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Numpy add with multiple arrays

Is there a way to add (as opposed to sum) multiple arrays together in a single operation? Obviously, np.sum and np.add are different operations, however, the problem I'm struggling with right now is that np.add only takes two arrays at once. I could utilize either
output = 0
for arr in arr_list:
output = output + array
or
output = 0
for arr in arr_list:
output = np.add(output, array)
and, yes, this is workable. However, it would be nice if I could simply do some variant of
output = np.add_multiple(arr_list)
Does this exist?
EDIT:
I failed to be clear initially. I specifically require a function that does not require an array of arrays to be constructed, as my arrays are not of equal dimensions and require broadcasting, for example:
a = np.arange(3).reshape(1,3)
b = np.arange(9).reshape(3,3)
a, b = a[:,:,None,None], b[None,None,:,:]
These work:
a + b # Works
np.add(a, b) # Works
These do not, and fail with the same exception:
np.sum([a, b], axis = 0)
np.add.reduce([a, b])
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: could not broadcast input array from shape (3,1,1) into shape (1)

You can just use the Python's sum built-in:
output = sum(arr_list)
For many other numpy functions, np.<ufunc>.reduce can be used as shown by #hpaulj.

You can use the sum() to add multiple arrays.
arr = np.array([[6,2,3,5,4,3],
[7,7,2,4,6,7],
[10,6,2,4,5,9]])
np.add(0, arr.sum(axis=0))

In [18]: alist = [np.arange(4),np.ones(4),np.array([10,1,11,2])]
In [19]: np.add.reduce(alist)
Out[19]: array([11., 3., 14., 6.])
In [20]: alist[0]+alist[1]+alist[2]
Out[20]: array([11., 3., 14., 6.])
And for more fun:
In [21]: np.add.accumulate(alist)
Out[21]:
array([[ 0., 1., 2., 3.],
[ 1., 2., 3., 4.],
[11., 3., 14., 6.]])
edit
In [53]: a.shape
Out[53]: (1, 1, 1, 3)
In [54]: b.shape
Out[54]: (3, 3, 1, 1)
Addition with broadcasting:
In [63]: sum([a,b]).shape
Out[63]: (3, 3, 1, 3)
In [64]: (a+b).shape
Out[64]: (3, 3, 1, 3)
In [66]: np.add.reduce([a,b]).shape
Out[66]: (3, 3, 1, 3)
For what it's worth, I was suggesting add.reduce because I thought you wanted to add more than 2 arrays.
All these work as long as the arrays broadcast together.

How to efficiently filter maximum elements of a matrix per row

Given a 2D array, I'm looking for a pythonic way to get an array of same shape, with only the maximum element per each row.
See max_row_filter function below
def max_row_filter(mat2d):
m = np.zeros(mat2d.shape)
for r in range(mat2d.shape[0]):
c = np.argmax(mat2d[r])
m[r,c]=mat2d[r,c]
return m
p = np.array([[1,2,3],[5,4,3,],[9,10,3]])
max_row_filter(p)
Out: array([[ 0., 0., 3.],
[ 5., 0., 0.],
[ 0., 10., 0.]])
I'm looking for an efficient way to do this, suitable to be done on big arrays.

Alternative answer (this will keep duplicates):
p * (p==p.max(axis=1, keepdims=True))

If there are no duplicates, you could use numpy.argmax:
import numpy as np
p = np.array([[1, 2, 3],
[5, 4, 3, ],
[9, 10, 3]])
result = np.zeros_like(p)
rows, cols = zip(*enumerate(np.argmax(p, axis=1)))
result[rows, cols] = p[rows, cols]
print(result)
Output
[[ 0 0 3]
[ 5 0 0]
[ 0 10 0]]
Note that, for multiple occurrences argmax return the first occurence.

Python : Mapping values to other values without gap

I have the following question. Is there somekind of method with numpy or scipy , which I can use to get an given unsorted array like this
a = np.array([0,0,1,1,4,4,4,4,5,1891,7]) #could be any number here
to something where the numbers are interpolated/mapped , there is no gap between the values and they are in the same order like before?:
[0,0,1,1,2,2,2,2,3,5,4]
EDIT
Is it furthermore possible to swap/shuffle the numbers after the mapping, so that
[0,0,1,1,2,2,2,2,3,5,4]
become something like:
[0,0,3,3,5,5,5,5,4,1,2]

Edit: I'm not sure what the etiquette is here (should this be a separate answer?), but this is actually directly obtainable from np.unique.
>>> u, indices = np.unique(a, return_inverse=True)
>>> indices
array([0, 0, 1, 1, 2, 2, 2, 2, 3, 5, 4])
Original answer: This isn't too hard to do in plain python by building a dictionary of what index each value of the array would map to:
x = np.sort(np.unique(a))
index_dict = {j: i for i, j in enumerate(x)}
[index_dict[i] for i in a]

Seems you need to rank (dense) your array, in which case use scipy.stats.rankdata:
from scipy.stats import rankdata
rankdata(a, 'dense')-1
# array([ 0., 0., 1., 1., 2., 2., 2., 2., 3., 5., 4.])

Unsuccessful append to an empty NumPy array

I am trying to fill an empty(not np.empty!) array with values using append but I am gettin error:
My code is as follows:
import numpy as np
result=np.asarray([np.asarray([]),np.asarray([])])
result[0]=np.append([result[0]],[1,2])
And I am getting:
ValueError: could not broadcast input array from shape (2) into shape (0)

I might understand the question incorrectly, but if you want to declare an array of a certain shape but with nothing inside, the following might be helpful:
Initialise empty array:
>>> a = np.zeros((0,3)) #or np.empty((0,3)) or np.array([]).reshape(0,3)
>>> a
array([], shape=(0, 3), dtype=float64)
Now you can use this array to append rows of similar shape to it. Remember that a numpy array is immutable, so a new array is created for each iteration:
>>> for i in range(3):
... a = np.vstack([a, [i,i,i]])
...
>>> a
array([[ 0., 0., 0.],
[ 1., 1., 1.],
[ 2., 2., 2.]])
np.vstack and np.hstack is the most common method for combining numpy arrays, but coming from Matlab I prefer np.r_ and np.c_:
Concatenate 1d:
>>> a = np.zeros(0)
>>> for i in range(3):
... a = np.r_[a, [i, i, i]]
...
>>> a
array([ 0., 0., 0., 1., 1., 1., 2., 2., 2.])
Concatenate rows:
>>> a = np.zeros((0,3))
>>> for i in range(3):
... a = np.r_[a, [[i,i,i]]]
...
>>> a
array([[ 0., 0., 0.],
[ 1., 1., 1.],
[ 2., 2., 2.]])
Concatenate columns:
>>> a = np.zeros((3,0))
>>> for i in range(3):
... a = np.c_[a, [[i],[i],[i]]]
...
>>> a
array([[ 0., 1., 2.],
[ 0., 1., 2.],
[ 0., 1., 2.]])

numpy.append is pretty different from list.append in python. I know that's thrown off a few programers new to numpy. numpy.append is more like concatenate, it makes a new array and fills it with the values from the old array and the new value(s) to be appended. For example:
import numpy
old = numpy.array([1, 2, 3, 4])
new = numpy.append(old, 5)
print old
# [1, 2, 3, 4]
print new
# [1, 2, 3, 4, 5]
new = numpy.append(new, [6, 7])
print new
# [1, 2, 3, 4, 5, 6, 7]
I think you might be able to achieve your goal by doing something like:
result = numpy.zeros((10,))
result[0:2] = [1, 2]
# Or
result = numpy.zeros((10, 2))
result[0, :] = [1, 2]
Update:
If you need to create a numpy array using loop, and you don't know ahead of time what the final size of the array will be, you can do something like:
import numpy as np
a = np.array([0., 1.])
b = np.array([2., 3.])
temp = []
while True:
rnd = random.randint(0, 100)
if rnd > 50:
temp.append(a)
else:
temp.append(b)
if rnd == 0:
break
result = np.array(temp)
In my example result will be an (N, 2) array, where N is the number of times the loop ran, but obviously you can adjust it to your needs.
new update
The error you're seeing has nothing to do with types, it has to do with the shape of the numpy arrays you're trying to concatenate. If you do np.append(a, b) the shapes of a and b need to match. If you append an (2, n) and (n,) you'll get a (3, n) array. Your code is trying to append a (1, 0) to a (2,). Those shapes don't match so you get an error.

This error arise from the fact that you are trying to define an object of shape (0,) as an object of shape (2,). If you append what you want without forcing it to be equal to result[0] there is no any issue:
b = np.append([result[0]], [1,2])
But when you define result[0] = b you are equating objects of different shapes, and you can not do this. What are you trying to do?

Here's the result of running your code in Ipython. Note that result is a (2,0) array, 2 rows, 0 columns, 0 elements. The append produces a (2,) array. result[0] is (0,) array. Your error message has to do with trying to assign that 2 item array into a size 0 slot. Since result is dtype=float64, only scalars can be assigned to its elements.
In [65]: result=np.asarray([np.asarray([]),np.asarray([])])
In [66]: result
Out[66]: array([], shape=(2, 0), dtype=float64)
In [67]: result[0]
Out[67]: array([], dtype=float64)
In [68]: np.append(result[0],[1,2])
Out[68]: array([ 1., 2.])
np.array is not a Python list. All elements of an array are the same type (as specified by the dtype). Notice also that result is not an array of arrays.
Result could also have been built as
ll = [[],[]]
result = np.array(ll)
while
ll[0] = [1,2]
# ll = [[1,2],[]]
the same is not true for result.
np.zeros((2,0)) also produces your result.
Actually there's another quirk to result.
result[0] = 1
does not change the values of result. It accepts the assignment, but since it has 0 columns, there is no place to put the 1. This assignment would work in result was created as np.zeros((2,1)). But that still can't accept a list.
But if result has 2 columns, then you can assign a 2 element list to one of its rows.
result = np.zeros((2,2))
result[0] # == [0,0]
result[0] = [1,2]
What exactly do you want result to look like after the append operation?

numpy.append always copies the array before appending the new values. Your code is equivalent to the following:
import numpy as np
result = np.zeros((2,0))
new_result = np.append([result[0]],[1,2])
result[0] = new_result # ERROR: has shape (2,0), new_result has shape (2,)
Perhaps you mean to do this?
import numpy as np
result = np.zeros((2,0))
result = np.append([result[0]],[1,2])

SO thread 'Multiply two arrays element wise, where one of the arrays has arrays as elements' has an example of constructing an array from arrays. If the subarrays are the same size, numpy makes a 2d array. But if they differ in length, it makes an array with dtype=object, and the subarrays retain their identity.
Following that, you could do something like this:
In [5]: result=np.array([np.zeros((1)),np.zeros((2))])
In [6]: result
Out[6]: array([array([ 0.]), array([ 0., 0.])], dtype=object)
In [7]: np.append([result[0]],[1,2])
Out[7]: array([ 0., 1., 2.])
In [8]: result[0]
Out[8]: array([ 0.])
In [9]: result[0]=np.append([result[0]],[1,2])
In [10]: result
Out[10]: array([array([ 0., 1., 2.]), array([ 0., 0.])], dtype=object)
However, I don't offhand see what advantages this has over a pure Python list or lists. It does not work like a 2d array. For example I have to use result[0][1], not result[0,1]. If the subarrays are all the same length, I have to use np.array(result.tolist()) to produce a 2d array.

Row-wise indexing in Numpy

I have two matrices, A and B:
A = array([[2., 13., 25., 1.], [ 18., 5., 1., 25.]])
B = array([[2, 1], [0, 3]])
I want to index each row of A with each row of B, producing the slice:
array([[25., 13.], [18., 25.]])
That is, I essentially want something like:
array([A[i,b] for i,b in enumerate(B)])
Is there a way to fancy-index this directly? The best I can do is this "flat-hack":
A.flat[B + arange(0,A.size,A.shape[1])[:,None]]

#Ophion's answer is great, and deserves the credit, but I wanted to add some explanation, and offer a more intuitive construction.
Instead of rotating B and then rotating the result back, it's better to just rotate the arange. I think this gives the most intuitive solution, even if it takes more characters:
A[((0,),(1,)), B]
or equivalently
A[np.arange(2)[:, None], B]
This works because what's really going on here, is you're making an i array and a j array, each of which have the same shape as your desired result.
i = np.array([[0, 0],
[1, 1]])
j = B
But you can use just
i = np.array([[0],
[1]])
Because it will broadcast to match B (this is what np.arange(2)[:,None] gives).
Finally, to make it more general (not knowing 2 as the arange size), you could also generate i from B with
i = np.indices(B.shape)[0]
however you build i and j, you just call it like
>>> A[i, j]
array([[ 25., 13.],
[ 18., 25.]])

Not pretty but:
A[np.arange(2),B.T].T
array([[ 25., 13.],
[ 18., 25.]])