Numpy.where() with an array in its conditional - python

I don't know how to describe this well so I'll just show it.
How do I do this...
for iy in random_y:
print(x[np.where(y == iy)], iy)
X y
[ 0.5] : 0.247403959255
[ 2.] : 0.841470984808
[ 49.5]: -0.373464754784
without for loops and I get a solution as a single array like when you use np.where() or array[cond]. Since you know, this is Python B)
NOTE: The reason why I want to do this is because I have a random subset of the Y values and I want to find the corresponding X values.

If you are looking for exact matches, you can simply use np.in1d as this is a perfect scenario for its usage, like so -
first_output = x[np.in1d(y,random_y)]
second_output = random_y[np.in1d(random_y,y)
If you are dealing with floating-point numbers, you might want to use some tolerance factor into the comparisons. So, for such cases, you can use NumPy broadcasting and then use np.where, like so -
tol = 1e-5 # Edit this to change tolerance
R,C = np.where(np.abs(random_y[:,None] - y)<=tol)
first_output = x[C]
second_output = random_y[R]

Maybe this could do the trick(not tested):
print(Str(x[np.where(y == iy)]) + " " + Str(iy) + "\n") for iy in random_y

Related

Find numpy array in bigger array, testing for approximate equality [duplicate]

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 )

What is an efficient way of writing sum from i = 1 to n, log(1 + exp(w_i)) in Numpy

Let w = (w_1, w_2, w_3, ...., w_n) be an array, n is large
Without using loops, I want to define the function
sum from i = 1 to i = n, log(1 + exp(w_i))
Is there a vector operation that handles this in Numpy? I was thinking of
np.dot(np.ones((n,)), np.log(1+np.exp(w))
but I don't know if that works.
You can use np.sum(...) to sum all elements of the array.
While np.log(1+np.exp(w)) should work fine, there's also np.log1p(...) which calculates the ln of one plus values with better precision in case of very small numbers.
Putting it all together:
result = np.sum(np.log1p(np.exp(w)))

How to find negative imaginary parts of values in an array then turning them to 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

Find float in ndarray

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)]

Numpy conditional arithmetic operations on two arrays

Still trying to earn my numpy stripes: I want to perform an arithmetic operation on two numpy arrays, which is simple enough:
return 0.5 * np.sum(((array1 - array2) ** 2) / (array1 + array2))
Problem is, I need to be able to specify the condition that, if both arrays are element-wise 0 at the same element i, don't perform the operation at all--would be great just to return 0 on this one--so as not to divide by 0.
However, I have no idea how to specify this condition without resorting to the dreaded nested for-loop. Thank you in advance for your assistance.
Edit: Would also be ideal not to have to resort to a pseudocount of +1.
Just replace np.sum() by np.nansum():
return 0.5 * np.nansum(((array1 - array2) ** 2) / (array1 + array2))
np.nansum() treats nans as zero.
return numpy.select([array1 == array2, array1 != array2], [0.5 * np.sum(((array1 - array2) ** 2) / (array1 + array2)), 0])
should do the trick... numpy.where might also be used.
You could also try post-applying numpy.nan_to_num:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.nan_to_num.html
although I found that when I have a divide by zero in your code, it gives a warning, but fills that element with zero (when doing integer math) and NaN when using floats.
If you want to skip the sum when you have a divide by zero, you could also just do the calculation and then test for the NaN before returning:
xx = np.sum(((array1 - array2) ** 2) / (array1 + array2))
if np.isnan(xx):
return 0
else:
return xx
Edit: To silence warnings you could try messing around with numpy.seterr:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.seterr.html

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