I have an array of arrays that I want to interpolate based on each array's min and max.
For a simple mxn array , with values ranging from 0 to 1, I can do this as follows:
x_inp=np.interp(x,(x.min(),x.max()),(0,0.7))
This suppresses every existing value to 0 to 0.7. However, if I have an array of dimension 100xmxn, the above method considers the global min/max and not the individual min/max of each of the mxn array.
Edit:
For example
x1=np.random.randint(0,5, size=(2, 4))
x2=np.random.randint(6,10, size=(2, 4))
my_list=[x1,x2]
my_array=np.asarray(my_list)
print(my_array)
>> array([[[1, 4, 3, 4],
[3, 2, 0, 0]],
[9, 6, 8, 6],
8, 7, 6, 7]]])
my_array is now of dimension 2x2x4 and my_array.min() and my_array.max() would give me 0 and 9. So If I interpolate, it won't work based on the min/max of the individual 2x4 arrays. What I want is, to have the interpolation work based on min/max of 0/4 for the 1st array and 6/9 for the second.
Related
Using numpy, given a sorted 1D array, how to efficiently obtain a 1D array with equal size where the value at each position is the number of preceding equal elements? I have very large arrays and processing each element in Python code one way or another is not acceptable.
Example:
input = [0, 0, 4, 4, 4, 5, 5, 5, 5, 6]
output = [0, 1, 0, 1, 2, 0, 1, 2, 3, 0]
import numpy as np
A=np.array([0, 0, 4, 4, 4, 5, 5, 5, 5, 6])
uni,counts=np.unique(A, return_counts=True)
out=np.concatenate([np.arange(n) for n in counts])
print(out)
Not certain about the efficiency (probably better way to form the out array rather than concatenating), but a very straightforward way to get the result you are looking for. Counts the unique elements, then does np.arange on each count to get the ascending sequence, then concatenates these arrays together.
I applied pandas.cut on Series. if I don't use customised Interval Index, label=False works as expected (returns only integer indicators of the bins). However, when I used customised Interval Index, even I set label=False, it returns interval of each bin. I guess this is probably because I used interval index in stead of number of bins.
Is there anyway to use customised interval index, also only return integer indicator of the bins?
bins = pd.interval_range(start=0, end=10, periods=5, closed='left')
pd.cut([1, 3, 5, 7, 9], bins, labels=False)
codes attribute
pd.cut([1, 3, 5, 7, 9], bins, labels=False).codes
array([0, 1, 2, 3, 4], dtype=int8)
The pd.cut function returns a Categorical type object. What gets displayed are the various categories for each element. However, the Categorical object has two attributes codes and categories. The categories are what you'd think. It is an array of your unique categories in the proper order. The codes are the positions of that categories array that each element of the Categorical object is referencing.
You can produce the Categorical values by slicing the categories array with the codes array like so:
mycut = pd.cut([1, 3, 5, 7, 9], bins, labels=False)
mycut.categories[mycut.codes]
IntervalIndex([[0, 2), [2, 4), [4, 6), [6, 8), [8, 10)],
closed='left',
dtype='interval[int64]')
However, the codes are the exact thing you were looking for... so take it.
I am doing a Computer Vision project in which I am getting an error 'setting an array element with a sequence' when I am trying to change the data type of input image matrix.
I realized this is happening because the input image matrix I am having does not have the same number of elements in all of its array. Is there any way I can convert that input image into the 2D array with the same number of elements in each of its array?
I am getting an error when I am trying to execute the following line:
X_train = X_train.astype('float32')
Any help would be appreciated.
Cheers.
You need to pad the rows with less elements with zeros to make their lengths equal to the length of the longest array (or list) in the list of lists (matrix).
Below's a code snippet to pad a list of lists of unequal lengths to a matrix of same row-lengths:
import numpy as np
unpadded_matrix = np.array([[1, 2], [3, 4, 5], [6, 7, 8, 9]])
max_len = max([len(row) for row in unpadded_matrix])
np.array([row + [0]*(max_len-len(row)) for row in unpadded_matrix])
o/p:
array([[1, 2, 0, 0],
[3, 4, 5, 0],
[6, 7, 8, 9]])
This question already has answers here:
Index n dimensional array with (n-1) d array
(3 answers)
Closed 4 years ago.
I have a 3-d array. I find the indexes of the maxima along an axis using argmax. How do I now use these indexes to obtain the maximal values?
2nd part: How to do this for arrays of N-d?
Eg:
u = np.arange(12).reshape(3,4,1)
In [125]: e = u.argmax(axis=2)
Out[130]: e
array([[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]])
It would be nice if u[e] produced the expected results, but it doesn't work.
The return value of argmax along an axis can't be simply used as an index. It only works in a 1d case.
In [124]: u = np.arange(12).reshape(3,4,1)
In [125]: e = u.argmax(axis=2)
In [126]: u.shape
Out[126]: (3, 4, 1)
In [127]: e.shape
Out[127]: (3, 4)
e is (3,4), but its values only index the last dimension of u.
In [128]: u[e].shape
Out[128]: (3, 4, 4, 1)
Instead we have to construct indices for the other 2 dimensions, ones which broadcast with e. For example:
In [129]: I,J=np.ix_(range(3),range(4))
In [130]: I
Out[130]:
array([[0],
[1],
[2]])
In [131]: J
Out[131]: array([[0, 1, 2, 3]])
Those are (3,1) and (1,4). Those are compatible with (3,4) e and the desired output
In [132]: u[I,J,e]
Out[132]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
This kind of question has been asked before, so probably should be marked as a duplicate. The fact that your last dimension is size 1, and hence e is all 0s, distracting readers from the underlying issue (using a multidimensional argmax as index).
numpy: how to get a max from an argmax result
Get indices of numpy.argmax elements over an axis
Assuming you've taken the argmax on the last dimension
In [156]: ij = np.indices(u.shape[:-1])
In [157]: u[(*ij,e)]
Out[157]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
or:
ij = np.ix_(*[range(i) for i in u.shape[:-1]])
If the axis is in the middle, it'll take a bit more tuple fiddling to arrange the ij elements and e.
so for general N-d array
dims = np.ix_(*[range(x) for x in u.shape[:-1]])
u.__getitem__((*dims,e))
You can't write u[*dims,e], that's a syntax error, so I think you must use getitem directly.
I am having a lot of trouble understanding numpy indexing for multidimensional arrays. In this example that I am working with, let's say that I have a 2D array, A, which is 100x10. Then I have another array, B, which is a 100x1 1D array of values between 0-9 (indices for A). In MATLAB, I would use A(sub2ind(size(A), 1:size(A,1)', B) to return for each row of A, the value at the index stored in the corresponding row of B.
So, as a test case, let's say I have this:
A = np.random.rand(100,10)
B = np.int32(np.floor(np.random.rand(100)*10))
If I print their shapes, I get:
print A.shape returns (100L, 10L)
print B.shape returns (100L,)
When I try to index into A using B naively (incorrectly)
Test1 = A[:,B]
print Test1.shape returns (100L, 100L)
but if I do
Test2 = A[range(A.shape[0]),B]
print Test2.shape returns (100L,)
which is what I want. I'm having trouble understanding the distinction being made here. In my mind, A[:,5] and A[range(A.shape[0]),5] should return the same thing, but it isn't here. How is : different from using range(sizeArray) which just creates an array from [0:sizeArray] inclusive, to use an indices?
Let's look at a simple array:
In [654]: X=np.arange(12).reshape(3,4)
In [655]: X
Out[655]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
With the slice we can pick 3 columns of X, in any order (and even repeated). In other words, take all the rows, but selected columns.
In [656]: X[:,[3,2,1]]
Out[656]:
array([[ 3, 2, 1],
[ 7, 6, 5],
[11, 10, 9]])
If instead I use a list (or array) of 3 values, it pairs them up with the column values, effectively picking 3 values, X[0,3],X[1,2],X[2,1]:
In [657]: X[[0,1,2],[3,2,1]]
Out[657]: array([3, 6, 9])
If instead I gave it a column vector to index rows, I get the same thing as with the slice:
In [659]: X[[[0],[1],[2]],[3,2,1]]
Out[659]:
array([[ 3, 2, 1],
[ 7, 6, 5],
[11, 10, 9]])
This amounts to picking 9 individual values, as generated by broadcasting:
In [663]: np.broadcast_arrays(np.arange(3)[:,None],np.array([3,2,1]))
Out[663]:
[array([[0, 0, 0],
[1, 1, 1],
[2, 2, 2]]),
array([[3, 2, 1],
[3, 2, 1],
[3, 2, 1]])]
numpy indexing can be confusing. But a good starting point is this page: http://docs.scipy.org/doc/numpy/reference/arrays.indexing.html