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Group a numpy array

I have an one-dimensional array A, such that 0 <= A[i] <= 11, and I want to map A to an array B such that
for i in range(len(A)):
if 0 <= A[i] <= 2: B[i] = 0
elif 3 <= A[i] <= 5: B[i] = 1
elif 6 <= A[i] <= 8: B[i] = 2
elif 9 <= A[i] <= 11: B[i] = 3
How can implement this efficiently in numpy?

You need to use an int division by //3, and that is the most performant solution
A = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11])
B = A // 3
print(A) # [0 1 2 3 4 5 6 7 8 9 10 11]
print(B) # [0 0 0 1 1 1 2 2 2 3 3 3]

I would do something like dividing the values of the A[i] by 3 'cause you're sorting out them 3 by 3, 0-2 divided by 3 go answer 0, 3-5 go answer 1, 6-8 divided by 3 is equal to 2, and so on
I built a little schema here:
A[i] --> 0-2. divided by 3 = 0, what you wnat in array B[i] is 0, so it's ok
A[i] --> 3-5. divided by 3 = 1, and so on. Just use a method to make floor the value, so that it don't become float type.

Answers provided by others are valid, however I find this function from numpy quite elegant, plus it allows you to avoid for loop which could be quite inefficient for large arrays
import numpy as np
bins = [3, 5, 8, 9, 11]
B = np.digitize(A, bins)

Something like this might work:
C = np.zeros(12, dtype=np.int)
C[3:6] = 1
C[6:9] = 2
C[9:12] = 3
B = C[A]

If you hope to expand this to a more complex example you can define a function with all your conditions:
def f(a):
if 0 <= a and a <= 2:
return 0
elif 3 <= a and a <= 5:
return 1
elif 6 <= a and a <= 8:
return 2
elif 9 <= a and a <= 11:
return 3
And call it on your array A:
A = np.array([0,1,5,7,8,9,10,10, 11])
B = np.array(list(map(f, A))) # array([0, 0, 1, 2, 2, 3, 3, 3, 3])

Is there multi arange in NumPy?

Numpy's arange accepts only single scalar values for start/stop/step. Is there a multi version of this function? Which can accept array inputs for start/stop/step? E.g. having input 2D array like:
[[1 5 1], # start/stop/step first
[3 8 2]] # start/stop/step second
should create array consisting of concatenation of aranges for every row of input (each start/stop/step), input above should create 1D array
1 2 3 4 3 5 7
i.e. we need to design such function that it does next:
print(np.multi_arange(np.array([[1,5,1],[3,8,2]])))
# prints:
# array([1, 2, 3, 4, 3, 5, 7])
And this function should be efficient (pure numpy), i.e. very fast process input array of shape (10000, 3) without pure-Python looping.
Of cause it is possible to create pure Python's loop (or listcomp) to create arange for each row and concatenate results of this loop. But I have very many rows with triples start/stop/step and need to have efficient and fast code, hence looking for pure numpy function.
Why do I need it. I needed this for several tasks. One of this is for indexing - suppose I have 1D array a and I need to extract many (possibly intersecting) subranges of this array. If I had that multi version of arange I would just do:
values = a[np.multi_arange(starts_stops_steps)]
Maybe it is possible to create multi arange function using some combinations of numpy functions? Can you suggest?
Also maybe there are some more efficient solutions for the specific case of extracting subranges of 1D array (see last line of code above) without creating all indexes using multi_arange?

Here's a vectorized one with cumsum that accounts for positive and negative stepsizes -
def multi_arange(a):
steps = a[:,2]
lens = ((a[:,1]-a[:,0]) + steps-np.sign(steps))//steps
b = np.repeat(steps, lens)
ends = (lens-1)*steps + a[:,0]
b[0] = a[0,0]
b[lens[:-1].cumsum()] = a[1:,0] - ends[:-1]
return b.cumsum()
If you need to validate for valid ranges : (start < stop when step > 0) and (start > stop when step < 0) , use a pre-processing step :
a = a[((a[:,1] > a[:,0]) & (a[:,2]>0) | (a[:,1] < a[:,0]) & (a[:,2]<0))]
Sample run -
In [17]: a
Out[17]:
array([[ 1, 5, 1],
[ 3, 8, 2],
[18, 6, -2]])
In [18]: multi_arange(a)
Out[18]: array([ 1, 2, 3, 4, 3, 5, 7, 18, 16, 14, 12, 10, 8])

In [1]: np.r_[1:5:1, 3:8:2]
Out[1]: array([1, 2, 3, 4, 3, 5, 7])
In [2]: np.hstack((np.arange(1,5,1),np.arange(3,8,2)))
Out[2]: array([1, 2, 3, 4, 3, 5, 7])
The r_ version is nice and compact, but not faster:
In [3]: timeit np.r_[1:5:1, 3:8:2]
23.9 µs ± 34.6 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [4]: timeit np.hstack((np.arange(1,5,1),np.arange(3,8,2)))
11.2 µs ± 19.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

I've just came up with my solution using numba. Still I prefer numpy-only solution if we find best one not to carry heavy numba JIT compiler.
I've also tested #Divakar solution in my code.
Next code output is:
naive_multi_arange 0.76601 sec
arty_multi_arange 0.01801 sec 42.52 speedup
divakar_multi_arange 0.05504 sec 13.92 speedup
Meaning my numba solution has 42x speedup, #Divakar's numpy solution has 14x speedup.
Next code can be also run online here.
import time, random
import numpy as np, numba
#numba.jit(nopython = True)
def arty_multi_arange(a):
starts, stops, steps = a[:, 0], a[:, 1], a[:, 2]
pos = 0
cnt = np.sum((stops - starts + steps - np.sign(steps)) // steps, dtype = np.int64)
res = np.zeros((cnt,), dtype = np.int64)
for i in range(starts.size):
v, stop, step = starts[i], stops[i], steps[i]
if step > 0:
while v < stop:
res[pos] = v
pos += 1
v += step
elif step < 0:
while v > stop:
res[pos] = v
pos += 1
v += step
assert pos == cnt
return res
def divakar_multi_arange(a):
steps = a[:,2]
lens = ((a[:,1]-a[:,0]) + steps-np.sign(steps))//steps
b = np.repeat(steps, lens)
ends = (lens-1)*steps + a[:,0]
b[0] = a[0,0]
b[lens[:-1].cumsum()] = a[1:,0] - ends[:-1]
return b.cumsum()
random.seed(0)
neg_prob = 0.5
N = 100000
minv, maxv, maxstep = -100, 300, 15
steps = [random.randrange(1, maxstep + 1) * ((1, -1)[random.random() < neg_prob]) for i in range(N)]
starts = [random.randrange(minv + 1, maxv) for i in range(N)]
stops = [random.randrange(*(((starts[i] + 1, maxv + 1), (minv, starts[i]))[steps[i] < 0])) for i in range(N)]
joined = np.array([starts, stops, steps], dtype = np.int64).T
tb = time.time()
aref = np.concatenate([np.arange(joined[i, 0], joined[i, 1], joined[i, 2], dtype = np.int64) for i in range(N)])
npt = time.time() - tb
print('naive_multi_arange', round(npt, 5), 'sec')
for func in ['arty_multi_arange', 'divakar_multi_arange']:
globals()[func](joined)
tb = time.time()
a = globals()[func](joined)
myt = time.time() - tb
print(func, round(myt, 5), 'sec', round(npt / myt, 2), 'speedup')
assert a.size == aref.size, (a.size, aref.size)
assert np.all(a == aref), np.vstack((np.flatnonzero(a != aref)[:5], a[a != aref][:5], aref[a != aref][:5])).T

Max value per diagonal in 2d array

I have array and need max of rolling difference with dynamic window.
a = np.array([8, 18, 5,15,12])
print (a)
[ 8 18 5 15 12]
So first I create difference by itself:
b = a - a[:, None]
print (b)
[[ 0 10 -3 7 4]
[-10 0 -13 -3 -6]
[ 3 13 0 10 7]
[ -7 3 -10 0 -3]
[ -4 6 -7 3 0]]
Then replace upper triangle matrix to 0:
c = np.tril(b)
print (c)
[[ 0 0 0 0 0]
[-10 0 0 0 0]
[ 3 13 0 0 0]
[ -7 3 -10 0 0]
[ -4 6 -7 3 0]]
Last need max values per diagonal, so it means:
max([0,0,0,0,0]) = 0
max([-10,13,-10,3]) = 13
max([3,3,-7]) = 3
max([-7,6]) = 6
max([-4]) = -4
So expected output is:
[0, 13, 3, 6, -4]
What is some nice vectorized solution? Or is possible some another way for expected output?

Use ndarray.diagonal
v = [max(c.diagonal(-i)) for i in range(b.shape[0])]
print(v) # [0, 13, 3, 6, -4]

Not sure exactly how efficient this is considering the advanced indexing involved, but this is one way to do that:
import numpy as np
a = np.array([8, 18, 5, 15, 12])
b = a[:, None] - a
# Fill lower triangle with largest negative
b[np.tril_indices(len(a))] = np.iinfo(b.dtype).min # np.finfo for float
# Put diagonals as rows
s = b.strides[1]
diags = np.ndarray((len(a) - 1, len(a) - 1), b.dtype, b, offset=s, strides=(s, (len(a) + 1) * s))
# Get maximum from each row and add initial zero
c = np.r_[0, diags.max(1)]
print(c)
# [ 0 13 3 6 -4]
EDIT:
Another alternative, which may not be what you were looking for though, is just using Numba, for example like this:
import numpy as np
import numba as nb
def max_window_diffs_jdehesa(a):
a = np.asarray(a)
dtinf = np.iinfo(b.dtype) if np.issubdtype(b.dtype, np.integer) else np.finfo(b.dtype)
out = np.full_like(a, dtinf.min)
_pwise_diffs(a, out)
return out
#nb.njit(parallel=True)
def _pwise_diffs(a, out):
out[0] = 0
for w in nb.prange(1, len(a)):
for i in range(len(a) - w):
out[w] = max(a[i] - a[i + w], out[w])
a = np.array([8, 18, 5, 15, 12])
print(max_window_diffs(a))
# [ 0 13 3 6 -4]
Comparing these methods to the original:
import numpy as np
import numba as nb
def max_window_diffs_orig(a):
a = np.asarray(a)
b = a - a[:, None]
out = np.zeros(len(a), b.dtype)
out[-1] = b[-1, 0]
for i in range(1, len(a) - 1):
out[i] = np.diag(b, -i).max()
return out
def max_window_diffs_jdehesa_np(a):
a = np.asarray(a)
b = a[:, None] - a
dtinf = np.iinfo(b.dtype) if np.issubdtype(b.dtype, np.integer) else np.finfo(b.dtype)
b[np.tril_indices(len(a))] = dtinf.min
s = b.strides[1]
diags = np.ndarray((len(a) - 1, len(a) - 1), b.dtype, b, offset=s, strides=(s, (len(a) + 1) * s))
return np.concatenate([[0], diags.max(1)])
def max_window_diffs_jdehesa_nb(a):
a = np.asarray(a)
dtinf = np.iinfo(b.dtype) if np.issubdtype(b.dtype, np.integer) else np.finfo(b.dtype)
out = np.full_like(a, dtinf.min)
_pwise_diffs(a, out)
return out
#nb.njit(parallel=True)
def _pwise_diffs(a, out):
out[0] = 0
for w in nb.prange(1, len(a)):
for i in range(len(a) - w):
out[w] = max(a[i] - a[i + w], out[w])
np.random.seed(0)
a = np.random.randint(0, 100, size=100)
r = max_window_diffs_orig(a)
print((max_window_diffs_jdehesa_np(a) == r).all())
# True
print((max_window_diffs_jdehesa_nb(a) == r).all())
# True
%timeit max_window_diffs_orig(a)
# 348 µs ± 986 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit max_window_diffs_jdehesa_np(a)
# 91.7 µs ± 1.3 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit max_window_diffs_jdehesa_nb(a)
# 19.7 µs ± 88.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
np.random.seed(0)
a = np.random.randint(0, 100, size=10000)
%timeit max_window_diffs_orig(a)
# 651 ms ± 26 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit max_window_diffs_jdehesa_np(a)
# 1.61 s ± 6.19 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit max_window_diffs_jdehesa_nb(a)
# 22 ms ± 967 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
The first one may be a bit better for smaller arrays, but doesn't work well for bigger ones. Numba on the other hand is pretty good in all cases.

You can use numpy.diagonal:
a = np.array([8, 18, 5,15,12])
b = a - a[:, None]
c = np.tril(b)
for i in range(b.shape[0]):
print(max(c.diagonal(-i)))
Output:
0
13
3
6
-4

Here's a vectorized solution with strides -
from skimage.util import view_as_windows
n = len(a)
z = np.zeros(n-1,dtype=a.dtype)
p = np.concatenate((a,z))
s = view_as_windows(p,n)
mask = np.tri(n,k=-1,dtype=bool)[:,::-1]
v = s[0]-s
out = np.where(mask,v.min()-1,v).max(1)
With one-loop for memory-efficiency -
n = len(a)
out = [max(a[:-i+n]-a[i:]) for i in range(n)]
Use np.max in place of max for better use of array-memory.

You can abuse the fact that reshaping non-square arrays of shape (N+1, N) to (N, N+1) will make diagonals appear as columns
from scipy.linalg import toeplitz
a = toeplitz([1,2,3,4], [1,4,3])
# array([[1, 4, 3],
# [2, 1, 4],
# [3, 2, 1],
# [4, 3, 2]])
a.reshape(3, 4)
# array([[1, 4, 3, 2],
# [1, 4, 3, 2],
# [1, 4, 3, 2]])
Which you can then use like (note that I've swapped the sign and set the lower triangle to zero)
smallv = -10000 # replace this with np.nan if you have floats
a = np.array([8, 18, 5,15,12])
b = a[:, None] - a
b[np.tril_indices(len(b), -1)] = smallv
d = np.vstack((b, np.full(len(b), smallv)))
d.reshape(len(d) - 1, -1).max(0)[:-1]
# array([ 0, 13, 3, 6, -4])

How much data are there in an interval?

I have a list object,i want to know that how many numbers are in a particular interval?The code is as follows
a = [1, 7, 4, 7, 4, 8, 5, 2, 17, 8, 3, 12, 9, 6, 28]
interval = 3
a = list(map(lambda x:int(x/interval),a))
for i in range(min(a),max(a)+1):
print(i*interval,(i+1)*interval,':',a.count(i))
Output
0 3 : 2
3 6 : 4
6 9 : 5
9 12 : 1
12 15 : 1
15 18 : 1
18 21 : 0
21 24 : 0
24 27 : 0
27 30 : 1
Is there a simple way to get this information?The simpler the better

Now that we're talking about performance, I'd like to offer my numpy solution using bincount:
import numpy as np
interval = 3
a = [1, 7, 4, 7, 4, 8, 5, 2, 17, 8, 3, 12, 9, 6, 28]
l = max(a) // interval + 1
b = np.bincount(a, minlength=l*interval).reshape((l,interval)).sum(axis=1)
(minlength is necessary just to be able to reshape if max(a) isn't a multiple of interval)
With the lables taken from Erfan's answer we get:
rnge = range(0, max(a) + interval + 1, interval)
lables = [f'[{i}-{j})' for i, j in zip(rnge[:-1], rnge[1:])]
for l,b in zip(lables,b):
print(l,b)
[0-3) 2
[3-6) 4
[6-9) 5
[9-12) 1
[12-15) 1
[15-18) 1
[18-21) 0
[21-24) 0
[24-27) 0
[27-30) 1
This is much faster than the pandas solution.
Performance and scaling comparison
In order to assess the scaling capability, I just replaced a = [1, ..., 28] * n and timed the execution (without imports and printing) for n = 1, 10, 100, 1K, 10K and 100K:
(python 3.7.3 on win32 / pandas 0.24.2 / numpy 1.16.2)

Pandas solution with pd.cut and groupby
s = pd.Series(a)
bins = pd.cut(s, range(0, s.max() + interval, interval), right=False)
s.groupby(bins).count()
[0, 3) 2
[3, 6) 4
[6, 9) 5
[9, 12) 1
[12, 15) 1
[15, 18) 1
[18, 21) 0
[21, 24) 0
[24, 27) 0
[27, 30) 1
dtype: int64
To get cleaner bins results, we can use this method from linked answer:
s = pd.Series(a)
rnge = range(0, s.max() + interval, interval)
labels = [f'{i}-{j}' for i, j in zip(rnge[:-1], rnge[1:])]
bins = pd.cut(s, range(0, s.max() + interval, interval), right=False, labels=labels)
s.groupby(bins).count()
0-3 2
3-6 4
6-9 5
9-12 1
12-15 1
15-18 1
18-21 0
21-24 0
24-27 0
27-30 1
dtype: int64

You can do it in one line using a dictionary comprehension :
a = [1, 7, 4, 7, 4, 8, 5, 2, 17, 8, 3, 12, 9, 6, 28]
{"[{};{}[".format(x, x+3) : len( [y for y in a if y >= x and y < x+3] )
for x in range(min(a), max(a), 3)}
Output :
{'[1;4[': 3,
'[4;7[': 4,
'[7;10[': 5,
'[10;13[': 1,
'[13;16[': 0,
'[16;19[': 1,
'[19;22[': 0,
'[22;25[': 0,
'[25;28[': 0}
Performance comparison :
Pandas solution with pd.cut and groupby : 8.51 ms ± 32 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Dictionary comprehension : 19.7 µs ± 37.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Using np.bincount : 22.4 µs ± 263 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

get the index of the last negative value in a 2d array per column

I'm trying to get the index of the last negative value of an array per column (in order to slice it after).
a simple working example on a 1d vector is :
import numpy as np
A = np.arange(10) - 5
A[2] = 2
print A # [-5 -4 2 -2 -1 0 1 2 3 4]
idx = np.max(np.where(A <= 0)[0])
print idx # 5
A[:idx] = 0
print A # [0 0 0 0 0 0 1 2 3 4]
Now I wanna do the same thing on each column of a 2D array :
A = np.arange(10) - 5
A[2] = 2
A2 = np.tile(A, 3).reshape((3, 10)) - np.array([0, 2, -1]).reshape((3, 1))
print A2
# [[-5 -4 2 -2 -1 0 1 2 3 4]
# [-7 -6 0 -4 -3 -2 -1 0 1 2]
# [-4 -3 3 -1 0 1 2 3 4 5]]
And I would like to obtain :
print A2
# [[0 0 0 0 0 0 1 2 3 4]
# [0 0 0 0 0 0 0 0 1 2]
# [0 0 0 0 0 1 2 3 4 5]]
but I can't manage to figure out how to translate the max/where statement to the this 2d array...

You already have good answers, but I wanted to propose a potentially quicker variation using the function np.maximum.accumulate. Since your method for a 1D array uses max/where, you may also find this approach quite intuitive. (Edit: quicker Cython implementation added below).
The overall approach is very similar to the others; the mask is created with:
np.maximum.accumulate((A2 < 0)[:, ::-1], axis=1)[:, ::-1]
This line of code does the following:
(A2 < 0) creates a Boolean array, indicating whether a value is negative or not. The index [:, ::-1] flips this left-to-right.
np.maximum.accumulate is used to return the cumulative maximum along each row (i.e. axis=1). For example [False, True, False] would become [False, True, True].
The final indexing operation [:, ::-1] flips this new Boolean array left-to-right.
Then all that's left to do is to use the Boolean array as a mask to set the True values to zero.
Borrowing the timing methodology and two functions from #Divakar's answer, here are the benchmarks for my proposed method:
# method using np.maximum.accumulate
def accumulate_based(A2):
A2[np.maximum.accumulate((A2 < 0)[:, ::-1], axis=1)[:, ::-1]] = 0
return A2
# large sample array
A2 = np.random.randint(-4, 10, size=(100000, 100))
A2c = A2.copy()
A2c2 = A2.copy()
The timings are:
In [47]: %timeit broadcasting_based(A2)
10 loops, best of 3: 61.7 ms per loop
In [48]: %timeit cumsum_based(A2c)
10 loops, best of 3: 127 ms per loop
In [49]: %timeit accumulate_based(A2c2) # quickest
10 loops, best of 3: 43.2 ms per loop
So using np.maximum.accumulate can be as much as 30% faster than the next fastest solution for arrays of this size and shape.
As #tom10 points out, each NumPy operation processes arrays in their entirety, which can be inefficient when multiple operations are needed to get a result. An iterative approach which works through the array just once may fare better.
Below is a naive function written in Cython which could more than twice as fast as a pure NumPy approach.
This function may be able to be sped up further using memory views.
cimport cython
import numpy as np
cimport numpy as np
#cython.boundscheck(False)
#cython.wraparound(False)
#cython.nonecheck(False)
def cython_based(np.ndarray[long, ndim=2, mode="c"] array):
cdef int rows, cols, i, j, seen_neg
rows = array.shape[0]
cols = array.shape[1]
for i in range(rows):
seen_neg = 0
for j in range(cols-1, -1, -1):
if seen_neg or array[i, j] < 0:
seen_neg = 1
array[i, j] = 0
return array
This function works backwards through each row and starts setting values to zero once it has seen a negative value.
Testing it works:
A2 = np.random.randint(-4, 10, size=(100000, 100))
A2c = A2.copy()
np.array_equal(accumulate_based(A2), cython_based(A2c))
# True
Comparing the performance of the function:
In [52]: %timeit accumulate_based(A2)
10 loops, best of 3: 49.8 ms per loop
In [53]: %timeit cython_based(A2c)
100 loops, best of 3: 18.6 ms per loop

Assuming that you are looking to set all elements for each row until the last negative element to be set to zero (as per the expected output listed in the question for a sample case), two approaches could be suggested here.
Approach #1
This one is based on np.cumsum to generate a mask of elements to be set to zeros as listed next -
# Get boolean mask with TRUEs for each row starting at the first element and
# ending at the last negative element
mask = (np.cumsum(A2[:,::-1]<0,1)>0)[:,::-1]
# Use mask to set all such al TRUEs to zeros as per the expected output in OP
A2[mask] = 0
Sample run -
In [280]: A2 = np.random.randint(-4,10,(6,7)) # Random input 2D array
In [281]: A2
Out[281]:
array([[-2, 9, 8, -3, 2, 0, 5],
[-1, 9, 5, 1, -3, -3, -2],
[ 3, -3, 3, 5, 5, 2, 9],
[ 4, 6, -1, 6, 1, 2, 2],
[ 4, 4, 6, -3, 7, -3, -3],
[ 0, 2, -2, -3, 9, 4, 3]])
In [282]: A2[(np.cumsum(A2[:,::-1]<0,1)>0)[:,::-1]] = 0 # Use mask to set zeros
In [283]: A2
Out[283]:
array([[0, 0, 0, 0, 2, 0, 5],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 3, 5, 5, 2, 9],
[0, 0, 0, 6, 1, 2, 2],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 9, 4, 3]])
Approach #2
This one starts with the idea of finding the last negative element indices from #tom10's answer and develops into a mask finding method using broadcasting to get us the desired output, similar to approach #1.
# Find last negative index for each row
last_idx = A2.shape[1] - 1 - np.argmax(A2[:,::-1]<0, axis=1)
# Find the invalid indices (rows with no negative indices)
invalid_idx = A2[np.arange(A2.shape[0]),last_idx]>=0
# Set the indices for invalid ones to "-1"
last_idx[invalid_idx] = -1
# Boolean mask with each row starting with TRUE as the first element
# and ending at the last negative element
mask = np.arange(A2.shape[1]) < (last_idx[:,None] + 1)
# Set masked elements to zeros, for the desired output
A2[mask] = 0
Runtime tests -
Function defintions:
def broadcasting_based(A2):
last_idx = A2.shape[1] - 1 - np.argmax(A2[:,::-1]<0, axis=1)
last_idx[A2[np.arange(A2.shape[0]),last_idx]>=0] = -1
A2[np.arange(A2.shape[1]) < (last_idx[:,None] + 1)] = 0
return A2
def cumsum_based(A2):
A2[(np.cumsum(A2[:,::-1]<0,1)>0)[:,::-1]] = 0
return A2
Runtimes:
In [379]: A2 = np.random.randint(-4,10,(100000,100))
...: A2c = A2.copy()
...:
In [380]: %timeit broadcasting_based(A2)
10 loops, best of 3: 106 ms per loop
In [381]: %timeit cumsum_based(A2c)
1 loops, best of 3: 167 ms per loop
Verify results -
In [384]: A2 = np.random.randint(-4,10,(100000,100))
...: A2c = A2.copy()
...:
In [385]: np.array_equal(broadcasting_based(A2),cumsum_based(A2c))
Out[385]: True

Finding the first is usually easier and faster than finding the last, so here I reverse the array and then find the first negative (using the OP's version of A2):
im = A2.shape[1] - 1 - np.argmax(A2[:,::-1]<0, axis=1)
# [4 6 3] # which are the indices of the last negative in A2
Also, though, note that if you have large arrays with many negative numbers, it might actually be faster to use a non-numpy approach so you can short circuit the search. That is, numpy will do the calculation on the entire array, so if you have 10000 elements in a row but typically will hit a negative number in the first 10 elements (of a reverse search), a pure Python approach might end up being faster.
Overall, iterating the rows might be faster for subsequent operations as well. For example, if your next step is multiplication, it could be faster to just multiply the slices at the ends that are non-zeros, or maybe find that longest non-zero section and just deal with the truncated array.
This basically comes down to number of negatives per row. If you have 1000 negatives per row you'll on average have non-zeros segments that are 1/1000th of your full row length, so you could get a 1000x speed-up by just looking at the ends. The short example given in the question is great for understanding and answering the basic question, but I wouldn't take timing tests too seriously when your end application is a very different use case; especially since your fractional time savings by using iteration improves in proportion to array size (assuming a constant ratio and random distribution of negative numbers).

You can access individual rows:
A2[0] == array([-5, -4, 2, -2, -1, 0, 1, 2, 3, 4])