Develop Reference

Is there a 2-D "where" in numpy?

This might seem an odd question, but it boils down to quite a simple operation that I can't find a numpy equivalent for. I've looked at np.where as well as many other operations but can't find anything that does this:
a = np.array([1,2,3])
b = np.array([1,2,3,4])
c = np.array([i<b for i in a])
The output is a 2-D array (3,4), of booleans comparing each value.

If you're asking how to get c without loop, try this
# make "a" a column vector
# > broadcasts to produce a len(a) x len(b) array
c = b > a[:, None]
c
array([[False, True, True, True],
[False, False, True, True],
[False, False, False, True]])

You can extend the approach in the other answer to get the values of a and b. Given a mask of
c = b > a[:, None]
You can extract the indices for each dimension using np.where or np.nonzero:
row, col = np.nonzero(c)
And use the indices to get the corresponding values:
ag = a[row]
bg = b[col]
Elements of a and b may be repeated in the result.

ndarray row-wise index of values greater than array

I have one array of shape (X, 5):
M = [[1,2,3,4,5],
[6,7,8,9,1],
[2,5,7,8,3]
...]
and one array of shape (X, 1):
n = [[3],
[7],
[100],
...]
Now I need to get the first index of M >= n for each row, or nan if there is no such index.
For example:
np.where([1,2,3,4,5] >= 3)[0][0] # Returns 2
np.searchsorted(np.array([1,2,3,4,5]), 3) # Returns 2
These examples are applied to each row individually (I could loop X times as both arrays have the length X).
I wonder, is there a way to do it in a multidimensional way to get an output of all indices at once?
Something like:
np.where(M>=n)
Thank you
Edit: Values in M are unsorted, I'm still looking for the first index/occurrence fitting M >= n (so probably not searchsorted)

You could start by checking which row indices are lower or equal than n and use argmax to get the first True for each row. For the rows where all columns are False, we can use np.where to set them to np.nan for instance:
M = np.array([[1,2,3,4,5],
[6,7,8,9,1],
[2,5,7,8,3]])
n = np.array([[3],[7],[100]])
le = n<=M
# array([[False, False, True, True, True],
# [False, True, True, True, False],
# [False, False, False, False, False]])
lea = le.argmax(1)
has_any = le[np.arange(len(le)), lea]
np.where(has_any, lea, np.nan)
# array([ 2., 1., nan])

Creating a "bitmask" from several boolean numpy arrays

I'm trying to convert several masks (boolean arrays) to a bitmask with numpy, while that in theory works I feel that I'm doing too many operations.
For example to create the bitmask I use:
import numpy as np
flags = [
np.array([True, False, False]),
np.array([False, True, False]),
np.array([False, True, False])
]
flag_bits = np.zeros(3, dtype=np.int8)
for idx, flag in enumerate(flags):
flag_bits += flag.astype(np.int8) << idx # equivalent to flag * 2 ** idx
Which gives me the expected "bitmask":
>>> flag_bits
array([1, 6, 0], dtype=int8)
>>> [np.binary_repr(bit, width=7) for bit in flag_bits]
['0000001', '0000110', '0000000']
However I feel that especially the casting to int8 and the addition with the flag_bits array is too complicated. Therefore I wanted to ask if there is any NumPy functionality that I missed that could be used to create such an "bitmask" array?
Note: I'm calling an external function that expects such a bitmask, otherwise I would stick with the boolean arrays.

>>> x = np.array(2**i for i in range(1, np.shape(flags)[1]+1))
>>> np.dot(flags, x)
array([1, 2, 2])
How it works: in a bit mask, every bit is effectively an original array element multiplied by a degree of 2 according to its position, e.g. 4 = False * 1 + True * 2 + False * 4. Effectively this can be represented as matrix multiplication, which is really efficient in numpy.
So, first line is a list comprehension to create these weights: x = [1, 2, 4, 8, ... 2^(n+1)].
Then, each line in flags is multiplied by the corresponding element in x and everything is summed up (this is how matrix multiplication works). At the end, we get the bitmask

How about this (added conversion to int8, if desired):
flag_bits = (np.transpose(flags) << np.arange(len(flags))).sum(axis=1)\
.astype(np.int8)
#array([1, 6, 0], dtype=int8)

Here's an approach to directly get to the string bitmask with boolean-indexing -
out = np.repeat('0000000',3).astype('S7')
out.view('S1').reshape(-1,7)[:,-3:] = np.asarray(flags).astype(int)[::-1].T
Sample run -
In [41]: flags
Out[41]:
[array([ True, False, False], dtype=bool),
array([False, True, False], dtype=bool),
array([False, True, False], dtype=bool)]
In [42]: out = np.repeat('0000000',3).astype('S7')
In [43]: out.view('S1').reshape(-1,7)[:,-3:] = np.asarray(flags).astype(int)[::-1].T
In [44]: out
Out[44]:
array([b'0000001', b'0000110', b'0000000'],
dtype='|S7')
Using the same matrix-multiplication strategy as dicussed in detail in #Marat's solution, but using a vectorized scaling array that gives us flag_bits -
np.dot(2**np.arange(3),flags)

Replace values in specific columns of a numpy array

I have a N x M numpy array (matrix). Here is an example with a 3 x 5 array:
x = numpy.array([[0,1,2,3,4,5],[0,-1,2,3,-4,-5],[0,-1,-2,-3,4,5]])
I'd like to scan all the columns of x and replace the values of each column if they are equal to a specific value.
This code for example aims to replace all the negative values (where the value is equal to the column number) to 100:
for i in range(1,6):
x[:,i == -(i)] = 100
This code obtains this warning:
DeprecationWarning: using a boolean instead of an integer will result in an error in the future
I'm using numpy 1.8.2. How can I avoid this warning without downgrade numpy?

I don't follow what your code is trying to do:
the i == -(i)
will evaluate to something like this:
x[:, True]
x[:, False]
I don't think this is what you want. You should try something like this:
for i in range(1, 6):
mask = x[:, i] == -i
x[:, i][mask] = 100
Create a mask over the whole column, and use that to change the values.

Even without the warning, the code you have there will not do what you want. i is the loop index and will equal minus itself only if i == 0, which is never. Your test will always return false, which is cast to 0. In other words your code will replace the first element of each row with 100.
To get this to work I would do
for i in range(1, 6):
col = x[:,i]
col[col == -i] = 100
Notice that you use the name of the array for the masking and that you need to separate the conventional indexing from the masking

If you are worried about the warning spewing out text, then ignore it as a Warning/Exception:
import numpy
import warnings
warnings.simplefilter('default') # this enables DeprecationWarnings to be thrown
x = numpy.array([[0,1,2,3,4,5],[0,-1,2,3,-4,-5],[0,-1,-2,-3,4,5]])
with warnings.catch_warnings():
warnings.simplefilter("ignore") # and this ignores them
for i in range(1,6):
x[:,i == -(i)] = 100
print(x) # just to show that you are actually changing the content
As you can see in the comments, some people are not getting DeprecationWarning. That is probably because python suppresses developer-only warnings since 2.7

As others have said, your loop isn't doing what you think it is doing. I would propose you change your code to use numpy's fancy indexing.
# First, create the "test values" (column index):
>>> test_values = numpy.arange(6)
# test_values is array([0, 1, 2, 3, 4, 5])
#
# Now, we want to check which columns have value == -test_values:
#
>>> mask = (x == -test_values) & (x < 0)
# mask is True wherever a value in the i-th column of x is negative i
>>> mask
array([[False, False, False, False, False, False],
[False, True, False, False, True, True],
[False, True, True, True, False, False]], dtype=bool)
#
# Now, set those values to 100
>>> x[mask] = 100
>>> x
array([[ 0, 1, 2, 3, 4, 5],
[ 0, 100, 2, 3, 100, 100],
[ 0, 100, 100, 100, 4, 5]])

Numpy: Subtract array element by element

The title might be ambiguous, didn't know how else to word it.
I have gotten a bit far with my particle simulator in python using numpy and matplotlib, I have managed to implement coloumb, gravity and wind, now I just want to add temperature and pressure but I have a pre-optimization question (root of all evil). I want to see when particles crash:
Q: Is it in numpy possible to take the difference of an array with each of its own element based on a bool condition? I want to avoid looping.
Eg: (x - any element in x) < a
Should return something like
[True, True, False, True]
If element 0,1 and 3 in x meets the condition.
Edit:
The loop quivalent would be:
for i in len(x):
for j in in len(x):
#!= not so important
##earlier question I asked lets me figure that one out
if i!=j:
if x[j] - x[i] < a:
True
I notice numpy operations are far faster than if tests and this has helped me speed up things ALOT.
Here is a sample code if anyone wants to play with it.
#Simple circular box simulator, part of part_sim
#Restructure to import into gravity() or coloumb () or wind() or pressure()
#Or to use all forces: sim_full()
#Note: Implement crashing as backbone to all forces
import numpy as np
import matplotlib.pyplot as plt
N = 1000 #Number of particles
R = 8000 #Radius of box
r = np.random.randint(0,R/2,2*N).reshape(N,2)
v = np.random.randint(-200,200,r.shape)
v_limit = 10000 #Speedlimit
plt.ion()
line, = plt.plot([],'o')
plt.axis([-10000,10000,-10000,10000])
while True:
r_hit = np.sqrt(np.sum(r**2,axis=1))>R #Who let the dogs out, who, who?
r_nhit = ~r_hit
N_rhit = r_hit[r_hit].shape[0]
r[r_hit] = r[r_hit] - 0.1*v[r_hit] #Get the dogs back inside
r[r_nhit] = r[r_nhit] +0.1*v[r_nhit]
#Dogs should turn tail before they crash!
#---
#---crash code here....
#---crash end
#---
vmin, vmax = np.min(v), np.max(v)
#Give the particles a random kick when they hit the wall
v[r_hit] = -v[r_hit] + np.random.randint(vmin, vmax, (N_rhit,2))
#Slow down honey
v_abs = np.abs(v) > v_limit
#Hit the wall at too high v honey? You are getting a speed reduction
v[v_abs] *=0.5
line.set_ydata(r[:,1])
line.set_xdata(r[:,0])
plt.draw()
I plan to add colors to the datapoints above once I figure out how...such that high velocity particles can easily be distinguished in larger boxes.

Eg: x - any element in x < a Should return something like
[True, True, False, True]
If element 0,1 and 3 in x meets the condition. I notice numpy operations are far faster than if tests and this has helped me speed up things ALOT.
Yes, it's just m < a. For example:
>>> m = np.array((1, 3, 10, 5))
>>> a = 6
>>> m2 = m < a
>>> m2
array([ True, True, False, True], dtype=bool)
Now, to the question:
Q: Is it in numpy possible to take the difference of an array with each of its own element based on a bool condition? I want to avoid looping.
I'm not sure what you're asking for here, but it doesn't seem to match the example directly below it. Are you trying to, e.g., subtract 1 from each element that satisfies the predicate? In that case, you can rely on the fact that False==0 and True==1 and just subtract the boolean array:
>>> m3 = m - m2
>>> m3
>>> array([ 0, 2, 10, 4])
From your clarification, you want the equivalent of this pseudocode loop:
for i in len(x):
for j in in len(x):
#!= not so important
##earlier question I asked lets me figure that one out
if i!=j:
if x[j] - x[i] < a:
True
I think the confusion here is that this is the exact opposite of what you said: you don't want "the difference of an array with each of its own element based on a bool condition", but "a bool condition based on the difference of an array with each of its own elements". And even that only really gets you to a square matrix of len(m)*len(m) bools, but I think the part left over is that the "any".
At any rate, you're asking for an implicit cartesian product, comparing each element of m to each element of m.
You can easily reduce this from two loops to one (or, rather, implicitly vectorize one of them, gaining the usual numpy performance benefits). For each value, create a new array by subtracting that value from each element and comparing the result with a, and then join those up:
>>> a = -2
>>> comparisons = np.array([m - x < a for x in m])
>>> flattened = np.any(comparisons, 0)
>>> flattened
array([ True, True, False, True], dtype=bool)
But you can also turn this into a simple matrix operation pretty easily. Subtracting every element of m from every other element of m is just m - m.T. (You can make the product more explicit, but the way numpy handles adding row and column vectors, it isn't necessary.) And then you just compare every element of that to the scalar a, and reduce with any, and you're done:
>>> a = -2
>>> m = np.matrix((1, 3, 10, 5))
>>> subtractions = m - m.T
>>> subtractions
matrix([[ 0, 2, 9, 4],
[-2, 0, 7, 2],
[-9, -7, 0, -5],
[-4, -2, 5, 0]])
>>> comparisons = subtractions < a
>>> comparisons
matrix([[False, False, False, False],
[False, False, False, False],
[ True, True, False, True],
[ True, False, False, False]], dtype=bool)
>>> np.any(comparisons, 0)
matrix([[ True, True, False, True]], dtype=bool)
Or, putting it all together in one line:
>>> np.any((m - m.T) < a, 0)
matrix([[ True, True, True, True]], dtype=bool)
If you need m to be an array rather than a matrix, you can replace the subtraction line with m - np.matrix(m).T.
For higher dimensions, you actually do need to work in arrays, because you're trying to cartesian-product a 2D array with itself to get a 4D array, and numpy doesn't do 4D matrices. So, you can't use the simple "row vector - column vector = matrix" trick. But you can do it manually:
>>> m = np.array([[1,2], [3,4]]) # 2x2
>>> m4d = m.reshape(1, 1, 2, 2) # 1x1x2x2
>>> m4d
array([[[[1, 2],
[3, 4]]]])
>>> mt4d = m4d.T # 2x2x1x1
>>> mt4d
array([[[[1]],
[[3]]],
[[[2]],
[[4]]]])
>>> subtractions = m - mt4d # 2x2x2x2
>>> subtractions
array([[[[ 0, 1],
[ 2, 3]],
[[-2, -1],
[ 0, 1]]],
[[[-1, 0],
[ 1, 2]],
[[-3, -2],
[-1, 0]]]])
And from there, the remainder is the same as before. Putting it together into one line:
>>> np.any((m - m.reshape(1, 1, 2, 2).T) < a, 0)
(If you remember my original answer, I'd somehow blanked on reshape and was doing the same thing by multiplying m by a column vector of 1s, which obviously is a much stupider way to proceed.)
One last quick thought: If your algorithm really is "the bool result of (for any element y of m, x - y < a) for each element x of m", you don't actually need "for any element y", you can just use "for the maximal element y". So you can simplify from O(N^2) to O(N):
>>> (m - m.max()) < a
Or, if a is positive, that's always false, so you can simplify to O(1):
>>> np.zeros(m.shape, dtype=bool)
But I'm guessing your real algorithm is actually using abs(x - y), or something more complicated, which can't be simplified in this way.