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Here is an example of what I would like to do:
Assume Array A
A = np.array([[0, 1, 3, 5, 9],
[2, 7, 5, 1, 4]])
And Array B
B = np.array([2, 4])
I am looking for an operation that will increment the element indexed by array B in each row of array A by 1.
So the result A is:
A = np.array([[0, 1, 4, 5, 9],
[2, 7, 5, 1, 5]])
The index 2 of first row is increased by 1, and the index 4 of second row is increased by 1
You can achieve this by using advanced indexing in numpy:
A[np.arange(len(B)), B] += 1
This works by creating a 2D array with dimensions (len(B), len(B)) using np.arange(len(B)), which represents the row indices. The second index of the advanced indexing, B, represents the column indices. By adding 1 to A[np.arange(len(B)), B], you increment the elements in each row specified by B.
In numpy you can do by using arrange and shape of an array
import numpy as np
A = np.array([[0, 1, 3, 5, 9],
[2, 7, 5, 1, 4]])
B = np.array([2, 4])
A[np.arange(A.shape[0]), B] += 1
print(A)
np.arange(A.shape[0]) generates an array of integers from 0 to A.shape[0] - 1. A.shape[0] is basically rows
you can do with looping also..
import numpy as np
A = np.array([[0, 1, 3, 5, 9],
[2, 7, 5, 1, 4]])
B = np.array([2, 4])
for i, index in enumerate(B):
A[i][index] += 1
print(A)
For example, let's consider the following numpy array:
[1, 5, 0, 5, 4, 6, 1, -1, 5, 10]
Also, let's suppose that the threshold is equal to 3.
That is to say that we are looking for sequences of at least two consecutive values that are all above the threshold.
The output would be the indices of those values, which in our case is:
[[3, 4, 5], [8, 9]]
If the output array was flattened that would work as well!
[3, 4, 5, 8, 9]
Output Explanation
In our initial array we can see that for index = 1 we have the value 5, which is greater than the threshold, but is not part of a sequence (of at least two values) where every value is greater than the threshold. That's why this index would not make it to our output.
On the other hand, for indices [3, 4, 5] we have a sequence of (at least two) neighboring values [5, 4, 6] where each and every of them are above the threshold and that's the reason that their indices are included in the final output!
My Code so far
I have approached the issue with something like this:
(arr > 3).nonzero()
The above command gathers the indices of all the items that are above the threshold. However, I cannot determine if they are consecutive or not. I have thought of trying a diff on the outcome of the above snippet and then may be locating ones (that is to say that indices are one after the other). Which would give us:
np.diff((arr > 3).nonzero())
But I'd still be missing something here.
If you convolve a boolean array with a window full of 1 of size win_size ([1] * win_size), then you will obtain an array where there is the value win_size where the condition held for win_size items:
import numpy as np
def groups(arr, *, threshold, win_size, merge_contiguous=False, flat=False):
conv = np.convolve((arr >= threshold).astype(int), [1] * win_size, mode="valid")
indexes_start = np.where(conv == win_size)[0]
indexes = [np.arange(index, index + win_size) for index in indexes_start]
if flat or merge_contiguous:
indexes = np.unique(indexes)
if merge_contiguous:
indexes = np.split(indexes, np.where(np.diff(indexes) != 1)[0] + 1)
return indexes
arr = np.array([1, 5, 0, 5, 4, 6, 1, -1, 5, 10])
threshold = 3
win_size = 2
print(groups(arr, threshold=threshold, win_size=win_size))
print(groups(arr, threshold=threshold, win_size=win_size, merge_contiguous=True))
print(groups(arr, threshold=threshold, win_size=win_size, flat=True))
[array([3, 4]), array([4, 5]), array([8, 9])]
[array([3, 4, 5]), array([8, 9])]
[3 4 5 8 9]
You can do what you want using simple numpy operations
import numpy as np
arr = np.array([1, 5, 0, 5, 4, 6, 1, -1, 5, 10])
arr_padded = np.concatenate(([0], arr, [0]))
a = np.where(arr_padded > 3, 1, 0)
da = np.diff(a)
idx_start = (da == 1).nonzero()[0]
idx_stop = (da == -1).nonzero()[0]
valid = (idx_stop - idx_start >= 2).nonzero()[0]
result = [list(range(idx_start[i], idx_stop[i])) for i in valid]
print(result)
Explanation
Array a is a padded binary version of the original array, with 1s where the original elements are greater than three. da contains 1s where "islands" of 1s begin in a, and -1 where the "islands" end in a. Due to the padding, there is guaranteed to be an equal number of 1s and -1s in da. Extracting their indices, we can calculate the length of the islands. Valid index pairs are those whose respective "islands" have length >= 2. Then, its just a matter of generating all numbers between the index bounds of the valid "islands".
I follow your original idea. You are almost done.
I use another diff2 to pick the index of the first value in a sequence. See comments in code for details.
import numpy as np
arr = np.array([ 1, 5, 0, 5, 4, 6, 1, -1, 5, 10])
threshold = 3
all_idx = (arr > threshold).nonzero()[0]
# array([1, 3, 4, 5, 8, 9])
result = np.empty(0)
if all_idx.size > 1:
diff1 = np.zeros_like(all_idx)
diff1[1:] = np.diff(all_idx)
# array([0, 2, 1, 1, 3, 1])
diff1[0] = diff1[1]
# array([2, 2, 1, 1, 3, 1])
# **Positions with a value 1 in diff1 should be reserved.**
# But we also want the position before each 1. Create another diff2
diff2 = np.zeros_like(all_idx)
diff2[:-1] = np.diff(diff1)
# array([ 2, -1, 0, 2, -2, 0])
# **Positions with a negative value in diff2 should be reserved.**
result = all_idx[(diff1==1) | (diff2<0)]
print(result)
# array([3, 4, 5, 8, 9])
I'll try something different using window views, I'm not sure this works all the time so counterexamples are welcome. It has the advantage of not requiring Python loops.
import numpy as np
from numpy.lib.stride_tricks import sliding_window_view as window
def consec_thresh(arr, thresh):
win = window(np.argwhere(arr > thresh), (2, 1))
return np.unique(win[np.diff(win, axis=2).ravel() == 1, :,:].ravel())
How does it work?
So we start with the array and gather the indices where the threshold is met:
In [180]: np.argwhere(arr > 3)
Out[180]:
array([[1],
[3],
[4],
[5],
[8],
[9]])
Then we build a sliding window that makes up pair of values along the column (which is the reason for the (2, 1) shape of the window).
In [181]: window(np.argwhere(arr > 3), (2, 1))
Out[181]:
array([[[[1],
[3]]],
[[[3],
[4]]],
[[[4],
[5]]],
[[[5],
[8]]],
[[[8],
[9]]]])
Now we want to take the difference inside each pair, if it's one then the indices are consecutive.
In [182]: np.diff(window(np.argwhere(arr > 3), (2, 1)), axis=2)
Out[182]:
array([[[[2]]],
[[[1]]],
[[[1]]],
[[[3]]],
[[[1]]]])
We can plug those values back in the windows we created above,
In [185]: window(np.argwhere(arr > 3), (2, 1))[np.diff(window(np.argwhere(arr > 3), (2, 1)), axis=2).ravel() == 1, :, :]
Out[185]:
array([[[[3],
[4]]],
[[[4],
[5]]],
[[[8],
[9]]]])
Then we can ravel (flatten without copy when possible), we have to get rid of the repeated indices created by windowing so I call np.unique. We ravel again and get:
array([3, 4, 5, 8, 9])
The below iteration code should help with O(n) complexity
arr = [1, 5, 0, 5, 4, 6, 1, -1, 5, 10]
threshold = 3
sequence = 2
output = []
temp_arr = []
for i in range(len(arr)):
if arr[i] > threshold:
temp_arr.append(i)
else:
if len(temp_arr) >= sequence:
output.append(temp_arr)
temp_arr = []
if len(temp_arr):
output.append(temp_arr)
temp_arr = []
print(output)
# Output
# [[3, 4, 5], [8, 9]]
I would suggest using a for loop with two indces. You will have one that starts at j=1 and the other at i=0, both stepping forward by 1.
You can then ask if the value at both is greater than the threshold, if so
add the indices to a list and keep moving forward with j until the threshold or .next() is not greater than threshhold.
values = [1, 5, 0, 5, 4, 6, 1, -1, 5, 10]
res=[]
threshold= 3
i=0
j=0
for _ in values:
j=i+1
lista=[]
try:
print(f"i: {i} j:{j}")
# check if condition is met
if(values[i] > threshold and values[j] > threshold):
lista.append(i)
# add sequence
while values[j] > threshold:
lista.append(j)
print(f"j while: {j}")
j+=1
if(j>=len(values)):
break
res.append(lista)
i=j
if(j>=len(values)):
break
except:
print("ex")
this works. but needs refactoring
Let's try the following code:
# Simple is better than complex
# Complex is better than complicated
arr = [1, 5, 0, 5, 4, 6, 1, -1, 5, 10]
arr_3=[i if arr[i]>3 else 'a' for i in range(len(arr))]
arr_4=''.join(str(x) for x in arr_3)
i=0
while i<len(arr_5):
if len(arr_5[i]) <=1:
del arr_5[i]
else:
i+=1
arr_6=[list(map(lambda x: int(x), list(x))) for x in arr_5]
print(arr_6)
Outputs:
[[3, 4, 5], [8, 9]]
Here is a solution that makes use of pandas Series:
thresh = 3
win_size = 2
s = pd.Series(arr)
# locating groups of values where there are at least (win_size) consecutive values above the threshold
groups = s.groupby(s.le(thresh).cumsum().loc[s.gt(thresh)]).transform('count').ge(win_size)
0 False
1 False
2 False
3 True
4 True
5 True
6 False
7 False
8 True
9 True
dtype: bool
We can now easily take their indices in a 1D array:
np.flatnonzero(groups)
# array([3, 4, 5, 8, 9], dtype=int64)
OR multiple lists:
[np.arange(index.start, index.stop) for index in np.ma.clump_unmasked(np.ma.masked_not_equal(groups.values, value=True))]
# [array([3, 4, 5], dtype=int64), array([8, 9], dtype=int64)]
Question:
Create a array x of shape (n_row.n_col), having first n natural numbers.
N = 30, n_row= 6, n_col=5
Print elements, overlapping first two rows and last three columns.
Expected output:
[[2 3 4]
[7 8 9]]
My output:
[2 3 7 8]
My approach:
x = np.arange (n)
x= x.reshape(n_row,n_col)
a= np. intersect1d(x[0:2,],x[:,-3:-1])
print (a)
I couldn't think of anything else, please help
The overlap of row and column slices of the same array is just the combined slice
import numpy as np
x = np.arange(30).reshape(6, 5)
x[:2,-3:]
Output
array([[2, 3, 4],
[7, 8, 9]])
To compute the overlap by finding same elements is odd but possible
r, c = np.where(np.isin(x, np.intersect1d(x[:2], x[:,-3:])))
x[np.ix_(np.unique(r), np.unique(c))]
Output
array([[2, 3, 4],
[7, 8, 9]])
I think the answers are a bit convoluted...
Personally from the original question:
Question: Create a array x of shape (n_row.n_col), having first n natural numbers. N = 30, n_row= 6, n_col=5
Print elements, overlapping first two rows and last three columns.
I understand "sub-indexing":
N, n_rows, n_cols = 30, 6, 5
a = np.arange(N).reshape(n_rows, n_cols)
print(a[:2, -3:])
Output:
[[2, 3, 4],
[7, 8, 9]]
I want to get the top N (maximal) args & values across an entire numpy matrix, as opposed to across a single dimension (rows / columns).
Example input (with N=3):
import numpy as np
mat = np.matrix([[9,8, 1, 2], [3, 7, 2, 5], [0, 3, 6, 2], [0, 2, 1, 5]])
print(mat)
[[9 8 1 2]
[3 7 2 5]
[0 3 6 2]
[0 2 1 5]]
Desired output: [9, 8, 7]
Since max isn't transitive across a single dimension, going by rows or columns doesn't work.
# by rows, no 8
np.squeeze(np.asarray(mat.max(1).reshape(-1)))[:3]
array([9, 7, 6])
# by cols, no 7
np.squeeze(np.asarray(mat.max(0)))[:3]
array([9, 8, 6])
I have code that works, but looks really clunky to me.
# reshape into single vector
mat_as_vector = np.squeeze(np.asarray(mat.reshape(-1)))
# get top 3 arg positions
top3_args = mat_as_vector.argsort()[::-1][:3]
# subset the reshaped matrix
top3_vals = mat_as_vector[top3_args]
print(top3_vals)
array([9, 8, 7])
Would appreciate any shorter way / more efficient way / magic numpy function to do this!
Using numpy.partition() is significantly faster than performing full sort for this purpose:
np.partition(np.asarray(mat), mat.size - N, axis=None)[-N:]
assuming N<=mat.size.
If you need the final result also be sorted (besides being top N), then you need to sort previous result (but presumably you will be sorting a smaller array than the original one):
np.sort(np.partition(np.asarray(mat), mat.size - N, axis=None)[-N:])
If you need the result sorted from largest to lowest, post-pend [::-1] to the previous command:
np.sort(np.partition(np.asarray(mat), mat.size - N, axis=None)[-N:])[::-1]
One way may be with flatten and sorted and slice top n values:
sorted(mat.flatten().tolist()[0], reverse=True)[:3]
Result:
[9, 8, 7]
The idea is from this answer: How to get indices of N maximum values in a numpy array?
import numpy as np
import heapq
mat = np.matrix([[9,8, 1, 2], [3, 7, 2, 5], [0, 3, 6, 2], [0, 2, 1, 5]])
ind = heapq.nlargest(3, range(mat.size), mat.take)
print(mat.take(ind).tolist()[0])
Output
[9, 8, 7]
I have a numpy array, say:
>>> a=np.array([[0,1,2],[4,3,6],[9,5,7],[8,9,8]])
>>> a
array([[0, 1, 2],
[4, 3, 6],
[9, 5, 7],
[8, 9, 8]])
I want to replace the second and third column elements with the minimum of them (row by row), except if one of these 2 elements is < 3.
The resulting array should be:
array([[0, 1, 2],# nothing changes since 1 and 2 are <3
[4, 3, 3], #min(3,6)=3 => 6 changed to 3
[9, 5, 5], #min(5,7)=5 => 7 changed to 5
[8, 8, 8]]) #min(9,8)=8 => 9 changed to 8
I know I can use clip, for instance a[:,1:3].clip(2,6,a[:,1:3]), but
1) clip will be applied to all elements, including those <3.
2) I don't know how to set the min and max values of clip to the minimum values of the 2 related elements of each row.
Just use the >= operator to first select what you are interested of:
b = a[:, 1:3] # select the columns
matching = numpy.all(b >= 3, axis=1) # find rows with all elements matching
b = b[matching, :] # select rows
Now you can replace the content with the minimum by e.g.:
# find row minimum and convert to a column vector
b[:, :] = b.min(1, keepdims=True)
We first defined a row_mask, depicting the <3 condition, and then apply a minimum along an axis to find the minimum (for rows in row_mask).
The newaxis part is required for the broadcasting of a 1dim array (of minimums) to the 2-dim target of the assignment.
a=np.array([[0,1,2],[4,3,6],[9,5,7],[8,9,8]])
row_mask = (a[:,0]>=3)
a[row_mask, 1:] = a[row_mask, 1:].min(axis=1)[...,np.newaxis]
a
=>
array([[0, 1, 2],
[4, 3, 3],
[9, 5, 5],
[8, 8, 8]])
Here's a one liner:
a[np.where(np.sum(a,axis=1)>3),1:3]=np.min(a[np.where(np.sum(a,axis=1)>3),1:3],axis=2).reshape(1,3,1)
Here's a breakdown:
>>> b = np.where(np.sum(a,axis=1)>3) # finds rows where, in a, row sums are > 3
(array([1, 2, 3]),)
>>> c = a[b,1:3] # the part of a that needs to change
array([[[3, 3],
[5, 5],
[8, 8]]])
>>> d = np.min(c,axis=2) # the minimum values in each row (cols 1 and 2)
array([[3, 5, 8]])
>>> e = d.reshape(1,3,1) # adjust shape for broadcast to a
array([[[3],
[5],
[8]]])
>>> a[np.where(np.sum(a,axis=1)>3),1:3] = e # set the values in a
>>> a
array([[0, 1, 2],
[4, 3, 3],
[9, 5, 5],
[8, 8, 8]])