I need a lot of operations to replace parts of the array repeatedly, I hope there is an efficient way to avoid loops, because I found that 2d array slicing does not support write operations.So I constructed a simple function to achieve this goal.
a = np.random.rand(4,4)
b = np.random.rand(4)
c = [1,1,1,1]
def ravel_index(a,b,row_index,col_index,order='c'):
rindex = row_index * a.shape[1] + col_index
lindex = rindex + b.ravel().shape[0]
return rindex,lindex
f,l = ravel_index(a,b,1,0)
a.ravel()[f:l]=c
print (a)
>>>[[ 0.013631517 0.81654666 0.96975073 0.832641632]
[ 1. 1. 1. 1. ]
[ 0.092047737 0.149801674 0.322049501 0.162026284]
[ 0.490197753 0.54935894 0.527087062 0.126544099]]
It looks ideal now, But when trying to write in column direction..
f,l = ravel_index(a,b,1,0)
a.ravel('F')[f:l]=c
print (a)
>>>[[ 0.306372691 0.586445896 0.052487946 0.864993735]
[ 0.873470159 0.762572666 0.986864265 0.803903923]
[ 0.000208709 0.579103322 0.811386673 0.196167481]
[ 0.928682626 0.707539068 0.752064295 0.564061717]]
Obviously the array is copied, I don't know how to solve this problem, hope to get help Thank you
The documentation for numpy.ravel mentions that
A copy is made only if needed
which means that ravel('F') won't be the solution you're looking for. If you assume instead that everything is in order 'C', you can modify your ravel_index function:
def ravel_index(a,c,row_index,col_index,order='c'):
rindex = row_index * a.shape[1] + col_index
if order == 'c':
lindex = rindex + b.ravel().shape[0]
return range(rindex, lindex)
elif order == 'f':
lindex = rindex + a.shape[1]*np.ravel(c).shape[0]
return (None, range(rindex, lindex, a.shape[1]))
E.g.
>>> a = np.random.rand(4,4)
>>> c = [1,1,1,1]
>>> ravel_slice = ravel_index(a,c,0,1,order='f')
>>> a.ravel()[ravel_slice]=c
>>>
>>> print a
[[0.56152208 1. 0.76850125 0.90981706]
[0.00753469 1. 0.33609404 0.01321701]
[0.36101786 1. 0.36610868 0.77170151]
[0.64812018 1. 0.33486985 0.58649772]]
I have a 3d mesh with points and the locations of the points are
in an array that looks like this:
mesh_vectors = np.array([[-0.85758871, 0.8965745 , -0.1427767 ],
[-0.23945311, 1.00544977, 1.45797086],
[-0.57341832, -1.07448494, -0.11827722],
[ 0.05894491, -0.97208506, 1.47583127],
[-0.71402085, -0.08872638, -0.12916484],
[-0.09181146, 1.01235461, 0.47418442],
[-0.09025362, 0.01668115, 1.46690106],
[ 0.19773833, -0.95349348, 0.49089319],
[ 0.05055711, 0.02909645, 0.48503664]])
I have two indexing arrays:
idx1 = np.array([4 2 1 6 5 0 1 5])
idx2 = np.array([6 3 0 4 7 2 3 7])
these translations correspond to the index arrays:
translate_1 = np.array([[ 0.00323021 0.00047712 -0.00422925]
[ 0.00153422 0.00022654 -0.00203258]
[ 0.00273207 0.00039626 0.00038201]
[ 0.0052439 0.00075993 0.00068843]
[-0.00414245 -0.00053918 0.00543974]
[-0.00681844 -0.00084955 0.00894626]
[ 0. 0. 0. ]
[-0.00672519 -0.00099897 -0.00090189]])
translate_2 = np.array([[ 0.00523871 0.00079512 0.00068814]
[ 0.00251901 0.00038234 0.00033379]
[ 0.00169134 0.00021078 -0.00218737]
[ 0.00324106 0.00040338 -0.00422859]
[-0.00413547 -0.00058669 0.00544016]
[-0.00681223 -0.0008921 0.00894669]
[ 0. 0. 0. ]
[-0.00672553 -0.00099677 -0.00090191]])
they are currently added to the mesh like this:
mesh_vectors[idx1] += translate_1
mesh_vectors[idx2] += translate_2
The trouble is, what I really need to add isn't the translations
but the mean of the translations where multiple translations are
applied to the same mesh point. The indexing arrays can have indices occurring in a variety of different frequencies. Could be [2,2,2,3,4,5] and [1,2,1,1,5,4] though they will always be the same size. I'm trying to do this with numpy for speed but I have the options of using loops on start to generate indexing arrays if needed.
Thanks in advance!
This works:
scaled_tr1 = translate_1 / np.bincount(idx1)[idx1,None]
np.add.at(mesh_vectors, idx1, scaled_tr1)
Note that the use of np.add.at instead of fancy indexing is required:
ufunc.at(a, indices, b=None)
Performs unbuffered in place operation on operand a for elements specified by indices. For addition ufunc, this method is equivalent to a[indices] += b, except that results are accumulated for elements that are indexed more than once. For example, a[[0,0]] += 1 will only increment the first element once because of buffering, whereas add.at(a, [0,0], 1) will increment the first element twice.
I have a numpy array which has only a few non-zero entries which can be either positive or negative. E.g. something like this:
myArray = np.array([[ 0. , 0. , 0. ],
[ 0.32, -6.79, 0. ],
[ 0. , 0. , 0. ],
[ 0. , 1.5 , 0. ],
[ 0. , 0. , -1.71]])
In the end, I would like to receive a list where each entry of this list corresponds to a row of myArray and is a cumulative product of function outputs which depend on the entries of the respective row of myArray and another list (in the example below it is called l).
The individual terms depend on the sign of the myArray entry: When it is positive, I apply "funPos", when it is negative, I apply "funNeg" and if the entry is 0, the term will be 1. So in the example array from above it would be:
output = [1*1*1 ,
funPos(0.32, l[0])*funNeg(-6.79,l[1])*1,
1*1*1,
1*funPos(1.5, l[1])*1,
1*1*funNeg(-1.71, l[2])]
I implemented this as shown below and it gives me the desired output (note: that is just a highly simplified toy example; the actual matrices are far bigger and the functions more complicated). I go through each row of the array, if the sum of the row is 0, I don't have to do any calculations and the output is just 1. If it is not equal 0, I go through this row, check the sign of each value and apply the appropriate function.
import numpy as np
def doCalcOnArray(Array1, myList):
output = np.ones(Array1.shape[0]) #initialize output
for indRow,row in enumerate(Array1):
if sum(row) != 0: #only then calculations are needed
tempProd = 1. #initialize the product that corresponds to the row
for indCol, valCol in enumerate(row):
if valCol > 0:
tempVal = funPos(valCol, myList[indCol])
elif valCol < 0:
tempVal = funNeg(valCol, myList[indCol])
elif valCol == 0:
tempVal = 1
tempProd = tempProd*tempVal
output[indRow] = tempProd
return output
def funPos(val1,val2):
return val1*val2
def funNeg(val1,val2):
return val1*(val2+1)
myArray = np.array([[ 0. , 0. , 0. ],
[ 0.32, -6.79, 0. ],
[ 0. , 0. , 0. ],
[ 0. , 1.5 , 0. ],
[ 0. , 0. , -1.71]])
l = [1.1, 2., 3.4]
op = doCalcOnArray(myArray,l)
print op
The output is
[ 1. -7.17024 1. 3. -7.524 ]
which is the desired one.
My question is whether there is a more efficient way for doing that since that is quite "expensive" for large arrays.
EDIT:
I accepted gabhijit's answer because the pure numpy solution he came up with seems to be the fastest one for the arrays I am dealing with. Please note, that there is also a nice working solution from RaJa that requires panda and also the solution from dave works fine which can serve as a nice example on how to use generators and numpy's "apply_along_axis".
Here's what I have tried - using reduce, map. I am not sure how fast this is - but is this what you are trying to do?
Edit 4: Simplest and most readable - Make l a numpy array and then greatly simplifies where.
import numpy as np
import time
l = np.array([1.0, 2.0, 3.0])
def posFunc(x,y):
return x*y
def negFunc(x,y):
return x*(y+1)
def myFunc(x, y):
if x > 0:
return posFunc(x, y)
if x < 0:
return negFunc(x, y)
else:
return 1.0
myArray = np.array([
[ 0.,0.,0.],
[ 0.32, -6.79, 0.],
[ 0.,0.,0.],
[ 0.,1.5,0.],
[ 0.,0., -1.71]])
t1 = time.time()
a = np.array([reduce(lambda x, (y,z): x*myFunc(z,l[y]), enumerate(x), 1) for x in myArray])
t2 = time.time()
print (t2-t1)*1000000
print a
Basically let's just look at last line it says cumulatively multiply things in enumerate(xx), starting with 1 (last parameter to reduce). myFunc simply takes the element in myArray(row) and element # index row in l and multiplies them as needed.
My output is not same as yours - so I am not sure whether this is exactly what you want, but may be you can follow the logic.
Also I am not so sure how fast this will be for huge arrays.
edit: Following is a 'pure numpy way' to do this.
my = myArray # just for brevity
t1 = time.time()
# First set the positive and negative values
# complicated - [my.itemset((x,y), posFunc(my.item(x,y), l[y])) for (x,y) in zip(*np.where(my > 0))]
# changed to
my = np.where(my > 0, my*l, my)
# complicated - [my.itemset((x,y), negFunc(my.item(x,y), l[y])) for (x,y) in zip(*np.where(my < 0))]
# changed to
my = np.where(my < 0, my*(l+1), my)
# print my - commented out to time it.
# Now set the zeroes to 1.0s
my = np.where(my == 0.0, 1.0, my)
# print my - commented out to time it
a = np.prod(my, axis=1)
t2 = time.time()
print (t2-t1)*1000000
print a
Let me try to explain the zip(*np.where(my != 0)) part as best as I can. np.where simply returns two numpy arrays first array is an index of row, second array is an index of column that matches the condition (my != 0) in this case. We take a tuple of those indices and then use array.itemset and array.item, thankfully, column index is available for free to us, so we can just take the element # that index in the list l. This should be faster than previous (and by orders of magnitude readable!!). Need to timeit to find out whether it indeed is.
Edit 2: Don't have to call separately for positive and negative can be done with one call np.where(my != 0).
So, let's see if I understand your question.
You want to map elements of your matrix to a new matrix such that:
0 maps to 1
x>0 maps to funPos(x)
x<0 maps to funNeg(x)
You want to calculate the product of all elements in the rows this new matrix.
So, here's how I would go about doing it:
1:
def myFun(a):
if a==0:
return 1
if a>0:
return funPos(a)
if a<0:
return funNeg(a)
newFun = np.vectorize(myFun)
newArray = newFun(myArray)
And for 2:
np.prod(newArray, axis = 1)
Edit: To pass the index to funPos, funNeg, you can probably do something like this:
# Python 2.7
r,c = myArray.shape
ctr = -1 # I don't understand why this should be -1 instead of 0
def myFun(a):
global ctr
global c
ind = ctr % c
ctr += 1
if a==0:
return 1
if a>0:
return funPos(a,l[ind])
if a<0:
return funNeg(a,l[ind])
I think this numpy function would be helpful to you
numpy.apply_along_axis
Here is one implementation. Also I would warn against checking if the sum of the array is 0. Comparing floats to 0 can give unexpected behavior due to machine accuracy constraints. Also if you have -5 and 5 the sum is zero and I'm not sure thats what you want. I used numpy's any() function to see if anything was nonzero. For simplicity I also pulled your list (my_list) into global scope.
import numpy as np
my_list = 1.1, 2., 3.4
def func_pos(val1, val2):
return val1 * val2
def func_neg(val1, val2):
return val1 *(val2 + 1)
def my_generator(row):
for i, a in enumerate(row):
if a > 0:
yield func_pos(a, my_list[i])
elif a < 0:
yield func_neg(a, my_list[i])
else:
yield 1
def reduce_row(row):
if not row.any():
return 1.0
else:
return np.prod(np.fromiter(my_generator(row), dtype=float))
def main():
myArray = np.array([
[ 0. , 0. , 0. ],
[ 0.32, -6.79, 0. ],
[ 0. , 0. , 0. ],
[ 0. , 1.5 , 0. ],
[ 0. , 0. , -1.71]])
return np.apply_along_axis(reduce_row, axis=1, arr=myArray)
There are probably faster implmentations, I think apply_along_axis is really just a loop under the covers.
I didn't test, but I bet this is faster than what you started with, and should be more memory efficient.
I've tried your example with the masking function of numpy arrays. However, I couldn't find a solution to replace the values in your array by funPos or funNeg.
So my suggestion would be to try this using pandas instead as it conserves indices while masking.
See my example:
import numpy as np
import pandas as pd
def funPos(a, b):
return a * b
def funNeg(a, b):
return a * (b + 1)
myPosFunc = np.vectorize(funPos) #vectorized form of funPos
myNegFunc = np.vectorize(funNeg) #vectorized form of funNeg
#Input
I = [1.0, 2.0, 3.0]
x = pd.DataFrame([
[ 0.,0.,0.],
[ 0.32, -6.79, 0.],
[ 0.,0.,0.],
[ 0.,1.5,0.],
[ 0.,0., -1.71]])
b = pd.DataFrame(myPosFunc(x[x>0], I)) #calculate all positive values
c = pd.DataFrame(myNegFunc(x[x<0], I)) #calculate all negative values
b = b.combineMult(c) #put values of c in b
b = b.fillna(1) #replace all missing values that were '0' in the raw array
y = b.product() #multiply all elements in one row
#Output
print ('final result')
print (y)
print (y.tolist())
I'm trying to vectorize a simple function in numpy and getting inconsistent behavior. I expect my code to return 0 for values < 0.5 and the unchanged value otherwise. Strangely, different runs of the script from the command line yield varying results: sometimes it works correctly, and sometimes I get all 0's. It doesn't matter which of the three lines I use for the case when d <= T. It does seem to be correlated with whether the first value to be returned is 0. Any ideas? Thanks.
import numpy as np
def my_func(d, T=0.5):
if d > T: return d
#if d <= T: return 0
else: return 0
#return 0
N = 4
A = np.random.uniform(size=N**2)
A.shape = (N,N)
print A
f = np.vectorize(my_func)
print f(A)
$ python x.py
[[ 0.86913815 0.96833127 0.54539153 0.46184594]
[ 0.46550903 0.24645558 0.26988519 0.0959257 ]
[ 0.73356391 0.69363161 0.57222389 0.98214089]
[ 0.15789303 0.06803493 0.01601389 0.04735725]]
[[ 0.86913815 0.96833127 0.54539153 0. ]
[ 0. 0. 0. 0. ]
[ 0.73356391 0.69363161 0.57222389 0.98214089]
[ 0. 0. 0. 0. ]]
$ python x.py
[[ 0.37127366 0.77935622 0.74392301 0.92626644]
[ 0.61639086 0.32584431 0.12345342 0.17392298]
[ 0.03679475 0.00536863 0.60936931 0.12761859]
[ 0.49091897 0.21261635 0.37063752 0.23578082]]
[[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]]
If this really is the problem you want to solve, then there's a much better solution:
A[A<=0.5] = 0.0
The problem with your code, however, is that if the condition passes, you are returning the integer 0, not the float 0.0. From the documentation:
The data type of the output of vectorized is determined by calling the function with the first element of the input. This can be avoided by specifying the otypes argument.
So when the very first entry is <0.5, it tries to create an integer, not float, array.
You should change return 0 to
return 0.0
Alternately, if you don't want to touch my_func, you can use
f = np.vectorize(my_func, otypes=[np.float])