Find second argument from first argument in a list - python-3.x

I have the following problem: let L be a list ( (i,j), w_i,j ) for i and j between 1 and N. Suppose we know the couple (i,j) (for example : (5,8)), is there a way to find w_5,8 only from the list L?
NB: my code is a priori better if I use the liste L instead of a matrix w[i][j], since it is mostly empty, so I would rather not use such a matrix.
Thanks in advance

There are many ways to store and lookup values of a (sparse) matrix, which I assume is the goal here.
Option 1
Do not attempt to optimize to early. Depending on what kind of computations you want to perform on the matrix you may be best advised to simply use a numpy array and do not worry about sparseness.
If your matrix is really large and indeed sparse you can use the specialised sparse matrix implementations of scipy.
Option 2
You already have the list L in the format you described and you do not want to use any of the numeric packages mentioned above. Then you can convert L into a dictionary for easy value lookup:
# example list
L = [((1,2), 456.5), ((5,4), 33.5)]
# convert to dictionary
D = dict(L)
# lookup
v = D[(1,2)] # v is now 456.5
# missing value
v = D[(3,3)] # throws KeyNotFound exception
# convert to dictionary with default value for missing keys
from collections import defaultdict
D = defaultdict(int, L)
# lookup
v = D[(1,2)] # v is now 456.5
# missing value
v = D[(3,3)] # v is now 0

Related

subtracting every element of a list from a previous element in a dictionary

I have a dictionary which has a 2D list (list of a list). This 2D list contains x and y coordinates [x,y] of a particle. Whenever the particle moves, its new coordinates are appended to this 2D list in a dictionary. I want to calculate the distance between every location and append the result to another list (can just be a normal list without dictionary). What I want is something like the following:
dist1 = sqrt((x1-x0)^2 + (y1-y0)^2)
dist2 = sqrt((x2-x1)^2 + (y2-y1)^2)
.....
distN = sqrt((xN-xN-1)^2 + (yN-yN-1)^2)
but I am having issues in accessing elements of a list in a dictionary. I have a very long 2D list but you can use the below example to give me some suggestions.
c = {"coordinates":[[1,2],[3,4],[5,6],[7,8]]}
for k, dk in c.items():
for x in dk:
print(x[0], x[1])
I can access one element in the dk at a time in a loop but how to get the previous one? There should be a nice way of doing it but I just don't know.
Any help will be appreciated.
Using a for loop (probably not the most efficient solution):
import numpy as np
c = {"coordinates":[[1,2],[3,4],[5,6],[7,8]]}
coordinates = np.array(c['coordinates'])
distances = []
for i in range(1, len(coordinates)):
distances.append(np.linalg.norm(coordinates[i-1] - coordinates[i]))
print(distances)
# [2.8284271247461903, 2.8284271247461903, 2.8284271247461903]
I also used numpy and its linalg.norm function to calculate the distance (How can the Euclidean distance be calculated with NumPy?), but you could ofcourse use your own function or calculation in case you'd want that.
I tried this and it also works:
c = {"coordinates":[[1,2],[3,4],[15,6],[7,8]]}
l1 = []
for k, dk in c.items():
for x in dk:
l1.append(x)
print(l1)
dist = [math.sqrt((p1[0]-p0[0])**2 + (p1[1]-p0[1])**2) for p1,p0 in zip(l1,l1[1:]
as others suggested in this question, better way to get l1 is to use the following command:
l1 = c["coordinates"]
dist = [math.sqrt((p1[0]-p0[0])**2 + (p1[1]-p0[1])**2) for p1,p0 in zip(l1,l1[1:]

Multiply every element of matrix with a vector to obtain a matrix whose elements are vectors themselves

I need help in speeding up the following block of code:
import numpy as np
x = 100
pp = np.zeros((x, x))
M = np.ones((x,x))
arrayA = np.random.uniform(0,5,2000)
arrayB = np.random.uniform(0,5,2000)
for i in range(x):
for j in range(x):
y = np.multiply(arrayA, np.exp(-1j*(M[j,i])*arrayB))
p = np.trapz(y, arrayB) # Numerical evaluation/integration y
pp[j,i] = abs(p**2)
Is there a function in numpy or another method to rewrite this piece of code with so that the nested for-loops can be omitted? My idea would be a function that multiplies every element of M with the vector arrayB so we get a 100 x 100 matrix in which each element is a vector itself. And then further each vector gets multiplied by arrayA with the np.multiply() function to then again obtain a 100 x 100 matrix in which each element is a vector itself. Then at the end perform numerical integration for each of those vectors with np.trapz() to obtain a 100 x 100 matrix of which each element is a scalar.
My problem though is that I lack knowledge of such functions which would perform this.
Thanks in advance for your help!
Edit:
Using broadcasting with
M = np.asarray(M)[..., None]
y = 1000*arrayA*np.exp(-1j*M*arrayB)
return np.trapz(y,B)
works and I can ommit the for-loops. However, this is not faster, but instead a little bit slower in my case. This might be a memory issue.
y = np.multiply(arrayA, np.exp(-1j*(M[j,i])*arrayB))
can be written as
y = arrayA * np.exp(-1j*M[:,:,None]*arrayB
producing a (x,x,2000) array.
But the next step may need adjustment. I'm not familiar with np.trapz.
np.trapz(y, arrayB)

Find n smallest values in a list of tensors

I am trying to find the indices of the n smallest values in a list of tensors in pytorch. Since these tensors might contain many non-unique values, I cannot simply compute percentiles to obtain the indices. The ordering of non-unique values does not matter however.
I came up with the following solution but am wondering if there is a more elegant way of doing it:
import torch
n = 10
tensor_list = [torch.randn(10, 10), torch.zeros(20, 20), torch.ones(30, 10)]
all_sorted, all_sorted_idx = torch.sort(torch.cat([t.view(-1) for t in tensor_list]))
cum_num_elements = torch.cumsum(torch.tensor([t.numel() for t in tensor_list]), dim=0)
cum_num_elements = torch.cat([torch.tensor([0]), cum_num_elements])
split_indeces_lt = [all_sorted_idx[:n] < cum_num_elements[i + 1] for i, _ in enumerate(cum_num_elements[1:])]
split_indeces_ge = [all_sorted_idx[:n] >= cum_num_elements[i] for i, _ in enumerate(cum_num_elements[:-1])]
split_indeces = [all_sorted_idx[:n][torch.logical_and(lt, ge)] - c for lt, ge, c in zip(split_indeces_lt, split_indeces_ge, cum_num_elements[:-1])]
n_smallest = [t.view(-1)[idx] for t, idx in zip(tensor_list, split_indeces)]
Ideally a solution would pick a random subset of the non-unique values instead of picking the entries of the first tensor of the list.
Pytorch does provide a more elegant (I think) way to do it, with torch.unique_consecutive (see here)
I'm going to work on a tensor, not a list of tensors because as you did yourself, there's just a cat to do. Unraveling the indices afterward is not hard either.
# We want to find the n=3 min values and positions in t
n = 3
t = torch.tensor([1,2,3,2,0,1,4,3,2])
# To get a random occurrence, we create a random permutation
randomizer = torch.randperm(len(t))
# first, we sort t, and get the indices
sorted_t, idx_t = t[randomizer].sort()
# small util function to extract only the n smallest values and positions
head = lambda v,w : (v[:n], w[:n])
# use unique_consecutive to remove duplicates
uniques_t, counts_t = head(*torch.unique_consecutive(sorted_t, return_counts=True))
# counts_t.cumsum gives us the position of the unique values in sorted_t
uniq_idx_t = torch.cat([torch.tensor([0]), counts_t.cumsum(0)[:-1]], 0)
# And now, we have the positions of uniques_t values in t :
final_idx_t = randomizer[idx_t[uniq_idx_t]]
print(uniques_t, final_idx_t)
#>>> tensor([0,1,2]), tensor([4,0,1])
#>>> tensor([0,1,2]), tensor([4,5,8])
#>>> tensor([0,1,2]), tensor([4,0,8])
EDIT : I think the added permutation solves your need-random-occurrence problem

user def. function modifying argument though it is not supposed to

Just for practice, I am using nested lists (for exaple, [[1, 0], [0, 1]] is the 2*2 identity matrix) as matrices. I am trying to compute determinant by reducing it to an upper triangular matrix and then by multiplying its diagonal entries. To do this:
"""adds two matrices"""
def add(A, B):
S = []
for i in range(len(A)):
row = []
for j in range(len(A[0])):
row.append(A[i][j] + B[i][j])
S.append(row)
return S
"""scalar multiplication of matrix with n"""
def scale(n, A):
return [[(n)*x for x in row] for row in A]
def detr(M):
Mi = M
#the loops below are supossed to convert Mi
#to upper triangular form:
for i in range(len(Mi)):
for j in range(len(Mi)):
if j>i:
k = -(Mi[j][i])/(Mi[i][i])
Mi[j] = add( scale(k, [Mi[i]]), [Mi[j]] )[0]
#multiplies diagonal entries of Mi:
k = 1
for i in range(len(Mi)):
k = k*Mi[i][i]
return k
Here, you can see that I have set M (argument) equal to Mi and and then operated on Mi to take it to upper triangular form. So, M is supposed to stay unmodified. But after using detr(A), print(A) prints the upper triangular matrix. I tried:
setting X = M, then Mi = X
defining kill(M): return M and then setting Mi = kill(M)
But these approaches are not working. This was causing some problems as I was trying to use detr(M) in another function, problems which I was able to bypass, but why is this happening? What is the compiler doing here, why was M modified even though I operated only on Mi?
(I am using Spyder 3.3.2, Python 3.7.1)
(I am sorry if this question is silly, but I have only started learning python and new to coding in general. This question means a lot to me because I still don't have a deep understanding of this language.)
See python documentation about assignment:
https://docs.python.org/3/library/copy.html
Assignment statements in Python do not copy objects, they create bindings between a target and an object. For collections that are mutable or contain mutable items, a copy is sometimes needed so one can change one copy without changing the other.
You need to import copy and then use Mi = copy.deepcopy(M)
See also
How to deep copy a list?

How do I call a list of numpy functions without a for loop?

I'm doing data analysis that involves minimizing the least-square-error between a set of points and a corresponding set of orthogonal functions. In other words, I'm taking a set of y-values and a set of functions, and trying to zero in on the x-value that gets all of the functions closest to their corresponding y-value. Everything is being done in a 'data_set' class. The functions that I'm comparing to are all stored in one list, and I'm using a class method to calculate the total lsq-error for all of them:
self.fits = [np.poly1d(np.polyfit(self.x_data, self.y_data[n],10)) for n in range(self.num_points)]
def error(self, x, y_set):
arr = [(y_set[n] - self.fits[n](x))**2 for n in range(self.num_points)]
return np.sum(arr)
This was fine when I had significantly more time than data, but now I'm taking thousands of x-values, each with a thousand y-values, and that for loop is unacceptably slow. I've been trying to use np.vectorize:
#global scope
def func(f,x):
return f(x)
vfunc = np.vectorize(func, excluded=['x'])
…
…
#within data_set class
def error(self, x, y_set):
arr = (y_set - vfunc(self.fits, x))**2
return np.sum(arr)
func(self.fits[n], x) works fine as long as n is valid, and as far as I can tell from the docs, vfunc(self.fits, x) should be equivalent to
[self.fits[n](x) for n in range(self.num_points)]
but instead it throws:
ValueError: cannot copy sequence with size 10 to array axis with dimension 11
10 is the degree of the polynomial fit, and 11 is (by definition) the number of terms in it, but I have no idea why they're showing up here. If I change the fit order, the error message reflects the change. It seems like np.vectorize is taking each element of self.fits as a list rather than a np.poly1d function.
Anyway, if someone could either help me understand np.vectorize better, or suggest another way to eliminate that loop, that would be swell.
As the functions in question all have a very similar structure we can "manually" vectorize once we've extracted the poly coefficients. In fact, the function is then a quite simple one-liner, eval_many below:
import numpy as np
def poly_vec(list_of_polys):
O = max(p.order for p in list_of_polys)+1
C = np.zeros((len(list_of_polys), O))
for p, c in zip(list_of_polys, C):
c[len(c)-p.order-1:] = p.coeffs
return C
def eval_many(x,C):
return C#np.vander(x,11).T
# make example
list_of_polys = [np.poly1d(v) for v in np.random.random((1000,11))]
x = np.random.random((2000,))
# put all coeffs in one master matrix
C = poly_vec(list_of_polys)
# test
assert np.allclose(eval_many(x,C), [p(x) for p in list_of_polys])
from timeit import timeit
print('vectorized', timeit(lambda: eval_many(x,C), number=100)*10)
print('loopy ', timeit(lambda: [p(x) for p in list_of_polys], number=10)*100)
Sample run:
vectorized 6.817315469961613
loopy 56.35076989419758

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