We Keep Coding

How to convert an array index to/from a mask

Say I have an array like:
a1 = np.array([.1, .2, 23., 4.3, 3.2, .1, .05, .2, .3, 4.2, 7.6])
And I filter out, and create a mask, of all values less than 1, like:
a2 = a1[a1 >= 1]
a2_mask = np.ma.masked_where(a1 < 1, a1)
And then search for a specific value:
a2_idx = np.where(a2==3.2)[0][0]
How would I convert that index to the corresponding index in the original array?
e.g.
>>> a2_idx
2
>>> a1_idx = reframe_index(a2_idx, a2_mask)
>>> a1_idx
4
My naive implementation would be:
def reframe_index(old_idx, mask):
cnt = 0
ref = 0
for v in mask:
if not isinstance(v, (int, float)):
cnt += 1
else:
if ref == old_idx:
return ref + cnt
ref += 1
Does Numpy have a more efficient way to do this?

a2 is a copy, so there's no link between it an a1 - except for some values.
In [19]: a2
Out[19]: array([23. , 4.3, 3.2, 4.2, 7.6])
In [20]: np.nonzero(a2 == 3.2)
Out[20]: (array([2]),)
In [21]: a2[2]
Out[21]: 3.2
The mask of a2_mask, just a1<1, does give us a way of finding the corresponding element of a1:
In [22]: a2_mask = np.ma.masked_where(a1 < 1, a1)
In [23]: a2_mask
Out[23]:
masked_array(data=[--, --, 23.0, 4.3, 3.2, --, --, --, --, 4.2, 7.6],
mask=[ True, True, False, False, False, True, True, True,
True, False, False],
fill_value=1e+20)
In [24]: a2_mask.compressed()
Out[24]: array([23. , 4.3, 3.2, 4.2, 7.6])
In [25]: a2_mask.mask
Out[25]:
array([ True, True, False, False, False, True, True, True, True,
False, False])
In [26]: np.nonzero(~a2_mask.mask)
Out[26]: (array([ 2, 3, 4, 9, 10]),)
In [27]: np.nonzero(~a2_mask.mask)[0][2]
Out[27]: 4
In [28]: a1[4]
Out[28]: 3.2
So you need the mask or indices used to select a2 in the first place. a2 itself does not have the information.
In [30]: np.nonzero(a1>=1)
Out[30]: (array([ 2, 3, 4, 9, 10]),)
In [31]: np.nonzero(a1 >= 1)[0][2]
Out[31]: 4

I had a similar problem recently, so I made haggis.npy_util.unmasked_index1. This function has a lot of overkill for your relatively simple case, because it's intended to operate on an arbitrary number of dimensions. That being said, given
>>> arr = np.array([.1, .2, 23., 4.3, 3.2, .1, .05, .2, .3, 4.2, 7.6])
and
>>> mask = arr >= 1
>>> mask
array([False, False, True, True, True, False, False, False, False,
True, True])
You can do something like
>>> idx = unmasked_index(np.flatnonzero(arr[mask] == 3.2), mask)
>>> idx
array([4])
If you ever need it, there is also an inverse function haggis.npy_util.masked_index that converts a location in a multidimensional input array into its index in the masked array.
1Disclaimer: I am the author of haggis.

NumPy masked operation?

Say there's a np.float32 matrix A of shape (N, M). Together with A, I possess another matrix B, of type np.bool, of the exact same shape (elements from A can be mapped 1:1 to B). Example:
A =
[
[0.1, 0.2, 0.3],
[4.02, 123.4, 534.65],
[2.32, 22.0, 754.01],
[5.41, 23.1, 1245.5],
[6.07, 0.65, 22.12],
]
B =
[
[True, False, True],
[False, False, True],
[True, True, False],
[True, True, True],
[True, False, True],
]
Now, I'd like to perform np.max, np.min, np.argmax and np.argmin on axis=1 of A, but only considering elements A[i,j] for which B[i,j] == True. Is it possible to do something like this in NumPy? The for-loop version is trivial, but I'm wondering whether I can get some of that juicy NumPy speed.
The result for A, B and np.max (for example) would be:
[ 0.3, 534.65, 22.0, 1245.5, 22.12 ]
I've avoided ma because I've heard that the computation gets very slow and I don't feel like specifying fill_value makes sense in this context. I just want the numbers to be ignored.
Also, if it matters at all in my case, N ranges in thousands and M ranges in units.

This is a textbook application for masked arrays. But as always there are other ways to do it.
import numpy as np
A = np.array([[ 0.1, 0.2, 0.3],
[ 4.02, 123.4, 534.65],
[ 2.32, 22.0, 754.01],
[ 5.41, 23.1, 1245.5],
[ 6.07, 0.65, 22.12]])
B = np.array([[ True, False, True],
[False, False, True],
[ True, True, False],
[ True, True, True],
[ True, False, True]])
With nanmax etc.
You could cast the 'invalid' values to NaN (say), then use NumPy's special NaN-ignoring functions:
>>> A[~B] = np.nan # <-- Note this mutates A
>>> np.nanmax(A, axis=1)
array([3.0000e-01, 5.3465e+02, 2.2000e+01, 1.2455e+03, 2.2120e+01])
The catch is that, while np.nanmax, np.nanmin, np.nanargmax, and np.nanargmin all exist, lots of functions don't have a non-NaN twin, so you might have to come up with something else eventually.
With ma
It seems weird not to mention masked arrays, which are straightforward. Notice that the mask is (to my mind anyway) 'backwards'. That is, True means the value is 'masked' or invalid and will be ignored. Hence having to negate B with the tilde. Then you can do what you want with the masked array:
>>> X = np.ma.masked_array(A, mask=~B) # <--- Note the tilde.
>>> np.max(X, axis=1)
masked_array(data=[0.3, 534.65, 22.0, 1245.5, 22.12],
mask=[False, False, False, False, False],
fill_value=1e+20)

Numpy array equality boolean

The following two approaches should yield an identical result, but it seems they do not:
Approach 1:
#Generate an array with 12 annual fractions corresponding to each month
ann_frac = np.arange(1,13,1).reshape([12,1])
ann_frac = ann_frac[:,0]/12
Output:
array([0.08333333, 0.16666667, 0.25 , 0.33333333, 0.41666667,
0.5 , 0.58333333, 0.66666667, 0.75 , 0.83333333,
0.91666667, 1. ])
Approach 2:
i = np.arange(1,13,1)
ann_frac2 = (i/12).reshape([12,1])
Output:
array([[0.08333333],
[0.16666667],
[0.25 ],
[0.33333333],
[0.41666667],
[0.5 ],
[0.58333333],
[0.66666667],
[0.75 ],
[0.83333333],
[0.91666667],
[1. ]])
Comparing approaches:
ann_frac2==ann_frac
Output:
array([[False],
[False],
[False],
[False],
[False],
[False],
[False],
[False],
[False],
[False],
[False],
[False]])
A bit strange it seems. Any explanation? Obviously, if one just wants to compare numbers in two different arrays for equlity, the example above demonstrates that even though the numbers can be identical, the way the numbers were created and stored can produce an unexpected behaviour.

I think you are comparing arrays with different dimensionality. What I guess you want is
ann_frac = np.arange(1,13,1).reshape([12,1])
ann_frac1 = ann_frac[:,0]/12
i = np.arange(1,13,1)
ann_frac2 = (i/12).reshape([12,1])
ann_frac3 = (i/12).reshape(12)
and the correct comparison should be
ann_frac3==ann_frac1
with output
array([ True, True, True, True, True, True, True, True, True,
True, True, True])
Even better would be
(ann_frac3==ann_frac1).all()
with output
True

how does numpy.where work?

I can understand following numpy behavior.
>>> a
array([[ 0. , 0. , 0. ],
[ 0. , 0.7, 0. ],
[ 0. , 0.3, 0.5],
[ 0.6, 0. , 0.8],
[ 0.7, 0. , 0. ]])
>>> argmax_overlaps = a.argmax(axis=1)
>>> argmax_overlaps
array([0, 1, 2, 2, 0])
>>> max_overlaps = a[np.arange(5),argmax_overlaps]
>>> max_overlaps
array([ 0. , 0.7, 0.5, 0.8, 0.7])
>>> gt_argmax_overlaps = a.argmax(axis=0)
>>> gt_argmax_overlaps
array([4, 1, 3])
>>> gt_max_overlaps = a[gt_argmax_overlaps,np.arange(a.shape[1])]
>>> gt_max_overlaps
array([ 0.7, 0.7, 0.8])
>>> gt_argmax_overlaps = np.where(a == gt_max_overlaps)
>>> gt_argmax_overlaps
(array([1, 3, 4]), array([1, 2, 0]))
I understood 0.7, 0.7 and 0.8 is a[1,1],a[3,2] and a[4,0] so I got the tuple (array[1,3,4] and array[1,2,0]) each array of which composed of 0th and 1st indices of those three elements. I then tried other examples to see my understanding is correct.
>>> np.where(a == [0.3])
(array([2]), array([1]))
0.3 is in a[2,1] so the outcome looks as I expected. Then I tried
>>> np.where(a == [0.3, 0.5])
(array([], dtype=int64),)
?? I expected to see (array([2,2]),array([2,3])). Why do I see the output above?
>>> np.where(a == [0.7, 0.7, 0.8])
(array([1, 3, 4]), array([1, 2, 0]))
>>> np.where(a == [0.8,0.7,0.7])
(array([1]), array([1]))
I can't understand the second result either. Could someone please explain it to me? Thanks.

The first thing to realize is that np.where(a == [whatever]) is just showing you the indices where a == [whatever] is True. So you can get a hint by looking at the value of a == [whatever]. In your case that "works":
>>> a == [0.7, 0.7, 0.8]
array([[False, False, False],
[False, True, False],
[False, False, False],
[False, False, True],
[ True, False, False]], dtype=bool)
You aren't getting what you think you are. You think that is asking for the indices of each element separately, but instead it's getting the positions where the values match at the same position in the row. Basically what this comparison is doing is saying "for each row, tell me whether the first element is 0.7, whether the second is 0.7, and whether the third is 0.8". It then returns the indices of those matching positions. In other words, the comparison is done between entire rows, not just individual values. For your last example:
>>> a == [0.8,0.7,0.7]
array([[False, False, False],
[False, True, False],
[False, False, False],
[False, False, False],
[False, False, False]], dtype=bool)
You now get a different result. It's not asking for "the indices where a has value 0.8", it's asking for only the indices where there is a 0.8 at the beginning of the row -- and likewise a 0.7 in either of the later two positions.
This type of row-wise comparison can only be done if the value you compare against matches the shape of a single row of a. So when you try it with a two-element list, it returns an empty set, because there it is trying to compare the list as a scalar value against individual values in your array.
The upshot is that you can't use == on a list of values and expect it to just tell you where any of the values occurs. The equality will match by value and position (if the value you compare against is the same shape as a row of your array), or it will try to compare the whole list as a scalar (if the shape doesn't match). If you want to search for the values independently, you need to do something like what Khris suggested in a comment:
np.where((a==0.3)|(a==0.5))
That is, you need to make two (or more) separate comparisons against separate values, not a single comparison against a list of values.

efficient numpy.cumsum and numpy.digitize

Given a matrix of values that represent probabilities I am trying to write an efficient process that returns the bin that the value belongs to. For example:
sample = 0.5
x = np.array([0.1]*10)
np.digitize( sample, np.cumsum(x))-1
#returns 5
is the result I am looking for.
According to timeit for x arrays with few elements it is more efficient to do it as:
cdf = 0
for key,val in enumerate(x):
cdf += val
if sample<=cdf:
print key
break
while for bigger x arrays the numpy solution is faster.
The question:
Is there a way to further accelerate it, e.g., a function that combines the steps?
Can we vectorize the process it for the case where sample is a list, whose each item is associated with its own x array (x will then be 2-D)?
In the application x contains the marginal probabilities; this is way I need to decrement the results of np.digitize

You could use some broadcasting magic there -
(x.cumsum(1) > sample[:,None]).argmax(1)-1
Steps involved :
I. Perform cumsum along each row.
II. Use broadcasted comparison for each cumsum row against each sample value and look for the first occurrence of sample being lesser than cumsum values, signalling that the element before that in x is the index we are looking for.
Step-by-step run -
In [64]: x
Out[64]:
array([[ 0.1 , 0.1 , 0.1 , 0.1 , 0.1 , 0.1 , 0.1 ],
[ 0.8 , 0.96, 0.88, 0.36, 0.5 , 0.68, 0.71],
[ 0.37, 0.56, 0.5 , 0.01, 0.77, 0.88, 0.36],
[ 0.62, 0.08, 0.37, 0.93, 0.65, 0.4 , 0.79]])
In [65]: sample # one elem per row of x
Out[65]: array([ 0.5, 2.2, 1.9, 2.2])
In [78]: x.cumsum(1)
Out[78]:
array([[ 0.1 , 0.2 , 0.3 , 0.4 , 0.5 , 0.6 , 0.7 ],
[ 0.8 , 1.76, 2.64, 2.99, 3.49, 4.18, 4.89],
[ 0.37, 0.93, 1.43, 1.45, 2.22, 3.1 , 3.47],
[ 0.62, 0.69, 1.06, 1.99, 2.64, 3.04, 3.83]])
In [79]: x.cumsum(1) > sample[:,None]
Out[79]:
array([[False, False, False, False, False, True, True],
[False, False, True, True, True, True, True],
[False, False, False, False, True, True, True],
[False, False, False, False, True, True, True]], dtype=bool)
In [80]: (x.cumsum(1) > sample[:,None]).argmax(1)-1
Out[80]: array([4, 1, 3, 3])
# A loopy solution to verify results against
In [81]: [np.digitize( sample[i], np.cumsum(x[i]))-1 for i in range(x.shape[0])]
Out[81]: [4, 1, 3, 3]
Boundary cases :
The proposed solution automatically handles the cases where sample values are lesser than smallest of cumulative summed values -
In [113]: sample[0] = 0.08 # editing first sample to be lesser than 0.1
In [114]: [np.digitize( sample[i], np.cumsum(x[i]))-1 for i in range(x.shape[0])]
Out[114]: [-1, 1, 3, 3]
In [115]: (x.cumsum(1) > sample[:,None]).argmax(1)-1
Out[115]: array([-1, 1, 3, 3])
For cases where a sample value is greater than largest of cumulative summed values, we need one extra step -
In [116]: sample[0] = 0.8 # editing first sample to be greater than 0.7
In [121]: mask = (x.cumsum(1) > sample[:,None])
In [122]: idx = mask.argmax(1)-1
In [123]: np.where(mask.any(1),idx,x.shape[1]-1)
Out[123]: array([6, 1, 3, 3])
In [124]: [np.digitize( sample[i], np.cumsum(x[i]))-1 for i in range(x.shape[0])]
Out[124]: [6, 1, 3, 3]