Currently I have a 2D tensor, for each row, I want to insert a new element e before the first index of a specified value v. Additional information: cannot guarantee each row could have a such value. If there isn't, just append the element
Example: Supporse e is 0, v is 10, Given a tensor
[[9, 6, 5, 4, 10],
[8, 7, 3, 5, 5],
[4, 9, 10, 10, 10]]
I want to get
[[9, 6, 5, 4, 0, 10],
[8, 7, 3, 5, 5, 0],
[4, 9, 0, 10, 10, 10]]
Are there some Torch-style ways to do this? The worst case I can treat this as a trivial Python problem but I think the corresponding solution is a little time-consuming.
I haven't yet found a full PyTorch solution. I'll keep looking, but here is somewhere to start:
>>> v, e = 10, 0
>>> v, e = torch.tensor([v]), torch.tensor([e])
>>> x = torch.tensor([[ 9, 6, 5, 4, 10],
[ 8, 7, 3, 5, 5],
[ 4, 9, 10, 10, 10],
[10, 9, 7, 10, 2]])
To deal with the edge case where v is not found in one of the rows you can add a temporary column to x. This will ensure every row has a value v in it. We will use x_ as a helper tensor:
>>> x_ = torch.cat([x, v.repeat(x.size(0))[:, None]], axis=1)
tensor([[ 9, 6, 5, 4, 10, 10],
[ 8, 7, 3, 5, 5, 10],
[ 4, 9, 10, 10, 10, 10],
[10, 9, 7, 10, 2, 10]])
Find the indices of the first value v on each row:
>>> bp = (x_ == v).int().argmax(axis=1)
tensor([4, 5, 2, 0])
Finally, the easiest way to insert values at different positions in each row is with a list comprehension:
>>> torch.stack([torch.cat([xi[:bpi], e, xi[bpi:]]) for xi, bpi in zip(x, bp)])
tensor([[ 9, 6, 5, 4, 0, 10],
[ 8, 7, 3, 5, 5, 0],
[ 4, 9, 0, 10, 10, 10],
[ 0, 10, 9, 7, 10, 2]])
Edit - If v cannot occur in the first position, then no need for x_:
>>> x
tensor([[ 9, 6, 5, 4, 10],
[ 8, 7, 3, 5, 5],
[ 4, 9, 10, 10, 10]])
>>> bp = (x == v).int().argmax(axis=1) - 1
>>> torch.stack([torch.cat([xi[:bpi], e, xi[bpi:]]) for xi, bpi in zip(x, bp)])
tensor([[ 9, 6, 5, 0, 4, 10],
[ 8, 7, 3, 5, 0, 5],
[ 4, 0, 9, 10, 10, 10]])
Related
I would like to append ids to a list which meet a specific condition.
output = []
areac = [4, 4, 4, 4, 1, 6, 7,8,9,6, 10, 11]
arean = [1, 1, 1, 4, 5, 6, 7,8,9,10, 10, 10]
id = [1, 2, 3, 4, 5, 6, 7,8,9,10, 11, 12]
dist = [2, 2, 2, 4, 5, 6, 7.2,5,5,5, 8.5, 9.1]
for a,b,c,d in zip(areac,arean,id,dist):
if a >= 5 and b==b and d >= 3:
output.append(c)
print(comp)
else:
pass
The condition is the following:
- areacount has to be >= 5
- At least 3 ids with a distance of >= 3 with the same area_number
So the id output should be [10,11,12].I already tried a different attempt with Counter that didn't work out. Thanks for your help!
Here you go:
I changed the list names to something more descriptive.
output = []
area_counts = [4, 4, 4, 4, 1, 6, 7, 8, 9, 6, 10, 11]
area_numbers = [1, 1, 1, 4, 5, 6, 7, 8, 9, 10, 10, 10]
ids = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
distances = [2, 2, 2, 4, 5, 6, 7.2, 5, 5, 5, 8.5, 9.1]
temp_numbers, temp_ids = [], []
for count, number, id, distance in zip(counts, numbers, ids, distances):
if count >= 5 and distance >= 3:
temp_numbers.append(number)
temp_ids.append(id)
for (number, id) in zip(temp_numbers, temp_ids):
if temp_numbers.count(number) == 3:
output.append(id)
output will be:
[10, 11, 12]
I'm doing some beginner python exercises and one of them is to remove duplicates from a list. I've successfully done it, but the strange thing is that it is returning a dictionary instead of a list.
This is my code.
import random
a = []
b = []
for i in range(0,20):
n = random.randint(0,10)
a.append(n)
for i in range(0,20):
n = random.randint(0,10)
b.append(n)
print(sorted(a))
print(sorted(b))
c = set(list(a+b))
print(c)
and this is what it's spitting out
[0, 0, 1, 1, 1, 1, 2, 3, 4, 4, 6, 6, 7, 7, 7, 8, 9, 9, 10, 10]
[0, 1, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 6, 7, 8, 9, 9, 10, 10, 10]
{0, 1, 2, 3, 4, 6, 7, 8, 9, 10}
thanks in advance!
{0, 1, 2, 3, 4, 6, 7, 8, 9, 10} is a set, not a dictionary, a dictionary would be printed as {key:value, key:value, ...}
Try print(type(c)) and you'll see it prints <class 'set'> rather than <class 'dict'>
Also try the following
s = {1,2,3}
print(type(s))
d = {'a':1,'b':2,'c':3}
print(type(d))
You'll see the type is different
Let's say I have defined a function to generate a list of 6 random integers ranging from 0 to 10
import random
def func():
randomlist = random.sample(range(11), 6)
return randomlist
Run:
func()
Output:
[3, 7, 10, 9, 4, 1]
Now I want to define another function which calls func() inside with a parameter n to generate a multidimensional matrix, where each element will be replaced by a newly created list generated by func(), and n is the times of the completed replacement (so that the total number of integers in the matrix would be 6^n) -- for example --
when n=1, expected result:
[3, 7, 10, 9, 4, 1]
when n=2, expected result:
[[6, 2, 9, 1, 4, 0],
[7, 8, 1, 9, 4, 1],
[1, 0, 4, 6, 3, 1],
[9, 4, 3, 8, 6, 7],
[2, 4, 3, 9, 5, 6],
[4, 7, 2, 0, 1, 8]]
when n=3, expected result:
[[[4, 7, 3, 0, 2, 8],[1, 5, 6, 5, 4, 8],[8, 9, 6, 5, 10, 4],[7, 8, 6, 6, 4, 10],[7, 8, 1, 0, 2, 3],[4, 5, 8, 5, 4, 6]],
[[1, 7, 2, 0, 2, 8],[4, 5, 8, 8, 4, 5],[9, 5, 6, 2, 1, 3],[5, 4, 1, 2, 6, 10],[7, 5, 4, 1, 1, 4],[9, 6, 5, 2, 2, 1]],
[[8, 2, 7, 10, 2, 7],[8, 9, 5, 4, 5, 5],[5, 8, 7, 7, 4, 6],[9, 5, 9, 10, 5, 4],[1, 4, 5, 6, 5, 7],[9, 8, 7, 6, 5, 1]],
[[0, 7, 4, 0, 1, 9],[4, 7, 3, 0, 2, 8],[8, 9, 6, 5, 10, 4],[8, 2, 7, 10, 2, 7],[[7, 5, 4, 1, 1, 4],[7, 8, 1, 9, 4, 1]],
[[9, 5, 3, 9, 2, 8],[8, 9, 6, 5, 10, 4],[9, 4, 3, 8, 6, 7],[3, 7, 10, 9, 4, 1],[4, 7, 3, 0, 2, 8],[9, 4, 3, 8, 6, 7]],
[[5, 3, 4, 5, 2, 10],[[7, 5, 4, 1, 1, 4],[4, 7, 3, 0, 2, 8],[4, 5, 8, 8, 4, 5],[7, 8, 1, 9, 4, 1],[8, 2, 7, 10, 2, 7]]]
I think I'd need a recursive function, but I'm totally out of a clue. Any insights? Thanks a lot.
Well base case of 1 is just to call your func, otherwise build a list of the recursion call on n-1 6^(n-1) times:
def recur_6(n):
if n <= 1:
return func()
return [recur_6(n-1) for _ in range(6**(n-1))]
Live example
I'm hoping to calculate the distances between two points in a (Nx1) numpy array, i.e.:
a = [2, 5, 5, 12, 5, 3, 10, 8, 1, 3, 1]
I'm hoping to get a square matrix with the (normed) distances between each point:
sq = [[0, |2-5|, |2-5|, |2-12|, |2-5|, ...],
[|5-2|, 0, ...], ...]
So far, what I have doesn't work, giving wrong values for the square distance matrix. Is there a way to (I'm not sure if it is the correct term?) vectorise my method too, but am unfamiliar with the advanced indexing.
What I currently have is the following:
sq = np.zero((len(a), len(a))
for i in a:
for j in len(a+1):
sq[i,j] = np.abs(a[:,0] - a[:,0])
Would appreciate any help!
I think that by exploiting numpy broadcasting, this is the faster solution:
a = [2, 5, 5, 12, 5, 3, 10, 8, 1, 3, 1]
a = np.array(a).reshape(-1,1)
sq = np.abs(a.T-a)
sq
array([[ 0, 3, 3, 10, 3, 1, 8, 6, 1, 1, 1],
[ 3, 0, 0, 7, 0, 2, 5, 3, 4, 2, 4],
[ 3, 0, 0, 7, 0, 2, 5, 3, 4, 2, 4],
[10, 7, 7, 0, 7, 9, 2, 4, 11, 9, 11],
[ 3, 0, 0, 7, 0, 2, 5, 3, 4, 2, 4],
[ 1, 2, 2, 9, 2, 0, 7, 5, 2, 0, 2],
[ 8, 5, 5, 2, 5, 7, 0, 2, 9, 7, 9],
[ 6, 3, 3, 4, 3, 5, 2, 0, 7, 5, 7],
[ 1, 4, 4, 11, 4, 2, 9, 7, 0, 2, 0],
[ 1, 2, 2, 9, 2, 0, 7, 5, 2, 0, 2],
[ 1, 4, 4, 11, 4, 2, 9, 7, 0, 2, 0]])
With numpy the following line might be the shortest to your result:
import numpy as np
a = np.array([2, 5, 5, 12, 5, 3, 10, 8, 1, 3, 1])
sq = np.array([np.array([(np.abs(i - j)) for j in a]) for i in a])
print(sq)
The following would give you the desired result without numpy.
a = [2, 5, 5, 12, 5, 3, 10, 8, 1, 3, 1]
sq = []
for i in a:
distances = []
for j in a:
distances.append(abs(i-j))
sq.append(distances)
print(sq)
With both, the result comes as:
[[0, 3, 3, 10, 3, 1, 8, 6, 1, 1, 1], [3, 0, 0, 7, 0, 2, 5, 3, 4, 2, 4], [3, 0, 0, 7, 0, 2, 5, 3, 4, 2, 4], [10, 7, 7, 0, 7, 9, 2, 4, 11, 9, 11], [3, 0, 0, 7, 0, 2, 5, 3, 4, 2, 4], [1, 2, 2, 9, 2, 0, 7, 5, 2, 0, 2], [8, 5, 5, 2, 5, 7, 0, 2, 9, 7, 9], [6, 3, 3, 4, 3, 5, 2, 0, 7, 5, 7], [1, 4, 4, 11, 4, 2, 9, 7, 0, 2, 0], [1, 2, 2, 9, 2, 0, 7, 5, 2, 0, 2], [1, 4, 4, 11, 4, 2, 9, 7, 0, 2, 0]]
There may be more than one way to do this but one way is to only use numpy operations instead of loops because internally python does lots of optimizations for numpy arrays.
One way to do only using array operations is to create an NxN matrix by repeating the original matrix (a) N times.
This will create a matrix N times.
E.g:
a = [1, 2, 3]
b = [[1 , 2, 3], [1 , 2, 3], [1 , 2, 3]]
Then you can do a matrix, array operation of
ans = abs(b - a)
Assuming a is numpy array, you can do:
b = np.repeat(a,a.shape).reshape((a.shape[0],a.shape[0]))
ans = np.abs(b - a)
I was wondering if there was a reasonably efficient way of sampling from a 2d numpy array. If I have a generic array:
dims = (4,4)
test_array = np.arange(np.prod(dims)).reshape(*dims)
test_array
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
Then I'd like to randomly set, say, two elements from it to a specific value (let's say 100). I've tried creating an indexing array and then applying that:
sample_from = np.random.randint(low=0, high=5, size=(2,2))
sample_from
array([[0, 2],
[1, 1]])
But if I try using this to index, it gives me a slightly unexpected answer:
test_array[sample_from]
array([[[ 0, 1, 2, 3],
[ 8, 9, 10, 11]],
[[ 4, 5, 6, 7],
[ 4, 5, 6, 7]]])
What I would have expected (and the kind of result I'd like) is if I'd just entered the indexing array directly:
test_array[[0,2],[1,1]] = 100
test_array
giving:
array([[ 0, 100, 2, 3],
[ 4, 5, 6, 7],
[ 8, 100, 10, 11],
[ 12, 13, 14, 15]])
Any help gratefully received.
You could use np.random.choice + np.unravel_index to assign directly to your array.
test_array[
np.unravel_index(np.random.choice(np.prod(dims), 2, replace=False), dims)
] = 100