I have an 2 dimensional array with np.shape(input)=(a,b) and that looks like
input=array[array_1[0,0,0,1,0,1,2,0,3,3,2,...,entry_b],...array_a[1,0,0,1,2,2,0,3,1,3,3,...,entry_b]]
Now I want to create an array np.shape(output)=(a,b,b) in which every entry that had the same value in the input get the value 1 and 0 otherwise
for example:
input=[[1,0,0,0,1,2]]
output=[array([[1., 0., 0., 0., 1., 0.],
[0., 1., 1., 1., 0., 0.],
[0., 1., 1., 1., 0., 0.],
[0., 1., 1., 1., 0., 0.],
[1., 0., 0., 0., 1., 0.],
[0., 0., 0., 0., 0., 1.]])]
My code so far is looking like:
def get_matrix(svdata,padding_size):
List=[]
for k in svdata:
matrix=np.zeros((padding_size,padding_size))
for l in range(padding_size):
for m in range(padding_size):
if k[l]==k[m]:
matrix[l][m]=1
List.append(matrix)
return List
But it takes 2:30 min for an input array of shape (2000,256). How can I become more effiecient by using built in numpy solutions?
res = input[:,:,None]==input[:,None,:]
Should give boolean (a,b,b) array
res = res.astype(int)
to get a 0/1 array
You're trying to create the array y where y[i,j,k] is 1 if input[i,j] == input[i, k]. At least that's what I think you're trying to do.
So y = input[:,:,None] == input[:,None,:] will give you a boolean array. You can then convert that to np.dtype('float64') using astype(...) if you want.
Related
I want to create m number of matrices, each of which is an n x 1 numpy arrays. Moreover those matrices should have only two nonzero entries in the two rows, all other rows should have 0 as their entries, meaning that matrix number m=1 should have entries m[0,:]=m[1,:]=1, rest elements are 0. And similarly the last matrix m=m should have entries like m[n-1,:]=m[n,:]=1, where rest of the elements in other rows are 0. So for consecutive two matrices, the nonzero elements shift by two rows. And finally, I would like them to be stored into a dictionary or in a file.
What would be a neat way to do this?
Is this what you're looking for?
In [2]: num_rows = 10 # should be divisible by 2
In [3]: np.repeat(np.eye(num_rows // 2), 2, axis=0)
Out[3]:
array([[1., 0., 0., 0., 0.],
[1., 0., 0., 0., 0.],
[0., 1., 0., 0., 0.],
[0., 1., 0., 0., 0.],
[0., 0., 1., 0., 0.],
[0., 0., 1., 0., 0.],
[0., 0., 0., 1., 0.],
[0., 0., 0., 1., 0.],
[0., 0., 0., 0., 1.],
[0., 0., 0., 0., 1.]])
In terms of storage in a file, you can use np.save and np.load.
Note that the default data type for np.eye will be float64. If you expect your values to be small when you begin integrating or whatever you're planning on doing with your state vectors, I'd recommend setting the data type appropriately (like np.uint8 for positive integers < 256 for example).
i am able to get the checkerboard pattern, the + pattern and the one with 1's on the border but i am not able to figure this out. Can somebody help?
If you're sticking with whole dimensions, then as #Péter Leéh pointed out:
>>> np.eye(n) + np.fliplr(np.eye(n))
array([[1., 0., 0., 1.],
[0., 1., 1., 0.],
[0., 1., 1., 0.],
[1., 0., 0., 1.]])
will suffice, np.fliplr(x) (horizontal flip) is identical to np.flip(x, axis=1).
However if n is odd, then you will have to replace the center element with a 1. e.g. n=5:
>>> x = np.eye(n) + np.fliplr(np.eye(n))
>>> x[n//2, n//2] = 1
array([[1., 0., 0., 0., 1.],
[0., 1., 0., 1., 0.],
[0., 0., 1., 0., 0.],
[0., 1., 0., 1., 0.],
[1., 0., 0., 0., 1.]])
I am working with numpy arrays and and I would like to move a certain value located at certain index by 2 positions on the right while shifting the others element involved on the left by one position each. To be more precise, here below is what I would like to achieve:
from:
start = [0., 1., 2.],
[0., 0., 0.],
[5., 0., 1.]
to:
end = [1., 2., 0.],
[0., 0., 0.],
[5., 0., 1.]
As you can see from the first row, 0 has been moved by two position on the right and 1 and 2 by one position on the left. So far, I succeded by moving an element by one position and moving the other by defining:
def right_move(arr, index, n_steps:int):
row = index[0]
col = index[1]
try:
arr[row,col], arr[row, col + n_steps] = arr[row, col + n_steps], arr[row, col]
except:
pass
return arr
where n_steps=1. If I input n_steps=2, it won't work since the element in the middle of the swapping remain unchanged.
Can someone help me? Other easier solution are more then welcome!
Thanks
You can use numpy.roll:
import numpy as np
arr = np.array([[0., 1., 2.],
[0., 0., 0.],
[5., 0., 1.]])
arr[0] = np.roll(arr[0], 2)
arr
# output:
# [[1., 2., 0.],
# [0., 0., 0.],
# [5., 0., 1.]]
It is supposed the functionality of np.roll. However, np.roll doesn't allow roll on individual row. It piqued my interest to implement a function using as_strided to extend this functionality to np.roll
from numpy.lib.stride_tricks import as_strided
def custom_roll(arr, r_tup):
m = np.asarray(r_tup)
arr_roll = arr[:, [*range(arr.shape[1]),*range(arr.shape[1]-1)]].copy() #need `copy`
strd_0, strd_1 = arr_roll.strides
n = arr.shape[1]
result = as_strided(arr_roll, (*arr.shape, n), (strd_0 ,strd_1, strd_1))
return result[np.arange(arr.shape[0]), (n-m)%n]
Just passing a tuple indicate number of steps for each row. In your case, it rolls only row 0 either 2 positions forward or -1 position backward. so the tuple is (2,0,0) or (-1,0,0)
start = np.array([[0., 1., 2.],
[0., 0., 0.],
[5., 0., 1.])
out = custom_roll(start, (2,0,0))
Out[780]:
array([[1., 2., 0.],
[0., 0., 0.],
[5., 0., 1.]])
Or
out = custom_roll(start, (-1,0,0))
Out[782]:
array([[1., 2., 0.],
[0., 0., 0.],
[5., 0., 1.]])
Roll first and last row: (2,0,2)
out = custom_roll(start, (2,0,2))
Out[784]:
array([[1., 2., 0.],
[0., 0., 0.],
[0., 1., 5.]])
Given a 3d tenzor, say:
batch x sentence length x embedding dim
a = torch.rand((10, 1000, 96))
and an array(or tensor) of actual lengths for each sentence
lengths = torch .randint(1000,(10,))
outputs tensor([ 370., 502., 652., 859., 545., 964., 566., 576.,1000., 803.])
How to fill tensor ‘a’ with zeros after certain index along dimension 1 (sentence length) according to tensor ‘lengths’ ?
I want smth like that :
a[ : , lengths : , : ] = 0
One way of doing it (slow if batch size is big enough):
for i_batch in range(10):
a[ i_batch , lengths[i_batch ] : , : ] = 0
You can do it using a binary mask.
Using lengths as column-indices to mask we indicate where each sequence ends (note that we make mask longer than a.size(1) to allow for sequences with full length).
Using cumsum() we set all entries in mask after the seq len to 1.
mask = torch.zeros(a.shape[0], a.shape[1] + 1, dtype=a.dtype, device=a.device)
mask[(torch.arange(a.shape[0]), lengths)] = 1
mask = mask.cumsum(dim=1)[:, :-1] # remove the superfluous column
a = a * (1. - mask[..., None]) # use mask to zero after each column
For a.shape = (10, 5, 96), and lengths = [1, 2, 1, 1, 3, 0, 4, 4, 1, 3].
Assigning 1 to respective lengths at each row, mask looks like:
mask =
tensor([[0., 1., 0., 0., 0., 0.],
[0., 0., 1., 0., 0., 0.],
[0., 1., 0., 0., 0., 0.],
[0., 1., 0., 0., 0., 0.],
[0., 0., 0., 1., 0., 0.],
[1., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 1., 0.],
[0., 0., 0., 0., 1., 0.],
[0., 1., 0., 0., 0., 0.],
[0., 0., 0., 1., 0., 0.]])
After cumsum you get
mask =
tensor([[0., 1., 1., 1., 1.],
[0., 0., 1., 1., 1.],
[0., 1., 1., 1., 1.],
[0., 1., 1., 1., 1.],
[0., 0., 0., 1., 1.],
[1., 1., 1., 1., 1.],
[0., 0., 0., 0., 1.],
[0., 0., 0., 0., 1.],
[0., 1., 1., 1., 1.],
[0., 0., 0., 1., 1.]])
Note that it exactly has zeros where the valid sequence entries are and ones beyond the lengths of the sequences. Taking 1 - mask gives you exactly what you want.
Enjoy ;)
I am using h5py to build a dataset. Since I want to store arrays with different #of rows dimension, I use the h5py special_type vlen. However, I experience behavior I can't explain, maybe you can me help in understanding what is happening:
>>>> import h5py
>>>> import numpy as np
>>>> fp = h5py.File(datasource_fname, mode='w')
>>>> dt = h5py.special_dtype(vlen=np.dtype('float32'))
>>>> train_targets = fp.create_dataset('target_sequence', shape=(9549, 5,), dtype=dt)
>>>> test
Out[130]:
array([[ 0., 1., 1., 1., 0., 1., 1., 0., 1., 0., 0.],
[ 1., 0., 0., 0., 1., 0., 0., 1., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 1.]])
>>>> train_targets[0] = test
>>>> train_targets[0]
Out[138]:
array([ array([ 0., 1., 0., 0., 0., 1., 0., 0., 0., 0., 1.], dtype=float32),
array([ 1., 0., 0., 0., 1., 0., 0., 0., 0., 1., 0.], dtype=float32),
array([ 0., 0., 0., 1., 0., 0., 0., 0., 1., 0., 0.], dtype=float32),
array([ 0., 0., 1., 0., 0., 0., 0., 1., 0., 0., 0.], dtype=float32),
array([ 0., 1., 0., 0., 0., 0., 1., 0., 0., 0., 0.], dtype=float32)], dtype=object)
I do expect the train_targets[0] to be of this shape, however I can't recognize the rows in my array. They seem to be totally jumbled about, however it is consistent. By which I mean that every time I try the above code, train_targets[0] looks the same.
To clarify: the first element in my train_targets, in this case test, has shape (5,11), however the second element might be of shape (5,38) which is why I use vlen.
Thank you for your help
Mat
I think
train_targets[0] = test
has stored your (11,5) array as an F ordered array in a row of train_targets. According to the (9549,5) shape, that's a row of 5 elements. And since it is vlen, each element is a 1d array of length 11.
That's what you get back in train_targets[0] - an array of 5 arrays, each shape (11,), with values taken from test (order F).
So I think there are 2 issues - what a 2d shape means, and what vlen allows.
My version of h5py is pre v2.3, so I only get string vlen. But I suspect your problem may be that vlen only works with 1d arrays, an extension, so to speak, of byte strings.
Does the 5 in shape=(9549, 5,) have anything to do with 5 in the test.shape? I don't think it does, at least not as numpy and h5py see it.
When I make a file following the string vlen example:
>>> f = h5py.File('foo.hdf5')
>>> dt = h5py.special_dtype(vlen=str)
>>> ds = f.create_dataset('VLDS', (100,100), dtype=dt)
and then do:
ds[0]='this one string'
and look at ds[0], I get an object array with 100 elements, each being this string. That is, I've set a whole row of ds.
ds[0,0]='another'
is the correct way to set just one element.
vlen is 'variable length', not 'variable shape'. While the https://www.hdfgroup.org/HDF5/doc/TechNotes/VLTypes.html documentation is not entirely clear on this, I think you can store 1d arrays with shape (11,) and (38,) with vlen, but not 2d ones.
Actually, train_targets output is reproduced with:
In [54]: test1=np.empty((5,),dtype=object)
In [55]: for i in range(5):
test1[i]=test.T.flatten()[i:i+11]
It's 11 values taken from the transpose (F order), but shifted for each sub array.