I tried to find a float number in ndarray. Due to the software package I am using (Abaqus), the precision it outputs is a little bit low. For example, 10 is something like 10.00003. Therefore, I was wondering whether there is a "correct" way to do it, that is neater than my code.
Example code:
import numpy as np
array = np.arange(10)
number = 5.00001
if I do this：
idx = np.where(number==array)[0][0]
Then the result is empty because 5.00001 does not equal to 5.
Now I am doing:
atol = 1e-3 # Absolute tolerance
idx = np.where(abs(number-array) < atol)[0][0]
which works, and is not too messy... Yet I was wondering there would be a neater way to do it. Thanks!
PS: numpy.allclose() is another way to do it, but I need to use number * np.ones([array.shape[0], array.shape[1]]) and it still seems verbose to me...
Edit: Thank you all so much for the fantastic answers! np.isclose() is the exact function that I am looking for, and I missed it since it is not in the doc... I wouldn't have realized this until they update the doc, if it weren't you guys. Thank you again!
PS: numpy.allclose() is another way to do it, but I need to use number * np.ones([array.shape[0], array.shape[1]]) and it still seems verbose to me...
You almost never need to do anything like number * np.ones([array.shape[0], array.shape[1]]). Just as you can multiply that scalar number by that ones array to multiply all of its 1 values by number, you can pass that scalar number to allclose to compare all of the original array's values to number. For example:
>>> a = np.array([[2.000000000001, 2.0000000002], [2.000000000001, 1.999999999]])
>>> np.allclose(a, 2)
True
As a side note, if you really do need an array of all 2s, there's an easier way to do it than multiplying 2 by ones:
>>> np.tile(2, array.shape)
array([[2, 2], [2, 2]])
For that matter, I don't know why you need to do [array.shape[0], array.shape[1]]. If the array is 2D, that's exactly the same thing as array.shape. If the array might be larger, it's exactly the same as array.shape[:2].
I'm not sure this solves your actual problem, because it seems like you want to know which ones are close and not close, rather than just whether or not they all are. But the fact that you said you could use allclose if not for the fact that it's too verbose to create the array to compare with.
So, if you need whereclose rather than allclose… well, there's no such function. But it's pretty easy to build yourself, and you can always wrap it up if you're doing it repeatedly.
If you had an isclose method—like allclose, but returning a bool array instead of a single bool—you could just write:
idx = np.where(isclose(a, b, 0, atol))[0][0]
… or, if you're doing it over and over:
def whereclose(a, b, rtol=1e-05, atol=1e-08):
return np.where(isclose(a, b, rtol, atol))
idx = whereclose(a, b, 0, atol)[0][0]
As it turns out, version 1.7 of numpy does have exactly that function (see also here), but it doesn't appear to be in the docs. If you don't want to rely on a possibly-undocumented function, or need to work with numpy 1.6, you can write it yourself trivially:
def isclose(a, b, rtol=1e-05, atol=1e-08):
return np.abs(a-b) <= (atol + rtol * np.abs(b))
If you have up-to-date numpy (1.7), then the best way is to use np.isclose which will broadcast the shapes together automatically:
import numpy as np
a = np.arange(10)
n = 5.000001
np.isclose(a, n).nonzero()
#(array([5]),)
or, if you expect only one match:
np.isclose(a, n).nonzero()[0][0]
#5
(np.nonzero is basically the same thing as np.where except that it doesn't have the if condition then/else capability)
The method you use above, specifically abs(A - B) < atol, is standard for doing floating point comparisons across many languages. Obviously when using numpy A and/or B can be arrays or numbers.
Here is another approach that might be useful to look at. I'm not sure it applies to your case, but it could be very helpful if you're looking for more than one number in the array (which is a common use case). It's inspired by this question which is kind of similar.
import numpy as np
def find_close(a, b, rtol=1e-05, atol=1e-08):
tol = atol + abs(b) * rtol
lo = b - tol
hi = b + tol
order = a.argsort()
a_sorted = a[order]
left = a_sorted.searchsorted(lo)
right = a_sorted.searchsorted(hi, 'right')
return [order[L:R] for L, R in zip(left, right)]
a = np.array([2., 3., 3., 4., 0., 1.])
b = np.array([1.01, 3.01, 100.01])
print find_close(a, b, atol=.1)
# [array([5]), array([1, 2]), array([], dtype=int64)]
Related
I have two numpy arrays with floating point values and I am trying to find the indices where the numbers are approximately equal (floating point comparisons).
So something like:
x = np.random.rand(3)
y = np.random.rand(3)
x[2] = y[2]
# Do the comparison and it should return 2 as the index
I tried something like
np.where(np.allclose(x, y))
However, this returns an empty array. If I do:
np.where(x == y) # This is fine.
I tried using a combination of numpy.where and numpy.allclose but could not make it work. Of course, I can do it with a loop but that seems tedious and unpythonic.
What you look for is np.isclose:
np.where(np.isclose(x, y))
You can always use something relying on:
np.where( np.abs(x-y) < epsilon )
I am trying to expand the numpy "tensordot" such that things like:
K_ijklm = A_ki * B_jml can be written in a clear way like this: K = mytensordot(A,B,[2,0],[1,4,3])
To my understanding, numpy's tensordot (with optional argument 0) would be able to do something like this: K_kijml = A_ki * B_jml, i.e. keeping the order of the indexes. Therefore I would then have to do a number of np.swapaxes() to obtain the matrix `K_ijklm', which in a complicated case can be an easy source of errors (potentially very hard to debug).
The problem is that my implementation is slow (10x slower than tensordot [EDIT: It is actually MUCH slower than that]), even when using numba. I was wondering if anyone would have some insight on what could be done to improve the performance of my algorithm.
MWE
import numpy as np
import numba as nb
import itertools
import timeit
#nb.jit()
def myproduct(dimN):
N=np.prod(dimN)
L=len(dimN)
Product=np.zeros((N,L),dtype=np.int32)
rn=0
for n in range(1,N):
for l in range(L):
if l==0:
rn=1
v=Product[n-1,L-1-l]+rn
rn = 0
if v == dimN[L-1-l]:
v = 0
rn = 1
Product[n,L-1-l]=v
return Product
#nb.jit()
def mytensordot(A,B,iA,iB):
iA,iB = np.array(iA,dtype=np.int32),np.array(iB,dtype=np.int32)
dimA,dimB = A.shape,B.shape
NdimA,NdimB=len(dimA),len(dimB)
if len(iA) != NdimA: raise ValueError("iA must be same size as dim A")
if len(iB) != NdimB: raise ValueError("iB must be same size as dim B")
NdimN = NdimA + NdimB
dimN=np.zeros(NdimN,dtype=np.int32)
dimN[iA]=dimA
dimN[iB]=dimB
Out=np.zeros(dimN)
indexes = myproduct(dimN)
for nidxs in indexes:
idxA = tuple(nidxs[iA])
idxB = tuple(nidxs[iB])
v=A[(idxA)]*B[(idxB)]
Out[tuple(nidxs)]=v
return Out
A=np.random.random((4,5,3))
B=np.random.random((6,4))
def runmytdot():
return mytensordot(A,B,[0,2,3],[1,4])
def runtensdot():
return np.tensordot(A,B,0).swapaxes(1,3).swapaxes(2,3)
print(np.all(runmytdot()==runtensdot()))
print(timeit.timeit(runmytdot,number=100))
print(timeit.timeit(runtensdot,number=100))
Result:
True
1.4962144780438393
0.003484356915578246
You have run into a known issue. numpy.zeros requires a tuple when creating a multidimensional array. If you pass something other than a tuple, it sometimes works, but that's only because numpy is smart about converting the object into a tuple first.
The trouble is that numba does not currently support conversion of arbitrary iterables into tuples. So this line fails when you try to compile it in nopython=True mode. (A couple of others fail too, but this is the first.)
Out=np.zeros(dimN)
In theory you could call np.prod(dimN), create a flat array of zeros, and reshape it, but then you run into the very same problem: the reshape method of numpy arrays requires a tuple!
This is quite a vexing problem with numba -- I had not encountered it before. I really doubt the solution I have found is the correct one, but it is a working solution that allows us to compile a version in nopython=True mode.
The core idea is to avoid using tuples for indexing by directly implementing an indexer that follows the strides of the array:
#nb.jit(nopython=True)
def index_arr(a, ix_arr):
strides = np.array(a.strides) / a.itemsize
ix = int((ix_arr * strides).sum())
return a.ravel()[ix]
#nb.jit(nopython=True)
def index_set_arr(a, ix_arr, val):
strides = np.array(a.strides) / a.itemsize
ix = int((ix_arr * strides).sum())
a.ravel()[ix] = val
This allows us to get and set values without needing a tuple.
We can also avoid using reshape by passing the output buffer into the jitted function, and wrapping that function in a helper:
#nb.jit() # We can't use nopython mode here...
def mytensordot(A, B, iA, iB):
iA, iB = np.array(iA, dtype=np.int32), np.array(iB, dtype=np.int32)
dimA, dimB = A.shape, B.shape
NdimA, NdimB = len(dimA), len(dimB)
if len(iA) != NdimA:
raise ValueError("iA must be same size as dim A")
if len(iB) != NdimB:
raise ValueError("iB must be same size as dim B")
NdimN = NdimA + NdimB
dimN = np.zeros(NdimN, dtype=np.int32)
dimN[iA] = dimA
dimN[iB] = dimB
Out = np.zeros(dimN)
return mytensordot_jit(A, B, iA, iB, dimN, Out)
Since the helper contains no loops, it adds some overhead, but the overhead is pretty trivial. Here's the final jitted function:
#nb.jit(nopython=True)
def mytensordot_jit(A, B, iA, iB, dimN, Out):
for i in range(np.prod(dimN)):
nidxs = int_to_idx(i, dimN)
a = index_arr(A, nidxs[iA])
b = index_arr(B, nidxs[iB])
index_set_arr(Out, nidxs, a * b)
return Out
Unfortunately, this does not wind up generating as much of a speedup as we might like. On smaller arrays it's about 5x slower than tensordot; on larger arrays it's still 50x slower. (But at least it's not 1000x slower!) This is not too surprising in retrospect, since dot and tensordot are both using BLAS under the hood, as #hpaulj reminds us.
After finishing this code, I saw that einsum has solved your real problem -- nice!
But the underlying issue that your original question points to -- that indexing with arbitrary-length tuples is not possible in jitted code -- is still a frustration. So hopefully this will be useful to someone else!
tensordot with scalar axes values can be obscure. I explored it in
How does numpy.tensordot function works step-by-step?
There I deduced that np.tensordot(A, B, axes=0) is equivalent using axes=[[], []].
In [757]: A=np.random.random((4,5,3))
...: B=np.random.random((6,4))
In [758]: np.tensordot(A,B,0).shape
Out[758]: (4, 5, 3, 6, 4)
In [759]: np.tensordot(A,B,[[],[]]).shape
Out[759]: (4, 5, 3, 6, 4)
That in turn is equivalent to calling dot with a new size 1 sum-of-products dimenson:
In [762]: np.dot(A[...,None],B[...,None,:]).shape
Out[762]: (4, 5, 3, 6, 4)
(4,5,3,1) * (6,1,4) # the 1 is the last of A and 2nd to the last of B
dot is fast, using BLAS (or equivalent) code. Swapping axes and reshaping is also relatively fast.
einsum gives us a lot of control over axes
replicating the above products:
In [768]: np.einsum('jml,ki->jmlki',A,B).shape
Out[768]: (4, 5, 3, 6, 4)
and with swapping:
In [769]: np.einsum('jml,ki->ijklm',A,B).shape
Out[769]: (4, 4, 6, 3, 5)
A minor point - the double swap can be written as one transpose:
.swapaxes(1,3).swapaxes(2,3)
.transpose(0,3,1,2,4)
It seems that the np.where function evaluates all the possible outcomes first, then it evaluates the condition later. This means that, in my case, it will evaluate square root of -5, -4, -3, -2, -1 even though it will not be used later on.
My code runs and works. But my problem is the warning. I avoided using a loop to evaluate each element, because it will run much slower than np.where.
So, here, I am asking
Is there any way to make np.where evaluate the condition first?
Can I turn off just this specific warning? How?
Another better way to do it if you have a better suggestion.
Here just a short example code corresponding my real code which is gigantic. But essentially has the same problem.
Input:
import numpy as np
c=np.arange(10)-5
d=np.where(c>=0, np.sqrt(c) ,c )
Output:
RuntimeWarning: invalid value encountered in sqrt
d=np.where(c>=0,np.sqrt(c),c)
There is a much better way of doing this. Let's take a look at what your code is doing to see why.
np.where accepts three arrays as inputs. Arrays do not support lazy evaluation.
d = np.where(c >= 0, np.sqrt(c), c)
This line is therefore equivalent to doing
a = (c >= 0)
b = np.sqrt(c)
d = np.where(a, b, c)
Notice that the inputs are computed immediately, before where ever gets called.
Luckily, you don't need to use where at all. Instead, just use a boolean mask:
mask = (c >= 0)
d = np.empty_like(c)
d[mask] = np.sqrt(c[mask])
d[~mask] = c[~mask]
If you expect a lot of negatives, you can copy all the elements instead of just the negative ones:
d = c.copy()
d[mask] = np.sqrt(c[mask])
An even better solution might be to use masked arrays:
d = np.ma.masked_array(c, c < 0)
d = np.ma.sqrt(d)
To access the whole data array, with the masked portion unaltered, use d.data.
np.sqrt is a ufunc and accepts a where parameter. It can be used as a mask in this case:
In [61]: c = np.arange(10)-5.0
In [62]: d = c.copy()
In [63]: np.sqrt(c, where=c>=0, out=d);
In [64]: d
Out[64]:
array([-5. , -4. , -3. , -2. , -1. ,
0. , 1. , 1.41421356, 1.73205081, 2. ])
In contrast to the np.where case, this does not evaluate the function at the ~where elements.
This is answer to your 2nd question.
Yes you can turn off the warnings. Use warnings module.
import warnings
warnings.filterwarnings("ignore")
One solution is to not use np.where, and use indexing instead.
c = np.arange(10)-5
d = c.copy()
c_positive = c > 0
d[c_positive] = np.sqrt(c[c_positive])
I have a function a=x*V where x assumes thousands of values as x = arange(1,1000,0.1) and V is a combination of other constants. These make a always complex (has nonzero real and imaginary parts). However, because a depends on other values, the imag(a) can be negative for some x's.
For what I am doing, however, I need imag(a) to be always positive, so I need to take the negative values and turn them into positive.
I have tried doing
if imag(a)<0:
imag(a) = -1*imag(a)
That didn't seem to work because it gives me the error: SyntaxError: Can't assign to function call. I thought it was because it's an array so I tried any() and all(), but that didn't work either.
I'm out of options now.
IIUC:
In [35]: a = np.array([1+1j, 2-2j, 3+3j, 4-4j])
In [36]: a.imag *= np.where(a.imag < 0, -1, 1)
In [37]: a
Out[37]: array([ 1.+1.j, 2.+2.j, 3.+3.j, 4.+4.j])
You can't redefine a function that way. It would be like saying
sqrt(x) = 2*sqrt(x)
What you can do is reassign the value of a (not imag(a)).
if imag(a) < 0
a = a - 2*imag(a)*j
For example, if a = 3 - 5j, then it would give you
3 - 5j - 2(-5)j = 3 + 5j
It appears to be faster than doing subtraction. For a full function:
import numpy as np
def imag_abs(x):
mask = x.imag < 0
x[mask] = np.conj(x[mask])
return x
According to http://docs.scipy.org/doc/numpy/reference/generated/numpy.where.html, if x and y are given and input arrays are 1-D, where is equivalent to [xv if c else yv for (c,xv, yv) in zip(x!=0, 1/x, x)]. When doing runtime benchmarks, however, they have significantly different speeds:
x = np.array(range(-500, 500))
%timeit np.where(x != 0, 1/x, x)
10000 loops, best of 3: 23.9 µs per loop
%timeit [xv if c else yv for (c,xv, yv) in zip(x!=0, 1/x, x)]
1000 loops, best of 3: 232 µs per loop
Is there a way I can rewrite the second form so that it has a similar runtime to the first? The reason I ask is because I'd like to use a slightly modified version of the second case to avoid division by zero errors:
[1 / xv if c else xv for (c,xv) in zip(x!=0, x)]
Another question: the first case returns a numpy array while the second case returns a list. Is the most efficient way to have the second case return an array is to first make a list and then convert the list to an array?
np.array([xv if c else yv for (c,xv, yv) in zip(x!=0, 1/x, x)])
Thanks!
You just asked about 'delaying' the 'where':
numpy.where : how to delay evaluating parameters?
and someone else just asked about divide by zero:
Replace all elements of a matrix by their inverses
When people say that where is similar to the list comprehension, they attempt to describe the action, not the actual implementation.
np.where called with just one argument is the same as np.nonzero. This quickly (in compiled code) loops through the argument, and collects the indices of all non-zero values.
np.where when called with 3 arguments, returns a new array, collecting values from the 2 and 3rd arguments based on the nonzero values. But it's important to realize that those arguments must be other arrays. They are not functions that it evaluates element by element.
So the where is more like:
m1 = 1/xv
m2 = xv
[v1 if c else v2 for (c, v1, v2) in zip(x!=0, m1, m2)]
It's easy to run this iteration in compiled code because it just involves 3 arrays of matching size (matching via broadcasting).
np.array([...]) is a reasonable way of converting a list (or list comprehension) into an array. It may be a little slower than some alternatives because np.array is a powerful general purpose function. np.fromiter([], dtype) may be faster in some cases, because it isn't as general (you have to specify dtype, and it it only works with 1d).
There are 2 time proven strategies for getting more speed in element-by-element calculations:
use packages like numba and cython to rewrite the problem as c code
rework your calculations to use existing numpy methods. The use of masking to avoid divide by zero is a good example of this.
=====================
np.ma.where, the version for masked arrays is written in Python. Its code might be instructive. Note in particular this piece:
# Construct an empty array and fill it
d = np.empty(fc.shape, dtype=ndtype).view(MaskedArray)
np.copyto(d._data, xv.astype(ndtype), where=fc)
np.copyto(d._data, yv.astype(ndtype), where=notfc)
It makes a target, and then selectively copies values from the 2 inputs arrays, based on the condition array.
You can avoid division by zero while maintaining performance by using advanced indexing:
x = np.arange(-500, 500)
result = np.empty(x.shape, dtype=float) # set the dtype to whatever is appropriate
nonzero = x != 0
result[nonzero] = 1/x[nonzero]
result[~nonzero] = 0
If you for some reason want to bypass an error with numpy it might be worth looking into the errstate context:
x = np.array(range(-500, 500))
with np.errstate(divide='ignore'): #ignore zero-division error
x = 1/x
x[x!=x] = 0 #convert inf and NaN's to 0
Consider changing the array in place by using np.put():
In [56]: x = np.linspace(-1, 1, 5)
In [57]: x
Out[57]: array([-1. , -0.5, 0. , 0.5, 1. ])
In [58]: indices = np.argwhere(x != 0)
In [59]: indices
Out[59]:
array([[0],
[1],
[3],
[4]], dtype=int64)
In [60]: np.put(x, indices, 1/x[indices])
In [61]: x
Out[61]: array([-1., -2., 0., 2., 1.])
The approach above does not create a new array, which could be very convenient if x is a large array.