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How to plot with dtaidistance

I tried an example using dtaidistance and dtw, but it did not plot the result.
from dtaidistance import dtw
from dtaidistance import dtw_visualisation as dtwvis
import numpy as np
s1 = np.array([0., 0, 1, 2, 1, 0, 1, 0, 0, 2, 1, 0, 0])
s2 = np.array([0., 1, 2, 3, 1, 0, 0, 0, 2, 1, 0, 0, 0])
path = dtw.warping_path(s1, s2)
dtwvis.plot_warping(s1, s2, path, filename="warp.png")
It should have looked like this:
Unfortunately, it did not show up.
I tried to add "plt.show()" at the end of the code (previously importing pyplot of course). But, in this case, it has not helped.
What is the rason it does not plot the graph, like in the example above?

The picture in the example is not meant to be plotted within python but saved as .png to hard drive where the python code is stored.

How does scipy.ndimage.filters.convolve when the mode is reflective

I am trying to figure out how to do this with numpy, so I can then convert it to c++ from scratch. I have figured out how to do it when the mode is constant. The way that is done is shown below.
import numpy as np
from scipy import signal
a = np.array([[1, 2, 0, 0], [5, 3, 0, 4], [0, 0, 0, 7], [9, 3, 0, 0]])
k = np.array([[1,0,0],[0,1,0],[0,0,0]])
a = np.pad(a, 1)
k = np.flip(k)
output = signal.convolve(a, k, 'valid')
Which then comes out to the same output as scipy.ndimage.filters.convolve(a, k, mode='constant) So I thought that when the mode was reflect it would work the same way. Except, that the line a = np.pad(a, 1) would be changed to a = np.pad(a, 1, mode='reflect'). However, that does not seem to be the case. Could someone explain how it would work from scratch using numpy and scipy.signal.convolve? Thank you.

Optimizing execution time for mapping array to value with dictionary and numpy

I am trying to implement a simple mapping to a set of values from an array created with numpy of 2-D.
For each row in the array I need to choose the correct value corresponding with the set of values and add it to a array.
For example:
[0, 1, 0, 0] -> 3
...
[1, 0, 1, 0] -> 2
But, my first implementation made me wonder if I'm doing something really wrong or not efficient at all because of the size of my dataset, so I did this workaround without using for loops and optimize speed execution using dictionary lookup.
import numpy as np
# function to perform the search and return the index accordingly (it is supposed to be fast because of data structure)
def get_val(n):
map_list = {0: [0, 1, 0], 1: [0, 1, 0], 2: [1, 0, 0], 3: [0, 0, 1]}
map_vals = list(map_list.values())
index = map_vals.index(list(n))
return(index)
# set of arbitrary arrays
li = np.array([[0, 1, 0], [0, 0, 1]])
# here is the performance improvement attempt with the help of the function above
arr = [get_val(n) for n in li]
print(arr)
I'm not completely sure if this is the correct way to do it for getting the needed value for a set like this. If there is a better way, please let me know.
Otherwise, I refer to my main question:
what is the best way possible to optimize the code?
Thanks so much for your help.

You can try use matrix multiplication (dot product):
a=np.array([[0, 0, 0],[0, 1, 0], [1, 0, 0], [0, 0, 1]]) # dict values
c=np.array([0,1,2,3]) # dict keys
li = np.array([[0, 1, 0], [0, 0, 1]])
b=np.linalg.pinv(a)#c # decoding table
result=li#b
print(result)

Scikit image: proper way of counting cells in the objects of an image

Say you have an image in the form of a numpy.array:
vals=numpy.array([[3,24,25,6,2],[8,7,6,3,2],[1,4,23,23,1],[45,4,6,7,8],[17,11,2,86,84]])
And you want to compute how many cells are inside each object, given a threshold value of 17 (example):
from scipy import ndimage
from skimage.measure import regionprops
blobs = numpy.where(vals>17, 1, 0)
labels, no_objects = ndimage.label(blobs)
props = regionprops(blobs)
If you check, this gives an image with 4 distinct objects over the threshold:
In[1]: blobs
Out[1]:
array([[0, 1, 1, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 1, 1, 0],
[1, 0, 0, 0, 0],
[0, 0, 0, 1, 1]])
In fact:
In[2]: no_objects
Out[2]: 4
I want to compute the number of cells (or area) of each object. The intended outcome is a dictionary with the object ID: number of cells format:
size={0:2,1:2,2:1,3:2}
My attempt:
size={}
for label in props:
size[label]=props[label].area
Returns an error:
Traceback (most recent call last):
File "<ipython-input-76-e7744547aa17>", line 3, in <module>
size[label]=props[label].area
TypeError: list indices must be integers, not _RegionProperties
I understand I am using label incorrectly, but the intent is to iterate over the objects. How to do this?

A bit of testing and research sometimes goes a long way.
The problem is both with blobs, because it is not carrying the different labels but only 0,1 values, and label, which needs to be replaced by an iterator looping over range(0,no_objects).
This solution seems to be working:
import skimage.measure as measure
import numpy
from scipy import ndimage
from skimage.measure import regionprops
vals=numpy.array([[3,24,25,6,2],[8,7,6,3,2],[1,4,23,23,1],[45,4,6,7,8],[17,11,2,86,84]])
blobs = numpy.where(vals>17, 1, 0)
labels, no_objects = ndimage.label(blobs)
#blobs is not in an amicable type to be processed right now, so:
labelled=ndimage.label(blobs)
resh_labelled=labelled[0].reshape((vals.shape[0],vals.shape[1])) #labelled is a tuple: only the first element matters
#here come the props
props=measure.regionprops(resh_labelled)
#here come the sought-after areas
size={i:props[i].area for i in range (0, no_objects)}
Result:
In[1]: size
Out[1]: {0: 2, 1: 2, 2: 1, 3: 2}
And if anyone wants to check for the labels:
In[2]: labels
Out[2]:
array([[0, 1, 1, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 2, 2, 0],
[3, 0, 0, 0, 0],
[0, 0, 0, 4, 4]])
And if anyone wants to plot the 4 objects found:
import matplotlib.pyplot as plt
plt.set_cmap('OrRd')
plt.imshow(labels,origin='upper')

To answer the original question:
You have to apply regionprops to the labeled image: props = regionprops(labels)
You can then construct the dictionary using:
size = {r.label: r.area for r in props}
which yields
{1: 2, 2: 2, 3: 1, 4: 2}

That regionprops will generate a lot more information than just the area of each blob. So, if you are just looking to get the count of pixels for the blobs, as an alternative and with focus on performance, we can use np.bincount on labels obtained with ndimage.label, like so -
np.bincount(labels.ravel())[1:]
Thus, for the given sample -
In [53]: labeled_areas = np.bincount(labels.ravel())[1:]
In [54]: labeled_areas
Out[54]: array([2, 2, 1, 2])
To have these results in a dictionary, one additional step would be -
In [55]: dict(zip(range(no_objects), labeled_areas))
Out[55]: {0: 2, 1: 2, 2: 1, 3: 2}

Cannot create non-empty csr.matrix() in scipy

I need sparse matrix to solve problem and according to description of scr.matrix() in scipy here http://docs.scipy.org/doc/scipy/reference/generated/scipy.sparse.csr_matrix.html#scipy.sparse.csr_matrix it fits perfectly for my issue.
However I cannot even initialize it.
When I use empty matrix example from this doc http://docs.scipy.org/doc/scipy-0.15.1/reference/generated/scipy.sparse.csr_matrix.html it works fine, exactly as in doc
>>> import numpy as np
>>> from scipy.sparse import csr_matrix
>>> csr_matrix((3, 4), dtype=np.int8).toarray()
array([[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]], dtype=int8)
but when I use example of non-empty martix or try to fill it with my own data
>>> row = np.array([0, 0, 1, 2, 2, 2])
>>> col = np.array([0, 2, 2, 0, 1, 2])
>>> data = np.array([1, 2, 3, 4, 5, 6])
>>> csr_matrix((data, (row, col)), shape=(3, 3)).toarray()
I always got this message
/Library/Python/2.7/site-packages/numpy-1.9.2-py2.7-macosx-10.10-
intel.egg/numpy/core/fromnumeric.py:2507: VisibleDeprecationWarning:
`rank` is deprecated; use the `ndim` attribute or function instead.
To find the rank of a matrix see `numpy.linalg.matrix_rank`.
VisibleDeprecationWarning)
What does it mean? I completely stuck. Excuse me for that question I'm new to scipy and need help.

It is only a warning, your matrix I expect to be created.
Scipy is caling an old numpy function. It was fixed in April 2014 in scipy.
Scipy changes at:
https://github.com/scipy/scipy/commit/fa1782e04fdab91f672ccf7a4ebfb887de50f01c