non-broadcastable output operand numpy 2D cast into 3D - python

In NumPy,
foo = np.array([[i+10*j for i in range(10)] for j in range(3)])
array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29]])
filter = np.nonzero(foo > 100)#nothing matches
foo[:,filter]
array([], shape=(3, 2, 0), dtype=int64)
foo[:,0:0]
array([], shape=(3, 0), dtype=int64)
filter2 = np.nonzero(np.sum(foo,axis=0) < 47)
foo[:,filter2]
array([[[ 0, 1, 2, 3, 4, 5]],
[[10, 11, 12, 13, 14, 15]],
[[20, 21, 22, 23, 24, 25]]])
foo[:,filter2].shape
(3, 1, 6)
I have a 'filter' condition where I want to perform an operation on all rows for all matching columns, but if filter is an empty array, somehow my foo[:,filter] gets broadcast into a 3D array. Another example is with filter2 -> again, foo[:,filter2] gives me a 3D array when I am expecting the result of foo[:,(np.sum(foo,axis=0) < 47)]
Can someone explain what the proper use case of np.nonzero is compared to using booleans to find the correct columns/indices?

First, foo[filter] == foo[filter.nonzero()] when filter is a Boolean array.
To understand why you're getting unexpected results you have to understand a little about how python does indexing. To do multidimensional indexing in python you can either use indices in [], separated by commas or use a tuple. So foo[1, 2, 3] is the same as foo[(1, 2, 3)]. With this in mind take a look at what happens when you do foo[:, something]. I believe in your example you were trying to get foo[:, something[0], something[1]], but instead you got foo[(slice[None], (something[0], something[1]))].
This is all somewhat academic, because if you're just using filter for indexing you probably don't need to use nonzero, just use the boolean array as the index but if you need to, you can do something like:
foo[:, filter[0]]
# OR
index = (slice(None),) + filter.nonzero()
foo[index]

Related

How to change certain numpy array's target value to another value (if a list of target values is available)?

I have a following array:
arr = np.arange(24).reshape(4, 6)
and have a list of target values that should be changed:
target = [0, 2, 8]
Now I would like to change all target values in arr to 100.
I could perform that with:
arr[(arr == 0)|(arr == 2)|(arr == 8)] = 100
It is not practicable if I have long lists (with 10 elements).
Is there a better way to perform it without using for loop?
Thank you for any inputs :-)
Use np.isin
arr[np.isin(arr, target)] = 100
arr
array([[100, 1, 100, 3, 4, 5],
[ 6, 7, 100, 9, 10, 11],
[ 12, 13, 14, 15, 16, 17],
[ 18, 19, 20, 21, 22, 23]])

Iterating Over Rows in Python Array to Extract Column Data

I am using Python and looking to iterate through each row of an Nx9 array and extract certain values from the row to form another matrix with them. The N value can change depending on the file I am reading but I have used N=3 in my example. I only require the 0th, 1st, 3rd and 4th values of each row to form into an array which I need to store. E.g:
result = np.array([[1, 2, 3, 4, 5, 6, 7, 8, 9],
[11, 12, 13, 14, 15, 16, 17, 18, 19],
[21, 22, 23, 24, 25, 26, 27, 28, 29]])
#Output matrix of first row should be: ([[1,2],[4,5]])
#Output matrix of second row should be: ([[11,12],[14,15]])
#Output matrix of third row should be: ([[21,22],[24,25]])
I should then end up with N number of matrices formed with the extracted values - a 2D matrix for each row. However, the matrices formed appear 3D so when transposed and subtracted I receive the error ValueError: operands could not be broadcast together with shapes (2,2,3) (3,2,2). I am aware that a (3,2,2) matrix cannot be subtracted from a (2,2,3) so how do I obtain a 2D matrix N number of times? Would a loop be better suited? Any suggestions?
import numpy as np
result = np.array([[1, 2, 3, 4, 5, 6, 7, 8, 9],
[11, 12, 13, 14, 15, 16, 17, 18, 19],
[21, 22, 23, 24, 25, 26, 27, 28, 29]])
a = result[:, 0]
b = result[:, 1]
c = result[:, 2]
d = result[:, 3]
e = result[:, 4]
f = result[:, 5]
g = result[:, 6]
h = result[:, 7]
i = result[:, 8]
output = [[a, b], [d, e]]
output = np.array(output)
output_transpose = output.transpose()
result = 0.5 * (output - output_transpose)
In [276]: result = np.array(
...: [
...: [1, 2, 3, 4, 5, 6, 7, 8, 9],
...: [11, 12, 13, 14, 15, 16, 17, 18, 19],
...: [21, 22, 23, 24, 25, 26, 27, 28, 29],
...: ]
...: )
...:
...: a = result[:, 0]
...
...: i = result[:, 8]
...: output = [[a, b], [d, e]]
In [277]: output
Out[277]:
[[array([ 1, 11, 21]), array([ 2, 12, 22])],
[array([ 4, 14, 24]), array([ 5, 15, 25])]]
In [278]: arr = np.array(output)
In [279]: arr
Out[279]:
array([[[ 1, 11, 21],
[ 2, 12, 22]],
[[ 4, 14, 24],
[ 5, 15, 25]]])
In [280]: arr.shape
Out[280]: (2, 2, 3)
In [281]: arr.T.shape
Out[281]: (3, 2, 2)
transpose exchanges the 1st and last dimensions.
A cleaner way to make a (N,2,2) array from selected columns is:
In [282]: arr = result[:,[0,1,3,4]].reshape(3,2,2)
In [283]: arr.shape
Out[283]: (3, 2, 2)
In [284]: arr
Out[284]:
array([[[ 1, 2],
[ 4, 5]],
[[11, 12],
[14, 15]],
[[21, 22],
[24, 25]]])
Since the last 2 dimensions are 2, you could transpose them, and take the difference:
In [285]: arr-arr.transpose(0,2,1)
Out[285]:
array([[[ 0, -2],
[ 2, 0]],
[[ 0, -2],
[ 2, 0]],
[[ 0, -2],
[ 2, 0]]])
Another way to get the (N,2,2) array is with a matrix index:
In [286]: result[:,[[0,1],[3,4]]]
Out[286]:
array([[[ 1, 2],
[ 4, 5]],
[[11, 12],
[14, 15]],
[[21, 22],
[24, 25]]])
Ok, this is not a coding problem, but a math problem. I wrote some code for you, since it's pretty obvious you're a beginner, so there will be some unfamiliar syntax in there that you should look into so you can avoid problems like this in the future. You might not use them all that often, but it's good to know how to do it, because it expands your understanding of python syntax in general.
First up, the complete code for easy copy and pasting:
import numpy as np
result=np.array([[1,2,3,4,5,6,7,8,9],
[11,12,13,14,15,16,17,18,19],
[21,22,23,24,25,26,27,28,29]])
output = np.array(tuple(result[:,i] for i in (0,1,3)))
def Matrix_Operation(Matrix,Coefficient):
if (Matrix.shape == Matrix.shape[::-1]
and isinstance(Matrix,np.ndarray)
and isinstance(Coefficient,float)):
return Coefficient*(Matrix-Matrix.transpose())
else:
print('The shape of you Matrix is not palindromic')
print('You cannot substitute matrices of unequal shape')
print('Your shape: %s'%str(Matrix.shape))
print(Matrix_Operation(output,0.5))
Now let's talk about a step by step explanation of what's happening here:
import numpy as np
result=np.array([[1,2,3,4,5,6,7,8,9],
[11,12,13,14,15,16,17,18,19],
[21,22,23,24,25,26,27,28,29]])
Python uses indentation (alignment of whitespaces) as an integral part of it's syntax. However, if you provide brackets, a lot of the time you don't need aligning indentations in order for the interpreter to understand your code. If you provide a large array of values manually, it is usually adviseable to start new lines at the commas (here, the commas separating the sublists). It's just more readable and that way your data isn't off screen in your coding program.
output = np.array(tuple(result[:,i] for i in (0,1,3)))
List comprehensions are a big deal in python and really handy for dirty one liners. As far as I know, no other language gives you this option. That's one of the reasons why python is so great. I basically created a list of lists, where each sublist is result[:,i] for every i in (0,1,3). This is cast as a tuple (yes, list comprehensions can also be done with tuples, not just lists). Finally I wrapped it in the np.array function, since this is the type required for our mathematical operations later on.
def Matrix_Operation(Matrix,Coefficient):
if (Matrix.shape == Matrix.shape[::-1]
and isinstance(Matrix,np.ndarray)
and isinstance(Coefficient,(float,int))):
return Coefficient*(Matrix-Matrix.transpose())
else:
print('The shape of you Matrix is not palindromic')
print('You cannot substitute matrices of unequal shape')
print('Your shape: %s'%str(Matrix.shape))
print(Matrix_Operation(output,0.5))
If you're gonna create a complex formula in python code, why not wrap it inside an abstractable function? You can incorporate a lot of "quality control" into a function as well, to check if it is given the correct input for the task it is supposed to do.
Your code failed, because you were trying to subtract a (2,2,3) shaped matrix from a (3,2,2) matrix. So we'll need a code snippet to check, if our provided matrix has a palindromic shape. You can reverse the order of items in a container by doing Container[::-1] and so we ask, if Matrix.shape == Matrix.shape[::-1]. Further, it is necessary, that our Matrix is a np.ndarray and if our coefficient is a number. That's what I'm doing with the isinstance() function. You can check for multiple types at once, which is why the isinstance(Coefficient,(float,int)) contains a tuple with both int and float in it.
Now that we have ensured that our input makes sense, we can preform our Matrix_Operation.
So in conclusion: Check if your math is solid before asking SO for help, because people here can get pretty annoyed at that sort of thing. You probably noticed by now that someone has already downvoted your question. Personally, I believe it's necessary to let newbies ask a couple stupid questions before they get into the groove, but that's what the voting button is for, I guess.

Numpy reshape seems to output value error

I tried to use reshape
import numpy as np
d = np.arange(30).reshape(1,3)
It is not working cannot reshape array of size 30 into shape (1,3)
but when I tried to use
d = np.arange(30).reshape(-1,3) # This works
Why do we have to use -1?.
It's really confusing and I'm can't seem to figure out how reshape works. I would really appreciate if someone can help me figure out how this works. I tried docs and other posts in SO but it wasn't much helpful.
I am new to ML and Python.
A reshape means that you order your elements of the array, according to other dimensions. For example arange(27) will produce a vector containing 27 elements. But with .reshape(9, 3) you specify here that you want to transform it into a two dimensional array, where the first dimension contains 9 elements, and the second three elements. So the result will look like:
>>> np.arange(27).reshape(9, 3)
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11],
[12, 13, 14],
[15, 16, 17],
[18, 19, 20],
[21, 22, 23],
[24, 25, 26]])
But we can also make it a 3×3×3 array:
>>> np.arange(27).reshape(3, 3, 3)
array([[[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8]],
[[ 9, 10, 11],
[12, 13, 14],
[15, 16, 17]],
[[18, 19, 20],
[21, 22, 23],
[24, 25, 26]]])
-1 is used as a value that will numpy derive the dimension.
So if you have an array containing 30 elements, and you reshape these to m×3, then m is 10. -1 is thus not the real value, it is used for programmer convenience if for example you do not know the number of elements, but know that it is divisable by three.
The following two are (under the assumption that m contains 30 elements equivalent:
m.reshape(10, 3)
m.reshape(-1, 3)
Note that you can specify at most one -1, since otherwise there are multiple possibilities, and it becomes also harder to find a valid configuration.

Why np.concatenate changes dimension

In the following program, I am trying to understand how np.concatenate command works. After accessing each row of the array a by for loop, when I concatenate along row axis I expect a 2-dimensional array having the shape of (5,5) but it changes.
I want to have the same dimension (5,5) after concatenation. How can I do that?
I tried to repeat the above method for the 2-dimensional array by storing them in a list [(2,5),(2,5),(2,5)]. At the end when I concatenate it gives me the shape of (6,5) as expected but in the following case, it is different.
a = np.arange(25).reshape(5,5)
ind =[0,1,2,3,4]
list=[]
for i in ind:
list.append(a[i])
new= np.concatenate(list, axis=0)
print(list)
print(len(list))
print(new)
print(new.shape)
This gives the following results for new:
[ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24]
and for new.shape:
(25,)
To preface this you really should not be using concatenate here.
Setup
a = np.arange(25).reshape(5,5)
L = [i for i in a]
You're question asks:
Why is np.concatenate changing dimension?
It's not changing dimension, it is doing exactly what it is supposed to based on the input you are giving it. From the documentation:
Join a sequence of arrays along an existing axis
When you pass your list to concatenate, don't think of it as passing a (5, 5) list, think of it as passing 5 (5,) shape arrays, which are getting joined along axis 0, which will intuitively produce a (25,) shape output.
Now this behavior also gives insight on how to work around this. If passing 5 (5,) shape arrays produces a (25,) shape output, we just need to pass (1, 5) shape arrays to produce a (5, 5) shape output. We can accomplish this by simply adding a dimension to each element of L:
np.concatenate([[i] for i in L])
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]])
However, the much better way to approach this is to simply use stack, vstack, etc..
>>> np.stack(L)
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]])
>>> np.vstack(L)
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]])

Change shape of multi-dimensional NumPy array without changing dimensions

I am currently monkey-patching a Scikit-Learn function and one of the lines requires a NumPy array with 2 dimensions. However, the data I am working with is a NumPy array with 3 dimensions, which raises the error "too many values to unpack."
I am calling the K-Means function fit to cluster the data. My problem boils down to this following line of code, assuming X is the ndarray that I pass in:
n_samples, n_features = X.shape
X is an array with 3 dimensions, like the following:
X = np.array([[[1, 2, 3],
[4, 5, 6]],
[[7, 8, 9],
[10, 11, 12]],
[[13, 14, 15],
[16, 17, 18]]])
The data represents a group of time series of data points that have 6 dimensions. For example, the first element, [[1, 2, 3], [4, 5, 6]] would represent a time series with samples from 2 time periods, each sample with 3 dimensions.
And I have monkey-patched the k_means_ code to allow me to perform clustering on an ndarray of ndarrays. My goal is to perform k-means on 2D arrays.
Is it possible to set the shape of the 3D ndarray to 2 elements? For example, I tried converting the 3D array to a 2D array of objects but it ends up getting converted back to a 3D array.
np.array([[x.astype(object) for x in c] for c in combined])
Likewise, the following code is also converted back to a 3D array.
np.array([[np.array(x) for x in c] for c in combined])
The list comprehension [[x.astype(object) for x in c] for c in combined] looks like it creates the correct array, but because it is of type list, it no longer works in the function.
I am looking for some way to "convert" a 3D NumPy array into 2 dimensions. Any help would be greatly appreciated!
Note: I am not looking for a way to reshape the array. I need to keep all the dimensions but change the shape to ignore one of the dimensions.
To make an array of arrays, we have to play some tricks, because np.array tries to make an as-high dimensional array as it can. If the subarrays vary in size that is ok, but if they are all the same we have to fight that.
Here's one way:
start with a 3d array:
In [812]: arr = np.arange(24).reshape(2,3,4)
and an empty object array of the right size (but flattened)
In [813]: A = np.empty((6,),object)
copy values (again with flattening), and reshape to the target shape
In [814]: A[:]=list(arr.reshape(-1,4))
In [815]: A=A.reshape(2,3)
In [816]: A
Out[816]:
array([[array([0, 1, 2, 3]), array([4, 5, 6, 7]), array([ 8, 9, 10, 11])],
[array([12, 13, 14, 15]), array([16, 17, 18, 19]),
array([20, 21, 22, 23])]], dtype=object)
So now we have a (2,3) array, who's shape can be unpacked.
I tried to start with np.empty((2,3),object), but couldn't get the A[:,:]=... assignment to work. For this object reshaping to work we have to split arr into a list of arrays. An object array is, like a list, an array of pointers.
But will the scikit functions accept such an array? (after passing the shape hurdle). I suspect the object reshaping is a short sighted solution.
In [824]: [[x.astype(object) for x in c] for c in arr]
Out[824]:
[[array([0, 1, 2, 3], dtype=object),
array([4, 5, 6, 7], dtype=object),
array([8, 9, 10, 11], dtype=object)],
[array([12, 13, 14, 15], dtype=object),
array([16, 17, 18, 19], dtype=object),
array([20, 21, 22, 23], dtype=object)]]
In [825]: _[0][0].shape
Out[825]: (4,)
This creates a nested list of lists, with the inner elements being (4,) object array. Wrap that in np.array and it recreates a 3d array with dtype object.
reshaping, which for some unknown reason, you don't want to do preserves the numeric dtype
In [828]: arr.reshape(2,-1)
Out[828]:
array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
[12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]])
In [829]: arr.reshape(-1,4)
Out[829]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15],
[16, 17, 18, 19],
[20, 21, 22, 23]])

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