
,125.3359375 ,139.44140625 ,139.76953125 ,135.84765625 ,137.3671875
,141.67578125 ,139.53125 ,136.44921875 ,135.08203125 ,135.7890625
,137.58203125 ,138.7265625 ,154.33203125 ,172.01171875 ,152.24609375
,129.8046875 ,125.59375 ,125.234375 ,127.32421875 ,132.8984375
,147.98828125 ,152.328125 ,153.7734375 ,155.09765625 ,156.66796875
,159.0546875 ,151.83203125 ,138.91796875 ,138.0546875 ,140.671875
,143.48046875 ,143.99609375 ,146.875 ,146.7578125 ,141.15234375
,141.5 ,140.76953125 ,140.8828125 ,145.5625 ,150.78125
,148.89453125 ,150.02734375 ,150.70703125 ,152.24609375 ,148.47265625
,131.95703125 ,125.40625 ,123.265625 ,123.57421875 ,129.859375
,135.6484375 ,144.51171875 ,155.05078125 ,158.4453125 ,140.8125
,100.08984375 ,104.29296875 ,128.55078125 ,139.9921875 ,143.38671875
,143.69921875 ,137.734375 ,124.48046875 ,116.73828125 ,114.84765625
,113.85546875 ,117.45703125 ,122.859375 ,125.8515625 ,133.22265625
,139.484375 ,135.75 ,122.69921875 ,115.7734375 ,116.9375
,127.57421875]
y1 =cp.zeros(len(a), cp.complex128)
FT1=cp.zeros(len(a), cp.complex128)
for i in range(len(a)):
y1[i]=a[i] #convert to complex to feed the FFT
r=1000
series=sorted(timeit.repeat("fftbench(y1, False, FT1)", number=1, repeat=r, globals=globals()))
series=series[0:r-5]
print(str % (len(series), r, 1e6*np.min(series), 1e6*np.max(series), 1e6*np.mean(series), 1e6*np.std(series)));
a faster implementation t<<25usec

The drawback of your algorithm is that even on GPU it runs on a single-core.
In order to understand how to design algorithms on Nvidia GPGPU I recommend to look at :
the CUDA C Programming guide and to the numba documentation to apply the code in python.
Moreover to understand what's wrong with your code, I recommend to use Nvidia profiler.
The following parts of the answer will explained how to apply the basics on your example.
Run multiples threads
To improve performances, you will first need to launch multiples threads that can run in parallel, CUDA handle threads as follow:
Threads are grouped into blocs of n threads (n < 1024)
Each thread withing the same bloc can be synchronized and have access to a (fast) common memory space called "shared memory".
You can run multiples blocs in parallel in a "grid" but you will lose the synchronization mechanism.
The syntax to run multiples threads is the following:
fftbench[griddim, blockdim](y1, False, FT1)
to simplify, I will use only one bloc of size 256:
fftbench[1, 256](y1, False, FT1)
Memory
To improve GPU performances it's important to look where the data will be stored, their is three main spaces:
global memory: it's the "RAM" of your GPU, it's slow and have a high latency, this is where all your array are placed when you send them to the GPU.
shared memory: it's a little fast access memory, all the thread of a bloc have access to the same shared memory.
local memory: physically it's the same that global memory, but each thread access its own local memory.
Typically, if you use multiples times the sames data, you should try store them in shared memory to prevent latency from the global memory.
In your code, you can store exptable in shared memory:
exptable = cuda.shared.array(1024, dtype=complex128)
and if n is not too big, you may want to use a working instead of using out:
working = cuda.shared.array(256, dtype=complex128)
Assign tasks to each thread
Of course if you don't change your function, all thread will do the same job and it will just slow down your program.
In this example we will assign each thread to one cell of the array. To do so, we have to get the unique id of thread withing a bloc:
idx = cuda.threadIdx.x
Now we will be able to speed up the for loops, lets handle them one by one:
exptable = cuda.shared.array(1024, dtype=complex128)
...
for i in range(n // 2):
exptable[i] = cmath.exp(i * coef)
Here is the goal: we will want the n/2 first threads to fill this array, then all the thread will be able to use it.
So in this case just replace the for loop by a condition on the thread idx's:
if idx < n // 2:
exptable[idx] = cmath.exp(idx * coef)
For the two last loops it's easier, each thread will deal with one cell of the array:
for i in range(n):
x = i
y = 0
for j in range(levels):
y = (y << 1) | (x & 1)
x >>= 1
out[i] = vector[y]
become
x = idx
y = 0
for j in range(levels):
y = (y << 1) | (x & 1)
x >>= 1
working[idx] = vector[y]
and
for i in range(n):
out[i]=out[i]/scale # the inverse requires a scaling
become
out[idx]=working[idx]/scale # the inverse requires a scaling
I use the shared array working but you can replace it by out if you want to use global memory.
Now, lets look at the while loop, we said that we want each thread to only deal with one cell of the array. So we can try to parallelize the two for loops inside.
...
for i in range(0, n, size):
k = 0
for j in range(i, i + halfsize):
temp = out[j + halfsize] * exptable[k]
out[j + halfsize] = out[j] - temp
out[j] += temp
k += tablestep
...
To simplify I will only use half of the threads, we will take the 128 first threads and determine j as follow:
...
if idx < 128:
j = (idx%halfsize) + size*(idx//halfsize)
...
k is:
k = tablestep*(idx%halfsize)
so we got the loop:
size = 2
while size <= n:
halfsize = size // 2
tablestep = n // size
if idx < 128:
j = (idx%halfsize) + size*(idx//halfsize)
k = tablestep*(idx%halfsize)
temp = working[j + halfsize] * exptable[k]
working[j + halfsize] = working[j] - temp
working[j] += temp
size *= 2
Synchronization
Last but not least, we need to synchronize all theses threads. In fact the program will not work if we do not synch. On the GPU thread may not run at the same time so you can get issues when data are produced by one thread and used by another one, for example:
exptable[0] is used by thread_2 before thread_0 fill store its value
working[j + halfsize] is moddified by another thread before you store it in temp
to prevent this we can use the function:
cuda.syncthreads()
All the threads in the same bloc will finish this line before execution the rest of the code.
In this example, you need to synchronize at two point, after the working initialization and after each iteration of the while loop.
then your code look like:
def _transform_radix2(vector, inverse, out):
n = len(vector)
levels = int32(math.log(float32(n))/math.log(float32(2)))
assert 2**levels==n # error: Length is not a power of 2
exptable = cuda.shared.array(1024, dtype=complex128)
working = cuda.shared.array(256, dtype=complex128)
assert (n // 2) <= len(exptable) # error: FFT length > MAXFFTSIZE
coef = complex128((2j if inverse else -2j) * math.pi / n)
if idx < n // 2:
exptable[idx] = cmath.exp(idx * coef)
x = idx
y = 0
for j in range(levels):
y = (y << 1) | (x & 1)
x >>= 1
working[idx] = vector[y]
cuda.syncthreads()
size = 2
while size <= n:
halfsize = size // 2
tablestep = n // size
if idx < 128:
j = (idx%halfsize) + size*(idx//halfsize)
k = tablestep*(idx%halfsize)
temp = working[j + halfsize] * exptable[k]
working[j + halfsize] = working[j] - temp
working[j] += temp
size *= 2
cuda.syncthreads()
scale=float64(n if inverse else 1)
out[idx]=working[idx]/scale # the inverse requires a scaling
I feel like your question is a good way to introduce some basics about GPGPU computing and I try to answer it in a didactic way. The final code is far from perfect and can be optimized a lot, I highly recommend you to read this Programming guide if you want to learn more about GPU optimizations.

Nested Numba function performance

currently I am trying to improve the performance of my python code. To do so I successfully use numba. In order to improve the structure of my code I create functions. Now I have noticed to my surprise that if I split the code into different numba functions, the code is significantly slower than if I put the whole code in one function with a numba decorator.
An example would be:
#nb.njit
def fct_4(a, b):
x = a ^ b
setBits = 0
while x > 0:
setBits += x & 1
x >>= 1
return setBits
#nb.njit
def fct_3(c, set_1, set_2):
h = 2
if c not in set_1 and c not in set_2:
if fct_4(0, c) <= h:
set_1.add(c)
else:
set_2.add(c)
#nb.njit
def fct_2(c, set_1, set_2):
fct_3(c, set_1, set_2)
#nb.njit
def fct_1(set_1, set_2):
for x1 in range(1000):
c = 2
fct_2(c, set_1, set_2)
is slower than
#nb.njit
def fct_1(set_1, set_2):
for x1 in range(1000):
c = 2
h = 2
if c not in set_1 and c not in set_2:
if fct_4(0, c) <= h:
set_1.add(c)
else:
set_2.add(c)
with
#nb.njit
def main_fct(set_1, set_2):
for i in range(50):
for x in range(1000):
fct_1(set_1, set_2)
set_1 = {0}
set_2 = {47}
start = timeit.default_timer()
main_fct(set_1, set_2)
stop = timeit.default_timer()
(2.70 seconds vs 0.46 seconds). I thought this shouldn't make a difference. Could you enlighten me?

Since python is a dynamically typed language, its function call overhead is quite high.
On top of that you are looping over the function calls, so the execution time incurred in calling the function and checking the arguments is multiplied 1000 times.

Can a Cython/Numba compiled function improve on numpy.max(numpy.abs(a-b))?

I am optimizing a bottleneck section of my code--iterating on a function a' = f(a), where a and a' are N by 1 vectors, until max(abs(a' - a)) is sufficiently small.
I have put a Numba wrapper on f(a), and got a nice speedup over the most optimized pure NumPy version I was able to produe (cut runtime by about 50%).
I tried writing a C-compatible version of numpy.max(numpy.abs(aprime - a)), but it turns out this is slower! I actually lose back ALL of the gains I got from Numba-fying the first portion of the iteration!
Is there likely to be a way for Numba or Cython to improve upon numpy.max(numpy.abs(aprime - a))? I reproduce my code below for reference, where a is P0 and a' is Pprime:
EDIT: For me, it seems that it is important to "flatten()" the inputs to "maxabs()". When I do this, the performance is no worse than NumPy. Then, when I do a "dry run" of the function outside the timing brackets as JoshAdel suggested, the loop with "maxabs" does slightly better than the loop with numpy.max(numpy.abs()).
from numba import jit
import numpy as np
### Preliminaries, to make the working example fully functional
n = 1200
Gammer = np.exp(-np.random.rand(n,n))
alpher = np.ones((n,1))
xxer = 10000*np.random.rand(n,1)
chii = 6.5
varkappa = 6.5
phi3 = 1.5
A = .5
sig = .2
mmer = np.dot(Gammer,xxer**phi3)
totalprod = A*alpher + (1-A)*mmer
Gammerchii = Gammer**chii
Gammerrats = Gammerchii[:,0].flatten()/Gammerchii[0,:].flatten()
Gammerrats[(Gammerchii[0,:].flatten() == 0) | (Gammerchii[:,0].flatten() == 0)] = 1.
P0 = (Gammerrats*(xxer[0]/totalprod[0])*(totalprod/xxer).flatten())**(1/(1+2*chii))
P0 *= n/np.sum(P0)
### End of preliminaries
### This is the function to produce a' = f(a)
#jit
def Piteration(P0, chii, sig, n, xxer, totalprod, Gammerrats, Gammerchii):
Mac = np.zeros((n,))
Pprime = np.zeros((n,))
themacpow = 1-(1/chii)*(sig/(1-sig))
specialchiipow = 1/(1+2*chii)
Psum = 0.
for i in range(n):
for j in range(n):
Mac[j] += ((P0[i]/P0[j])**chii)*Gammerchii[i,j]*totalprod[j]
for i in range(n):
Pprime[i] = (Gammerrats[i]*(xxer[0]/totalprod[0])*(totalprod[i]/xxer[i])*((Mac[i]/Mac[0])**themacpow))**specialchiipow
Psum += Pprime[i]
Psum = n/Psum
for i in range(n):
Pprime[i] *= Psum
return Pprime
### This is the function to find max(abs(aprime - a))
#jit
def maxabs(vec1,vec2,n):
themax = 0.
curdiff = 0.
for i in range(n):
curdiff = vec1[i] - vec2[i]
if curdiff < 0:
curdiff *= -1
if curdiff > themax:
themax = curdiff
return themax
### This is the main loop
diff = 1000.
while diff > 1e-2:
Pprime = Piteration(P0.flatten(), chii, sig, n, xxer.flatten(), totalprod.flatten(), Gammerrats.flatten(), Gammerchii)
diff = maxabs(P0.flatten(),Pprime.flatten(),n)
P0 = 1.*Pprime

When I time your maxabs function vs np.max(np.abs(vec1 - vec2)) for an array of shape (1200,), the numba version is ~2.6x faster using numba 0.32.0.
When you time the code, make sure you run your function once before you time it so that you don't include the time it takes to jit the code, which you only pay the first time. In general using timeit and running multiple times takes care of this. I'm not sure how you did the timing though since I see almost no difference in using maxabs vs the numpy call, most of the runtime seems to be in the call to Piteration.