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Get indices greater than value and keep value

I have a 2D array that looks like this:
[[0.1, 0.2, 0.4, 0.6, 0.9]
[0.3, 0.7, 0.8, 0.3, 0.9]
[0.7, 0.9, 0.4, 0.6, 0.9]
[0.1, 0.2, 0.6, 0.6, 0.9]]
And I want to save the indices where the array is higher than 0.6 but I also want to keep the value of that position, so the output would be:
[0, 3, 0.6]
[0, 4, 0.9]
[1, 2, 0.7]
and so on.
To get the indices I did this:
x = np.where(PPCF> 0.6)
high_pc = np.asarray(x).T.tolist()
but how do I keep the value in a third position?

Simple, no loops:
x = np.where(PPCF > 0.6) # condition to screen values
vals = PPCF[x] # find values by indices
np.concatenate((np.array(x).T, vals.reshape(vals.size, 1)), axis = 1) # resulting array
Feel free to convert it to a list.

This should work :
x = np.where(PPCF> 0.6)
high_pc = np.asarray(x).T.tolist()
for i in high_pc:
i.append(float(PPCF[i[0],i[1]]))

You could just run a loop along the columns and rows and check if each element is greater than the threshold and save them in a list.
a = [[0.1, 0.2, 0.4, 0.6, 0.9],
[0.3, 0.7, 0.8, 0.3, 0.9],
[0.7, 0.9, 0.4, 0.6, 0.9],
[0.1, 0.2, 0.6, 0.6, 0.9]]
def find_ix(a, threshold = 0.6):
res_list = []
for i in range(len(a)):
for j in range(len(a[i])):
if a[i][j] >= threshold:
res_list.append([i, j, a[i][j]])
return res_list
print("Resulting list = \n ", find_ix(a))

import numpy as np
arr = np.array([[0.1, 0.2, 0.4, 0.6, 0.9],
[0.3, 0.7, 0.8, 0.3, 0.9],
[0.7, 0.9, 0.4, 0.6, 0.9],
[0.1, 0.2, 0.6, 0.6, 0.9]])
rows, cols = np.where(arr > 0.6) # Get rows and columns where arr > 0.6
values = arr[rows, cols] # Get all values > 0.6 in arr
result = np.column_stack((rows, cols, values)) # Stack three columns to create final array
"""
Result -
[ 0. 4. 0.9]
[ 1. 1. 0.7]
[ 1. 2. 0.8]
[ 1. 4. 0.9]
[ 2. 0. 0.7]
[ 2. 1. 0.9]
[ 2. 4. 0.9]
[ 3. 4. 0.9]]
"""
You can convert result into a list.

Why does random.shuffle fail on numpy lists?

I have an array of row vectors, upon which I run random.shuffle:
#!/usr/bin/env python
import random
import numpy as np
zzz = np.array([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.6, 0.7, 0.8, 0.9, 1. ]])
iterations = 100000
f = 0
for _ in range(iterations):
random.shuffle(zzz)
if np.array_equal(zzz[0], zzz[1]):
print(zzz)
f += 1
print(float(f)/float(iterations))
Between 99.6 and 100% of the time, using random.shuffle on zzz returns a list with the same elements in it, e.g.:
$ ./test.py
...
[[ 0.1 0.2 0.3 0.4 0.5]
[ 0.1 0.2 0.3 0.4 0.5]]
0.996
Using numpy.random.shuffle appears to pass this test and shuffle row vectors correctly. I'm curious to know why random.shuffle fails.

If you look at the code of random.shuffle it performs swaps in the following way:
x[i], x[j] = x[j], x[i]
which for a numpy.array would fail, without raising any error. Example:
>>> zzz[1], zzz[0] = zzz[0], zzz[1]
>>> zzz
array([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.1, 0.2, 0.3, 0.4, 0.5]])
The reason is that Python first evaluates the right hand side completely and then make the assignment (this is why with Python single line swap is possible) but for a numpy array this is not True.
numpy
>>> arr = np.array([[1],[1]])
>>> arr[0], arr[1] = arr[0]+1, arr[0]
>>> arr
array([[2],
[2]])
Python
>>> l = [1,1]
>>> l[0], l[1] = l[0]+1, l[0]
>>> l
[2, 1]

Try it like this :
#!/usr/bin/env python
import random
import numpy as np
zzz = np.array([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.6, 0.7, 0.8, 0.9, 1. ]])
iterations = 100000
f = 0
for _ in range(iterations):
random.shuffle(zzz[0])
random.shuffle(zzz[1])
if np.array_equal(zzz[0], zzz[1]):
print(zzz)
f += 1
print(float(f)/float(iterations))

In [200]: zzz = np.array([[0.1, 0.2, 0.3, 0.4, 0.5],
...: [0.6, 0.7, 0.8, 0.9, 1. ]])
...:
In [201]: zl = zzz.tolist()
In [202]: zl
Out[202]: [[0.1, 0.2, 0.3, 0.4, 0.5], [0.6, 0.7, 0.8, 0.9, 1.0]]
random.random is probably using an in-place assignment like:
In [203]: zzz[0],zzz[1]=zzz[1],zzz[0]
In [204]: zzz
Out[204]:
array([[0.6, 0.7, 0.8, 0.9, 1. ],
[0.6, 0.7, 0.8, 0.9, 1. ]])
Note the replication.
But applied to a list of lists:
In [205]: zl[0],zl[1]=zl[1],zl[0]
In [206]: zl
Out[206]: [[0.6, 0.7, 0.8, 0.9, 1.0], [0.1, 0.2, 0.3, 0.4, 0.5]]
In [207]: zl[0],zl[1]=zl[1],zl[0]
In [208]: zl
Out[208]: [[0.1, 0.2, 0.3, 0.4, 0.5], [0.6, 0.7, 0.8, 0.9, 1.0]]
I tested zl = list(zzz) and still got the array behavior. This zl is a list with views of zzz. tolist makes a list of lists thats totally independent ofzzz`.
In short random.random cannot handle inplace modifications of a ndarray correctly. np.random.shuffle is designed to work with the 1st dim of an array, so it gets it right.
correct assignment for ndarray is:
In [211]: zzz = np.array([[0.1, 0.2, 0.3, 0.4, 0.5],
...: [0.6, 0.7, 0.8, 0.9, 1. ]])
...:
In [212]: zzz[[0,1]] = zzz[[1,0]]
In [213]: zzz
Out[213]:
array([[0.6, 0.7, 0.8, 0.9, 1. ],
[0.1, 0.2, 0.3, 0.4, 0.5]])
In [214]: zzz[[0,1]] = zzz[[1,0]]
In [215]: zzz
Out[215]:
array([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.6, 0.7, 0.8, 0.9, 1. ]])

How to sum/average a specific subset of columns or rows and return the new ndarray in numpy?

For the sake of illustration, imaging I have the following ndarray:
x = [[0.5, 0.3, 0.1, 0.1],
[0.4, 0.1, 0.3, 0.2],
[0.4, 0.3, 0.2, 0.1],
[0.6, 0.1, 0.1, 0.2]]
I want to sum the two vectors at columns 1 and 2 (starting the count from 0) so that the new ndarray would be:
y = [[0.5, 0.4, 0.1],
[0.4, 0.4, 0.2],
[0.4, 0.5, 0.1],
[0.6, 0.2, 0.2]]
And then, I want to average the vectors at rows 1 and 2 so that the final result would be:
z = [[0.5, 0.4, 0.1 ],
[0.4, 0.45, 0.15],
[0.6, 0.2, 0.2 ]]
Is there an efficient way to do that in numpy in one command? I really need efficiency as this operation is going to be applied in a nested loop.
Thanks in advance

#hpaulj s solution is very good, be sure to read it
You can sum columns quite easily:
a_summed = np.sum(a[:,1:3], axis=1)
You can also take the mean of multiple rows:
a_mean = np.mean(a[1:3], axis=0)
All you have to do is replace and delete the remaining columns, so it becomes:
import numpy as np
a_summed = np.sum(a[:,1:3], axis=1)
a[:, 1] = a_summed
a = np.delete(a, 2, 1)
a_mean = np.mean(a[1:3], axis=0)
a[1] = a_mean
a = np.delete(a, 2, 0)
print(a)

Since you are changing the original matrix size it would be better to do it in two steps as mentioned in the previous answers but, if you want to do it in one command, you could do it as follows and it makes for a nice generalized solution:
import numpy as np
x = np.array(([0.5, 0.3, 0.1, 0.1, 1],
[0.4, 0.1, 0.3, 0.2, 1],
[0.4, 0.3, 0.2, 0.1, 1],
[0.6, 0.1, 0.1, 0.2, 1]))
def sum_columns(matrix, col_start, col_end):
return np.column_stack((matrix[:, 0:col_start],
np.sum(matrix[:, col_start:col_end + 1], axis=1),
matrix[:, col_end + 1:]))
def avgRows_summedColumns(matrix, row_start, row_end):
return np.row_stack((matrix[0:row_start, :],
np.mean(matrix[row_start:row_end + 1, :], axis=0),
matrix[row_end:-1, :]))
# call the entire operation in one command
print(avgRows_summedColumns(sum_columns(x, 1, 2), 1, 2))
This way it doesn't matter how big your matrix is.

In [68]: x = [[0.5, 0.3, 0.1, 0.1],
...: [0.4, 0.1, 0.3, 0.2],
...: [0.4, 0.3, 0.2, 0.1],
...: [0.6, 0.1, 0.1, 0.2]]
In [69]: x=np.array(x)
ufunc like np.add have a reduceat method that lets us perform the action over groups of rows or columns. With that the first reduction is easy (but takes a little playing to understand the parameters):
In [70]: np.add.reduceat(x,[0,1,3], axis=1)
Out[70]:
array([[0.5, 0.4, 0.1],
[0.4, 0.4, 0.2],
[0.4, 0.5, 0.1],
[0.6, 0.2, 0.2]])
Apparently mean is not a ufunc, so I had to settle for add to reduce the rows:
In [71]: np.add.reduceat(Out[70],[0,1,3],axis=0)
Out[71]:
array([[0.5, 0.4, 0.1],
[0.8, 0.9, 0.3],
[0.6, 0.2, 0.2]])
and then divide by the row count to get the mean. I could generalize that to use the same [0,1,3] used in the reduceat, but for now just use a column array:
In [72]: np.add.reduceat(Out[70],[0,1,3],axis=0)/np.array([1,2,1])[:,None]
Out[72]:
array([[0.5 , 0.4 , 0.1 ],
[0.4 , 0.45, 0.15],
[0.6 , 0.2 , 0.2 ]])
and the whole thing in one expression:
In [73]: np.add.reduceat(np.add.reduceat(x,[0,1,3], axis=1),[0,1,3],axis=0)/ np.array([1,2,1])[:,None]
Out[73]:
array([[0.5 , 0.4 , 0.1 ],
[0.4 , 0.45, 0.15],
[0.6 , 0.2 , 0.2 ]])

How can a tensor in tensorflow be sliced ​using elements of another array as an index?

I'm looking for a similar function to tf.unsorted_segment_sum, but I don't want to sum the segments, I want to get every segment as a tensor.
So for example, I have this code:
(In real, I have a tensor with shapes of (10000, 63), and the number of segments would be 2500)
to_be_sliced = tf.constant([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.3, 0.2, 0.2, 0.6, 0.3],
[0.9, 0.8, 0.7, 0.6, 0.5],
[2.0, 2.0, 2.0, 2.0, 2.0]])
indices = tf.constant([0, 2, 0, 1])
num_segments = 3
tf.unsorted_segment_sum(to_be_sliced, indices, num_segments)
The output would be here
array([sum(row1+row3), row4, row2]
What I am looking for is 3 tensor with different shapes (maybe a list of tensors), first containing the first and third rows of the original (shape of (2, 5)), the second contains the 4th row (shape of (1, 5)), the third contains the second row, like this:
[array([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.9, 0.8, 0.7, 0.6, 0.5]]),
array([[2.0, 2.0, 2.0, 2.0, 2.0]]),
array([[0.3, 0.2, 0.2, 0.6, 0.3]])]
Thanks in advance!

You can do that like this:
import tensorflow as tf
to_be_sliced = tf.constant([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.3, 0.2, 0.2, 0.6, 0.3],
[0.9, 0.8, 0.7, 0.6, 0.5],
[2.0, 2.0, 2.0, 2.0, 2.0]])
indices = tf.constant([0, 2, 0, 1])
num_segments = 3
result = [tf.boolean_mask(to_be_sliced, tf.equal(indices, i)) for i in range(num_segments)]
with tf.Session() as sess:
print(*sess.run(result), sep='\n')
Output:
[[0.1 0.2 0.3 0.4 0.5]
[0.9 0.8 0.7 0.6 0.5]]
[[2. 2. 2. 2. 2.]]
[[0.3 0.2 0.2 0.6 0.3]]

For your case, you can do Numpy slicing in Tensorflow. So this will work:
sliced_1 = to_be_sliced[:3, :]
# [[0.4 0.5 0.5 0.7 0.8]
# [0.3 0.2 0.2 0.6 0.3]
# [0.3 0.2 0.2 0.6 0.3]]
sliced_2 = to_be_sliced[3, :]
# [0.3 0.2 0.2 0.6 0.3]
Or a more general option, you can do it in the following way:
to_be_sliced = tf.constant([[0.1, 0.2, 0.3, 0.4, 0.5],
[0.3, 0.2, 0.2, 0.6, 0.3],
[0.9, 0.8, 0.7, 0.6, 0.5],
[2.0, 2.0, 2.0, 2.0, 2.0]])
first_tensor = tf.gather_nd(to_be_sliced, [[0], [2]])
second_tensor = tf.gather_nd(to_be_sliced, [[3]])
third_tensor = tf.gather_nd(to_be_sliced, [[1]])
concat = tf.concat([first_tensor, second_tensor, third_tensor], axis=0)

multiplying an array of matrices with a vector

I have an array of matrices in which I would like to multiply each matrix by a different number. I tried it this way:
>>> import numpy as np
>>> c = np.array([[[1, 2],[3, 4]],[[1, 2],[3, 4]]])
>>> d = np.array([0.1, 0.2])
>>> d*c
array([[[ 0.1, 0.4],
[ 0.3, 0.8]],
[[ 0.1, 0.4],
[ 0.3, 0.8]]])
While my intention is to get this result -
>>> np.array([d[0]*c[0], d[1]*c[1]])
array([[[ 0.1, 0.2],
[ 0.3, 0.4]],
[[ 0.2, 0.4],
[ 0.6, 0.8]]])
What is the NumPy'iest way to do it?

You need an extra couple of axes:
In [22]: d[:,None,None] * c
Out[22]:
array([[[ 0.1, 0.2],
[ 0.3, 0.4]],
[[ 0.2, 0.4],
[ 0.6, 0.8]]])
d[:,None,None] has shape (2,1,1) which is broadcast across your c array of shape (2,2,2) to multiply each block of c by the corresponding element of d.

You can do this with normal broadcasting:
>>> c*d[:, np.newaxis, np.newaxis]
array([[[ 0.1, 0.2],
[ 0.3, 0.4]],
[[ 0.2, 0.4],
[ 0.6, 0.8]]])