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.
Related
I have a problem I'm working on where I have to produce a function which mirrors a mathematical one given:
Probability = e^beta / 1 + e^beta. So far I produced code that works when I feed it an integer, but I need to use the function to calculate the probabilities of an array.
My code so far:
import math
e = math.e
def likelihood(beta):
for i in range(beta):
return (e**(beta)/(1+ e**(beta)))
beta_candidate = np.random.uniform(-5, 5, 50)
likelihood_candidate = likelihood(beta_candidate)
Whenever I run the code I'm met with an error stating: only integer scalar arrays can be converted to a scalar index.
In [3]: import math
In [4]: e = math.e
In [5]: def likelihood(beta):
...: return [e**i/(1+e**i) for i in beta]
...:
In [7]: likelihood_candidate = likelihood(beta_candidate)
Since you have your beta_candidate as numpy array, you can just do vectorized numpy operations:
l = np.exp(beta_candidate)/(1+np.exp(beta_candidate))
I want to verify if a numpy array is a continuous sequence in another array.
E.g.
a = np.array([1,2,3,4,5,6,7])
b = np.array([3,4,5])
c = np.array([2,3,4,6])
The expected result would be:
is_sequence_of(b, a) # should return True
is_sequence_of(c, a) # should return False
I want to know if there is a numpy method that does this.
Approach #1
We can use one with np.searchsorted -
def isin_seq(a,b):
# Look for the presence of b in a, while keeping the sequence
sidx = a.argsort()
idx = np.searchsorted(a,b,sorter=sidx)
idx[idx==len(a)] = 0
ssidx = sidx[idx]
return (np.diff(ssidx)==1).all() & (a[ssidx]==b).all()
Note that this assumes that the input arrays have no duplicates.
Sample runs -
In [42]: isin_seq(a,b) # search for the sequence b in a
Out[42]: True
In [43]: isin_seq(c,b) # search for the sequence b in c
Out[43]: False
Approach #2
Another with skimage.util.view_as_windows -
from skimage.util import view_as_windows
def isin_seq_v2(a,b):
return (view_as_windows(a,len(b))==b).all(1).any()
Approach #3
This could also be considered as a template-matching problem and hence, for int numbers, we can use OpenCV's built-in function for template-matching : cv2.matchTemplate (inspired by this post), like so -
import cv2
from cv2 import matchTemplate as cv2m
def isin_seq_v3(arr,seq):
S = cv2m(arr.astype('uint8'),seq.astype('uint8'),cv2.TM_SQDIFF)
return np.isclose(S,0).any()
Approach #4
Our methods could benefit with a short-circuiting based one. So, we will use one with numba for performance-efficiency, like so -
from numba import njit
#njit
def isin_seq_numba(a,b):
m = len(a)
n = len(b)
for i in range(m-n+1):
for j in range(n):
if a[i+j]!=b[j]:
break
if j==n-1:
return True
return False
I'm trying to insert a small multidimensional array into a larger one inside a numba jitclass. The small array is set specific positions of the larger array defined by an index list.
The following MWE shows the problem without numba - everything works as expected
import numpy as np
class NumbaClass(object):
def __init__(self, n, m):
self.A = np.zeros((n, m))
# solution 1 using pure python
def nonNumbaFunction1(self, idx, values):
self.A[idx[:, None], idx] = values
# solution 2 using pure python
def nonNumbaFunction2(self, idx, values):
self.A[np.ix_(idx, idx)] = values
if __name__ == "__main__":
n = 6
m = 8
obj = NumbaClass(n, m)
print(f'A =\n{obj.A}')
idx = np.array([0, 2, 5])
values = np.arange(len(idx)**2).reshape(len(idx), len(idx))
print(f'values =\n{values}')
obj.nonNumbaFunction1(idx, values)
print(f'A =\n{obj.A}')
obj.nonNumbaFunction2(idx, values)
print(f'A =\n{obj.A}')
Both functions nonNumbaFunction1 and nonNumbaFunction2 do not work inside a numba class. So my current solution looks like this which is not really nice in my opinion
import numpy as np
from numba import jitclass
from numba import int64, float64
from collections import OrderedDict
specs = OrderedDict()
specs['A'] = float64[:, :]
#jitclass(specs)
class NumbaClass(object):
def __init__(self, n, m):
self.A = np.zeros((n, m))
# solution for numba jitclass
def numbaFunction(self, idx, values):
for i in range(len(values)):
idxi = idx[i]
for j in range(len(values)):
idxj = idx[j]
self.A[idxi, idxj] = values[i, j]
if __name__ == "__main__":
n = 6
m = 8
obj = NumbaClass(n, m)
print(f'A =\n{obj.A}')
idx = np.array([0, 2, 5])
values = np.arange(len(idx)**2).reshape(len(idx), len(idx))
print(f'values =\n{values}')
obj.numbaFunction(idx, values)
print(f'A =\n{obj.A}')
So my questions are:
Does anyone know a solution to this indexing in numba or is there another vectorized solution?
Is there a faster solution for nonNumbaFunction1?
It might be useful to know that inserted array is small (4x4 to 10x10), but this indexing appears in nested loops so it has to be quiet fast as well! Later I need a similar indexing for three dimensional objects too.
Because of limitations on numba's indexing support, I don't think you can do any better than writing out the for loops yourself. To make it generic across dimensions, you could use the generated_jit decorator to specialize. Something like this:
def set_2d(target, values, idx):
for i in range(values.shape[0]):
for j in range(values.shape[1]):
target[idx[i], idx[j]] = values[i, j]
def set_3d(target, values, idx):
for i in range(values.shape[0]):
for j in range(values.shape[1]):
for k in range(values.shape[2]):
target[idx[i], idx[j], idx[k]] = values[i, j, l]
#numba.generated_jit
def set_nd(target, values, idx):
if target.ndim == 2:
return set_2d
elif target.ndim == 3:
return set_3d
Then, this could be used in your jitclass
specs = OrderedDict()
specs['A'] = float64[:, :]
#jitclass(specs)
class NumbaClass(object):
def __init__(self, n, m):
self.A = np.zeros((n, m))
def numbaFunction(self, idx, values):
set_nd(self.A, values, idx)
I programmed class which looks something like this:
import numpy as np
class blank():
def __init__(self,a,b,c):
self.a=a
self.b=b
self.c=c
n=5
c=a/b*8
if (a>b):
y=c+a*b
else:
y=c-a*b
p = np.empty([1,1])
k = np.empty([1,1])
l = np.empty([1,1])
p[0]=b
k[0]=b*(c-1)
l[0]=p+k
for i in range(1, n, 1):
p=np.append(p,l[i-1])
k=np.append(k,(p[i]*(c+1)))
l=np.append(l,p[i]+k[i])
komp = np.zeros(shape=(n, 1))
for i in range(0, n):
pl_avg = (p[i] + l[i]) / 2
h=pl_avg*3
komp[i]=pl_avg*h/4
self.tot=komp+l
And when I call it like this:
from ex1 import blank
import numpy as np
res=blank(1,2,3)
print(res.tot)
everything works well.
BUT I want to call it like this:
res = blank(np.array([1,2,3]), np.array([3,4,5]), 3)
Is there an easy way to call it for each i element of this two arrays without editing class code?
You won't be able to instantiate a class with NumPy arrays as inputs without changing the class code. #PabloAlvarez and #NagaKiran already provided alternative: iterate with zip over arrays and instantiate class for each pair of elements. While this is pretty simple solution, it defeats the purpose of using NumPy with its efficient vectorized operations.
Here is how I suggest you to rewrite the code:
from typing import Union
import numpy as np
def total(a: Union[float, np.ndarray],
b: Union[float, np.ndarray],
n: int = 5) -> np.array:
"""Calculates what your self.tot was"""
bc = 8 * a
c = bc / b
vectorized_geometric_progression = np.vectorize(geometric_progression,
otypes=[np.ndarray])
l = np.stack(vectorized_geometric_progression(bc, c, n))
l = np.atleast_2d(l)
p = np.insert(l[:, :-1], 0, b, axis=1)
l = np.squeeze(l)
p = np.squeeze(p)
pl_avg = (p + l) / 2
komp = np.array([0.75 * pl_avg ** 2]).T
return komp + l
def geometric_progression(bc, c, n):
"""Calculates array l"""
return bc * np.logspace(start=0,
stop=n - 1,
num=n,
base=c + 2)
And you can call it both for sole numbers and NumPy arrays like that:
>>> print(total(1, 2))
[[2.6750000e+01 6.6750000e+01 3.0675000e+02 1.7467500e+03 1.0386750e+04]
[5.9600000e+02 6.3600000e+02 8.7600000e+02 2.3160000e+03 1.0956000e+04]
[2.1176000e+04 2.1216000e+04 2.1456000e+04 2.2896000e+04 3.1536000e+04]
[7.6205600e+05 7.6209600e+05 7.6233600e+05 7.6377600e+05 7.7241600e+05]
[2.7433736e+07 2.7433776e+07 2.7434016e+07 2.7435456e+07 2.7444096e+07]]
>>> print(total(3, 4))
[[1.71000000e+02 3.39000000e+02 1.68300000e+03 1.24350000e+04 9.84510000e+04]
[8.77200000e+03 8.94000000e+03 1.02840000e+04 2.10360000e+04 1.07052000e+05]
[5.59896000e+05 5.60064000e+05 5.61408000e+05 5.72160000e+05 6.58176000e+05]
[3.58318320e+07 3.58320000e+07 3.58333440e+07 3.58440960e+07 3.59301120e+07]
[2.29323574e+09 2.29323590e+09 2.29323725e+09 2.29324800e+09 2.29333402e+09]]
>>> print(total(np.array([1, 3]), np.array([2, 4])))
[[[2.67500000e+01 6.67500000e+01 3.06750000e+02 1.74675000e+03 1.03867500e+04]
[1.71000000e+02 3.39000000e+02 1.68300000e+03 1.24350000e+04 9.84510000e+04]]
[[5.96000000e+02 6.36000000e+02 8.76000000e+02 2.31600000e+03 1.09560000e+04]
[8.77200000e+03 8.94000000e+03 1.02840000e+04 2.10360000e+04 1.07052000e+05]]
[[2.11760000e+04 2.12160000e+04 2.14560000e+04 2.28960000e+04 3.15360000e+04]
[5.59896000e+05 5.60064000e+05 5.61408000e+05 5.72160000e+05 6.58176000e+05]]
[[7.62056000e+05 7.62096000e+05 7.62336000e+05 7.63776000e+05 7.72416000e+05]
[3.58318320e+07 3.58320000e+07 3.58333440e+07 3.58440960e+07 3.59301120e+07]]
[[2.74337360e+07 2.74337760e+07 2.74340160e+07 2.74354560e+07 2.74440960e+07]
[2.29323574e+09 2.29323590e+09 2.29323725e+09 2.29324800e+09 2.29333402e+09]]]
You can see that results are in compliance.
Explanation:
First of all I'd like to note that your calculation of p, k, and l doesn't have to be in the loop. Moreover, calculating k is unnecessary. If you see carefully, how elements of p and l are calculated, they are just geometric progressions (except the 1st element of p):
p = [b, b*c, b*c*(c+2), b*c*(c+2)**2, b*c*(c+2)**3, b*c*(c+2)**4, ...]
l = [b*c, b*c*(c+2), b*c*(c+2)**2, b*c*(c+2)**3, b*c*(c+2)**4, b*c*(c+2)**5, ...]
So, instead of that loop, you can use np.logspace. Unfortunately, np.logspace doesn't support base parameter as an array, so we have no other choice but to use np.vectorize which is just a loop under the hood...
Calculating of komp though is easily vectorized. You can see it in my example. No need for loops there.
Also, as I already noted in a comment, your class doesn't have to be a class, so I took a liberty of changing it to a function.
Next, note that input parameter c is overwritten, so I got rid of it. Variable y is never used. (Also, you could calculate it just as y = c + a * b * np.sign(a - b))
And finally, I'd like to remark that creating NumPy arrays with np.append is very inefficient (as it was pointed out by #kabanus), so you should always try to create them at once - no loops, no appending.
P.S.: I used np.atleast_2d and np.squeeze in my code and it could be unclear why I did it. They are necessary to avoid if-else clauses where we would check dimensions of array l. You can print intermediate results to see what is really going on there. Nothing difficult.
if it is just calling class with two different list elements, loop can satisfies well
res = [blank(i,j,3) for i,j in zip(np.array([1,2,3]),np.array([3,4,5]))]
You can see list of values for res variable
The only way I can think of iterating lists of arrays is by using a function on the main program for iteration and then do the operations you need to do inside the loop.
This solution works for each element of both arrays (note to use zip function for making the iteration in both lists if they have a small size as listed in this answer here):
for n,x in zip(np.array([1,2,3]),np.array([3,4,5])):
res=blank(n,x,3)
print(res.tot)
Hope it is what you need!
I've written a simple function
def rect(t):
field = np.zeros((len(t),3))
for i in range(len(t)):
if t[i] >=-10 and t[i] <=10:
field[i,1] = 1
else:
field[i,1] = 0
return field
and want to integrate it
from scipy import integrate
def S_elem(t):
return rect(t)[:,1]*integr
integrate.quad(S_elem, -10, 10)
I get this error
TypeError: object of type 'float' has no len()
I know I may not use numpy array but I need it for the other purpose. How can I
perform integration without the removal of the numpy array type?
You have to convert your float to a numpy array:
def rect(t):
t = np.atleast_1d(t)
field = np.zeros((len(t), 3))
field[:, 1] = abs(t) <= 10
return field