I'm using numpy in python , in order to create a nx1 matrix . I want the 1st element of the matrix to be 3 , the 2nd -1 , then the n-1 element -1 again and at the end the n element 3. All the in between elements , i.e. from element 3 to element n-2 should be 0. I've made a drawing of the mentioned matrix , is like this :
I'm fairly new to python and using numpy but seems like a great tool for managing matrices. What I've tried so far is creating the nx1 array (giving n some value) and initializing it to 0 .
import numpy as np
n = 100
I = np.arange(n)
matrix = np.row_stack(0*I)
print("\Matrix is \n",matrix)
Any clues to how i proceed? Or what routine to use ?
Probably the simplest way is to just do the following:
import numpy as np
n = 10
a = np.zeros(n)
a[0] = 3
a[1] = -1
a[len(a)-1] = 3
a[len(a)-2] = -1
>>print(a)
output: [ 3. -1. 0. 0. 0. 0. 0. 0. -1. 3.]
Hope this helps ;)
In [97]: n=10
In [98]: arr = np.zeros(n,int)
In [99]: arr[[0,-1]]=3; arr[[1,-2]]=-1
In [100]: arr
Out[100]: array([ 3, -1, 0, 0, 0, 0, 0, 0, -1, 3])
Easily changed to (n,1):
In [101]: arr[:,None]
Out[101]:
array([[ 3],
[-1],
[ 0],
[ 0],
[ 0],
[ 0],
[ 0],
[ 0],
[-1],
[ 3]])
I guess something that works is :
import numpy as np
n = 100
I = np.arange(n)
matrix = np.row_stack(0*I)
matrix[0]=3
matrix[1]=-1
matrix[n-2]=-1
matrix[n-1]=3
print("\Matrix is \n",matrix)
Related
I have a numpy array named heartbeats with 100 rows. Each row has 5 elements.
I also have a single array named time_index with 5 elements.
I need to prepend the time index to each row of heartbeats.
heartbeats = np.array([
[-0.58, -0.57, -0.55, -0.39, -0.40],
[-0.31, -0.31, -0.32, -0.46, -0.46]
])
time_index = np.array([-2, -1, 0, 1, 2])
What I need:
array([-2, -0.58],
[-1, -0.57],
[0, -0.55],
[1, -0.39],
[2, -0.40],
[-2, -0.31],
[-1, -0.31],
[0, -0.32],
[1, -0.46],
[2, -0.46])
I only wrote two rows of heartbeats to illustrate.
Assuming you are using numpy, the exact output array you are looking for can be made by stacking a repeated version of time_index with the raveled version of heartbeats:
np.stack((np.tile(time_index, len(heartbeats)), heartbeats.ravel()), axis=-1)
Another approach, using broadcasting
In [13]: heartbeats = np.array([
...: [-0.58, -0.57, -0.55, -0.39, -0.40],
...: [-0.31, -0.31, -0.32, -0.46, -0.46]
...: ])
...: time_index = np.array([-2, -1, 0, 1, 2])
Make a target array:
In [14]: res = np.zeros(heartbeats.shape + (2,), heartbeats.dtype)
In [15]: res[:,:,1] = heartbeats # insert a (2,5) into a (2,5) slot
In [17]: res[:,:,0] = time_index[None] # insert a (5,) into a (2,5) slot
In [18]: res
Out[18]:
array([[[-2. , -0.58],
[-1. , -0.57],
[ 0. , -0.55],
[ 1. , -0.39],
[ 2. , -0.4 ]],
[[-2. , -0.31],
[-1. , -0.31],
[ 0. , -0.32],
[ 1. , -0.46],
[ 2. , -0.46]]])
and then reshape to 2d:
In [19]: res.reshape(-1,2)
Out[19]:
array([[-2. , -0.58],
[-1. , -0.57],
[ 0. , -0.55],
[ 1. , -0.39],
[ 2. , -0.4 ],
[-2. , -0.31],
[-1. , -0.31],
[ 0. , -0.32],
[ 1. , -0.46],
[ 2. , -0.46]])
[17] takes a (5,), expands it to (1,5), and then to (2,5) for the insert. Read up on broadcasting.
As an alternative way, you can repeat time_index by np.concatenate based on the specified times:
concatenated = np.concatenate([time_index] * heartbeats.shape[0])
# [-2 -1 0 1 2 -2 -1 0 1 2]
# result = np.dstack((concatenated, heartbeats.reshape(-1))).squeeze()
result = np.array([concatenated, heartbeats.reshape(-1)]).T
Using np.concatenate may be faster than np.tile. This solution is faster than Mad Physicist, but the fastest is using broadcasting as hpaulj's answer.
I have a numpy array I'll use np.ones((2,3)) as a MWE:
arr = [[1,1,1],
[1,1,1],
[1,1,1]]
I wish to shift the rows by a set integer. This will increase with the row.
Shift the 1st row by 0
shift the 5th row by 4
I imagine the row length will have to be equal for all rows giving something list this:
to give this:
arr = [[1,1,1,0,0],
[0,1,1,1,0],
[0,0,1,1,1]]
This is a MWE and the actual arrays are taken from txt files and are up to (1000x96). The important values are not just 1 but any float from 0->inf.
Is there a way of doing this?
(Extra information: these data are for 2D heatmap plotting)
Assuming an array with arbitrary values, you could use:
# add enough "0" columns for the shift
arr2 = np.c_[arr, np.zeros((arr.shape[0], arr.shape[0]-1), dtype=arr.dtype)]
# get the indices as ogrid
r, c = np.ogrid[:arr2.shape[0], :arr2.shape[1]]
# roll the values
arr2 = arr2[r, c-r]
used input:
arr = np.arange(1,10).reshape(3,3)
# array([[1, 2, 3],
# [4, 5, 6],
# [7, 8, 9]])
output:
array([[1, 2, 3, 0, 0],
[0, 4, 5, 6, 0],
[0, 0, 7, 8, 9]])
I have the following solution:
import numpy as np
arr = [[1,1,1],
[1,1,1],
[1,1,1],
[1,1,1]]
arr = np.array(arr)
shift = 1
extend = shift*(np.shape(arr)[0]-1)
arr2 = np.zeros((np.shape(arr)[0],extend+np.shape(arr)[1]))
for i,row in enumerate(arr):
arr2[i,(i*shift):(i*shift)+3] = row
print(arr2)
[[1. 1. 1. 0. 0. 0.]
[0. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 0.]
[0. 0. 0. 1. 1. 1.]]
I have a matrix containing positive and negative numbers like this:
>>> source_matrix
array([[-4, -2, 0],
[-5, 0, 4],
[ 0, 6, 5]])
I'd like to had a copy of this matrix with inverted negatives:
>>> result
array([[-0.25, -0.5, 0],
[-0.2, 0, 4],
[ 0, 6, 5]])
Firstly, since your desired array is gonna contain float type you need to determine the array's dtype at creation time as float. The reason for that is because if you assign the float results of the inverted sub-array they'll automatically be casted to float. Secondly, you need to find the negative numbers in your array and then use a simple indexing in order to grab them and use np.true_divide() to perform the inversion.
In [25]: arr = np.array([[-4, -2, 0],
...: [-5, 0, 4],
...: [ 0, 6, 5]], dtype=np.float)
...:
...:
In [26]: mask = arr < 0
In [27]: arr[mask] = np.true_divide(1, arr[mask])
In [28]: arr
Out[28]:
array([[-0.25, -0.5 , 0. ],
[-0.2 , 0. , 4. ],
[ 0. , 6. , 5. ]])
You can also achieve this without masking, by using the where and out params of true_divide.
a = np.array([[-4, -2, 0],
[-5, 0, 4],
[ 0, 6, 5]], dtype=np.float)
np.true_divide(1, a, out=a, where=a<0)
Giving the result:
array([[-0.25, -0.5 , 0. ],
[-0.2 , 0. , 4. ],
[ 0. , 6. , 5. ]])
The where= parameter is passed an array of the same dimensions as your two inputs. Where this evaluates to True the divide is performed. Where it evaluates to False, the original input, passed in via out= is output into the result unchanged.
This question already has answers here:
Generalise slicing operation in a NumPy array
(4 answers)
Closed 5 years ago.
Here is some code I'm struggling with.
My goal is to create an array (db) from an existing one (t) , in db each line will represent a value of t. db will have 3 column, 1 for line index in t, 1 for column index in t and 1 for the value in t.
In my case, t was a distance matrix, thus diagonal was 0 and it was symetric, I replaced lower triangular values with 0. I don't need 0 values in the new array but I can just delete them in another step.
import numpy as np
t = np.array([[0, 2.5],
[0, 0]])
My goal is to obtain a new array such as :
db = np.array([[0, 0, 0],
[0, 1, 2.5],
[1, 0, 0],
[1, 1, 0]])
Thanks for your time.
You can create a meshgrid of 2D coordinates for the rows and columns, then unroll these into 1D arrays. You can then concatenate these two arrays as well as the unrolled version of t into one final matrix:
import numpy as np
(Y, X) = np.meshgrid(np.arange(t.shape[1]), np.arange(t.shape[0]))
db = np.column_stack((X.ravel(), Y.ravel(), t.ravel()))
Example run
In [9]: import numpy as np
In [10]: t = np.array([[0, 2.5],
...: [0, 0]])
In [11]: (Y, X) = np.meshgrid(np.arange(t.shape[1]), np.arange(t.shape[0]))
In [12]: db = np.column_stack((X.ravel(), Y.ravel(), t.ravel()))
In [13]: db
Out[13]:
array([[ 0. , 0. , 0. ],
[ 0. , 1. , 2.5],
[ 1. , 0. , 0. ],
[ 1. , 1. , 0. ]])
I've got a simple problem and I can't figure out how to solve it.
Here is a matrix: A = np.array([[1,0,3],[0,7,9],[0,0,8]]).
I want to find a quick way to replace all elements of this matrix by their inverses, excluding of course the zero elements.
I know, thanks to the search engine of Stackoverflow, how to replace an element by a given value with a condition. On the contrary, I do not figure out how to replace elements by new elements depending on the previous ones (e.g. squared elements, inverses, etc.)
Use 1. / A (notice the dot for Python 2):
>>> A
array([[1, 0, 3],
[0, 7, 9],
[0, 0, 8]], dtype)
>>> 1./A
array([[ 1. , inf, 0.33333333],
[ inf, 0.14285714, 0.11111111],
[ inf, inf, 0.125 ]])
Or if your array has dtype float, you can do it in-place without warnings:
>>> A = np.array([[1,0,3], [0,7,9], [0,0,8]], dtype=np.float64)
>>> A[A != 0] = 1. / A[A != 0]
>>> A
array([[ 1. , 0. , 0.33333333],
[ 0. , 0.14285714, 0.11111111],
[ 0. , 0. , 0.125 ]])
Here we use A != 0 to select only those elements that are non-zero.
However if you try this on your original array you'd see
array([[1, 0, 0],
[0, 0, 0],
[0, 0, 0]])
because your array could only hold integers, and inverse of all others would have been rounded down to 0.
Generally all of the numpy stuff on arrays does element-wise vectorized transformations so that to square elements,
>>> A = np.array([[1,0,3],[0,7,9],[0,0,8]])
>>> A * A
array([[ 1, 0, 9],
[ 0, 49, 81],
[ 0, 0, 64]])
And just a note on Antti Haapala's answer, (Sorry, I can't comment yet)
if you wanted to keep the 0's, you could use
B=1./A #I use the 1. to make sure it uses floats
B[B==np.inf]=0