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I have a list of numpy arrays and want to modify some numbers of arrays. This is my simplified list:
first_list=[np.array([[1.,2.,0.], [2.,1.,0.], [6.,8.,3.], [8.,9.,7.]]),
np.array([[1.,0.,2.], [0.,0.,2.], [5.,5.,1.], [0.,6.,2.]])]
I have a factor which defines how many splits I have in each arrays:
spl_array=2.
it means each array of the list can be splited into 2 ones. I want to add a fixed value (3.) into last column of each split of each array and also copy the last split and subtract this value (3.) from the third column of this copied split. Finally I want to have it as following:
final_list=[np.array([[1.,2.,3.], [2.,1.,3.], [6.,8.,6.], [8.,9.,10.], \
[6.,8.,0.], [8.,9.,4.]]), # copied and subtracted
np.array([[1.,0.,5.], [0.,0.,5.], [5.,5.,4.], [0.,6.,5.], \
[5.,5.,-2.], [0.,6.,-1.]])] # copied and subtracted
I tried some for loops but I totaly lost. In advance , I do appreciate any help.
final_list=[]
for i in first_list:
each_lay=np.split (i, spl_array)
for j in range (len(each_lay)):
final_list.append([each_lay[j][:,0], each_lay[j][:,1], each_lay[j][:,2]+3])
Is it what you expect:
m = np.asarray(first_list)
m = np.concatenate((m, m[:, 2:]), axis=1)
m[:, :4, 2] += 3
m[:, 4:, 2] -= 3
final_list = m.tolist()
>>> m
array([[[ 1., 2., 3.],
[ 2., 1., 3.],
[ 6., 8., 6.],
[ 8., 9., 10.],
[ 6., 8., 0.],
[ 8., 9., 4.]],
[[ 1., 0., 5.],
[ 0., 0., 5.],
[ 5., 5., 4.],
[ 0., 6., 5.],
[ 5., 5., -2.],
[ 0., 6., -1.]]])
This question already has answers here:
Find the min/max excluding zeros in a numpy array (or a tuple) in python
(6 answers)
Closed 2 years ago.
I'm trying to find the smallest non-zero value in each row of a 2d numpy array but haven't been to find an elegant solution. I've looked at some other posts but none address the exact same problem e.g.
Minimum value in 2d array or Min/Max excluding zeros but in 1d array.
For example for the given array:
x = np.array([[3., 2., 0., 1., 6.], [8., 4., 5., 0., 6.], [0., 7., 2., 5., 0.]])
the answer would be:
[1., 4., 2.]
One way to do this is to re-assign the zeros to the np.inf, then take min per row:
np.where(x>0, x, np.inf).min(axis=1)
Output:
array([1., 4., 2.])
Masked arrays are designed exactly for these kind of purposes. You can leverage masking zeros from array (or ANY other kind of mask you desire) and do pretty much most of the stuff you do on regular arrays on your masked array now:
import numpy.ma as ma
mx = ma.masked_array(x, mask=x==0)
mx.min(1)
output:
[1.0 4.0 2.0]
I solved this way that's time complexity is o(n^2) .
import numpy as np
x = np.array([[3., 2., 0., 1., 6.], [8., 4., 5., 0., 6.], [0., 7., 2., 5., 0.]])
for i in range(len(x)) :
small=x[i][i]
for j in x[i] :
if (j!=0 and j<small):
small=j
print(small)
# example data
x = np.array([[3., 2., 0., 1., 6.], [8., 4., 5., 0., 6.], [0., 7., 2., 5., 0.]])
# set all the values inside the maxtrix which are equal to 0, to *inf*
# np.inf represents a very large number
# inf, stands for infinity
x[x==0] = np.inf
# grep the lowest value, in each array (now that there is no 0 value anymore)
np.min(x, axis=1)
Suppose we create an numpy array like this:
x = np.linspace(1,5,5).reshape(-1,1)
which results in this:
array([[ 1.],
[ 2.],
[ 3.],
[ 4.],
[ 5.]])
now we add the transpose of this array to it:
x + x.T
which results in this:
array([[ 2., 3., 4., 5., 6.],
[ 3., 4., 5., 6., 7.],
[ 4., 5., 6., 7., 8.],
[ 5., 6., 7., 8., 9.],
[ 6., 7., 8., 9., 10.]])
I don't understand this because the two arrays have different dimensions (5x1 and 1x5) and I learned in linear algebra that we can only sum up matrices when they have the same dimension.
Edit: OK thanks, got it
Here, we have
x = array([[ 1.],[ 2.],[ 3.],[ 4.],[ 5.]])
x.T = array([[ 1., 2., 3., 4., 5.]])
Now you are trying to add two matrices of different dimensions (1 X 5) and (5 X 1).
The way numpy handles this is by copying elements in each row of 1st matrix across its columns to match a number of columns of 2nd matrix and copying elements in each column of 2nd matrix across its rows to match no. of rows of 1st matrix. This gives you 2 5 X 5 matrix which can be added together.
The element wise addition is done as
array([[ 1., 1., 1., 1., 1.],[ 2., 2., 2., 2., 2.,],[ 3., 3., 3., 3., 3.,],[4., 4., 4., 4., 4.],[ 5., 5., 5., 5., 5.,]]) + array([[ 1., 2., 3., 4., 5.],[ 1., 2., 3., 4., 5.],[ 1., 2., 3., 4., 5.],[ 1., 2., 3., 4., 5.],[ 1., 2., 3., 4., 5.]])
which produces the result
array([[ 2., 3., 4., 5., 6.],
[ 3., 4., 5., 6., 7.],
[ 4., 5., 6., 7., 8.],
[ 5., 6., 7., 8., 9.],
[ 6., 7., 8., 9., 10.]])
Consider the following numpy.arrays:
a = np.array([1., 2., 3.])
b = np.array([4., 5.])
c = np.array([6., 7.])
I need to combine these so I end up with the following:
[(1., 4., 6.), (1., 5., 7.), (2., 4., 6.), (2., 5., 7.), (3., 4., 6.), (3., 5., 7.)]
Note that in this case, the array a happens to be the largest array. This is not guaranteed however. Nor is the length guaranteed. In other words, any array could be the longest and each array is of arbitrary length.
I tried using itertools.izip_longest but I can only use fillvalue for the tuple with 3. which will not work. I tried itertools.product also but my result is not a true cartesian product.
You can transpose b and c and then create a product of the a with the transposed array using itertools.product:
>>> from itertools import product
>>> [np.insert(j,0,i) for i,j in product(a,np.array((b,c)).T)]
[array([ 1., 4., 6.]), array([ 1., 5., 7.]), array([ 2., 4., 6.]), array([ 2., 5., 7.]), array([ 3., 4., 6.]), array([ 3., 5., 7.])]
>>>
Let's say you have:
a = np.array([4., 5.])
b = np.array([1., 2., 3.])
c = np.array([6., 7.])
d = np.array([5., 1])
e = np.array([3., 2.])
Now, if you know before-hand which one is the longest array, which is b in this case, you can use an approach based upon np.meshgrid -
# Concatenate elements from identical positions from the equal arrays
others = np.vstack((a,c,d,e)).T # If you have more arrays, edit this line
# Get grided version of the longest array and
# grided-indices for indexing into others array
X,Y = np.meshgrid(np.arange(others.shape[0]),b)
# Concatenate grided longest array and grided indexed others for final output
out = np.hstack((Y.ravel()[:,None],others[X.ravel()]))
Sample run -
In [47]: b
Out[47]: array([ 1., 2., 3.])
In [48]: a
Out[48]: array([ 4., 5.])
In [49]: c
Out[49]: array([ 6., 7.])
In [50]: d
Out[50]: array([ 5., 1.])
In [51]: e
Out[51]: array([ 3., 2.])
In [52]: out
Out[52]:
array([[ 1., 4., 6., 5., 3.],
[ 1., 5., 7., 1., 2.],
[ 2., 4., 6., 5., 3.],
[ 2., 5., 7., 1., 2.],
[ 3., 4., 6., 5., 3.],
[ 3., 5., 7., 1., 2.]])
If the length differences are not extreme (check inputs first) I'd be tempted to pad out the shorter lists to the length of the longest with None and generate all the permutations (27 of them for 3 lists of 3 elements). Then
results = []
for candidate in possibles:
if not (None in candidate): results.append(candidate)
Reasons not to do this: if the cube of the length of the longest list is significant in terms of memory usage (space to store N cubed possibles) or CPU usage.
In my Python application I have a 3D matrix (array) such this:
array([[[ 1., 2., 3.]], [[ 4., 5., 6.]], [[ 7., 8., 9.]]])
and I would like to add, in a particular "line", for example, in the middle, zero arrays. At the end I would like to end with the following matrix:
array([[[ 1., 2., 3.]],
[[ 4., 5., 6.]],
[[ 0., 0., 0.]],
[[ 0., 0., 0.]],
[[ 7., 8., 9.]]])
Anybody knows how to solve this issue? I tried to use "numpy.concatenate", but it allow me only to add more "lines".
Thanks in advance!
Possible duplicate of
Inserting a row at a specific location in a 2d array in numpy?
For example:
a = array([[[ 1., 2., 3.]], [[ 4., 5., 6.]], [[ 7., 8., 9.]]])
output = np.insert(a, 2, np.array([0,0,0]), 0)
output:
array([[[ 1., 2., 3.]],
[[ 4., 5., 6.]],
[[ 0., 0., 0.]],
[[ 7., 8., 9.]]])
Why this works on 3D array?
See doc here.
It says:
numpy.insert(arr, obj, values, axis=None)
...
Parameters :
values : array_like
Values to insert into arr.
If the type of values is different from that of arr,
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
values is converted to the type of arr. values should be shaped so that
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
arr[...,obj,...] = values is legal.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
...
So it's very wise function!!
Is this what you want?
result = np.r_[ a[:2], np.zeros(1,2,3), a[2][None] ]
I'd do it this way:
>>> a = np.array([[[ 1., 2., 3.]], [[ 4., 5., 6.]], [[ 7., 8., 9.]]])
>>> np.concatenate((a[:2], np.tile(np.zeros_like(a[0]), (2,1,1)), a[2:]))
array([[[ 1., 2., 3.]],
[[ 4., 5., 6.]],
[[ 0., 0., 0.]],
[[ 0., 0., 0.]],
[[ 7., 8., 9.]]])
The 2 in (2,1,1) given to tile() is how many zero "rows" to insert. The 2 in the slice indexes is of course where to insert.
If you're going to insert a large amount of zeros, it may be more efficient to just create a big array of zeros first and then copy in the parts you need from the original array.