Simple question here:
I'm trying to get an array that alternates values (1, -1, 1, -1.....) for a given length. np.repeat just gives me (1, 1, 1, 1,-1, -1,-1, -1). Thoughts?
I like #Benjamin's solution. An alternative though is:
import numpy as np
a = np.empty((15,))
a[::2] = 1
a[1::2] = -1
This also allows for odd-length lists.
EDIT: Also just to note speeds, for a array of 10000 elements
import numpy as np
from timeit import Timer
if __name__ == '__main__':
setupstr="""
import numpy as np
N = 10000
"""
method1="""
a = np.empty((N,),int)
a[::2] = 1
a[1::2] = -1
"""
method2="""
a = np.tile([1,-1],N)
"""
method3="""
a = np.array([1,-1]*N)
"""
method4="""
a = np.array(list(itertools.islice(itertools.cycle((1,-1)), N)))
"""
nl = 1000
t1 = Timer(method1, setupstr).timeit(nl)
t2 = Timer(method2, setupstr).timeit(nl)
t3 = Timer(method3, setupstr).timeit(nl)
t4 = Timer(method4, setupstr).timeit(nl)
print 'method1', t1
print 'method2', t2
print 'method3', t3
print 'method4', t4
Results in timings of:
method1 0.0130500793457
method2 0.114426136017
method3 4.30518102646
method4 2.84446692467
If N = 100, things start to even out but starting with the empty numpy arrays is still significantly faster (nl changed to 10000)
method1 0.05735206604
method2 0.323992013931
method3 0.556654930115
method4 0.46702003479
Numpy arrays are special awesome objects and should not be treated like python lists.
use resize():
In [38]: np.resize([1,-1], 10) # 10 is the length of result array
Out[38]: array([ 1, -1, 1, -1, 1, -1, 1, -1, 1, -1])
it can produce odd-length array:
In [39]: np.resize([1,-1], 11)
Out[39]: array([ 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1])
Use numpy.tile!
import numpy
a = numpy.tile([1,-1], 15)
use multiplication:
[1,-1] * n
If you want a memory efficient solution, try this:
def alternator(n):
for i in xrange(n):
if i % 2 == 0:
yield 1
else:
yield -1
Then you can iterate over the answers like so:
for i in alternator(n):
# do something with i
Maybe you're looking for itertools.cycle?
list_ = (1,-1,2,-2) # ,3,-3, ...
for n, item in enumerate(itertools.cycle(list_)):
if n==30:
break
print item
I'll just throw these out there because they could be more useful in some circumstances.
If you just want to alternate between positive and negative:
[(-1)**i for i in range(n)]
or for a more general solution
nums = [1, -1, 2]
[nums[i % len(nums)] for i in range(n)]
Related
If I have:
x = np.array(([1,4], [2,5], [2,6], [3,4], [3,6], [3,7], [4,3], [4,5], [5,2]))
for item in range(3):
choice = random.choice(x)
How can I get the index number of the random choice taken from the array?
I tried:
indexNum = np.where(x == choice)
print(indexNum[0])
But it didn't work.
I want the output, for example, to be something like:
chosenIndices = [1 5 8]
Another possibility is using np.where and np.intersect1d. Here random choice without repetition.
x = np.array(([1,4], [2,5], [2,6], [3,4], [3,6], [3,7], [4,3], [4,5], [5,2]))
res=[]
cont = 0
while cont<3:
choice = random.choice(x)
ind = np.intersect1d(np.where(choice[0]==x[:,0]),np.where(choice[1]==x[:,1]))[0]
if ind not in res:
res.append(ind)
cont+=1
print (res)
# Output [8, 1, 5]
You can achieve this by converting the numpy array to list of tuples and then apply the index function.
This would work:
import random
import numpy as np
chosenIndices = []
x = np.array(([1,4], [2,5], [2,6], [3,4], [3,6], [3,7], [4,3], [4,5], [5,2]))
x = x.T
x = list(zip(x[0],x[1]))
item = 0
while len(chosenIndices)!=3:
choice = random.choice(x)
indexNum = x.index(choice)
if indexNum in chosenIndices: # if index already exist, then it will rerun that particular iteration again.
item-=1
else:
chosenIndices.append(indexNum)
print(chosenIndices) # Thus all different results.
Output:
[1, 3, 2]
To find all possible combinations that can be added to make given sum.
Combinations can be formed with multiple elements and also if any single element exists.
Input:
l1 = [9,1, 2, 7, 6, 1, 5]
target = 8
**Constraints**
1<=(len(l1))<=500000
1<=each_list_element<=1000
Output:
Format : {index:element}
{1:1, 5:1, 4:6} #Indices : 1,5,4 Elements : 1,1,6
{1:1, 2:2, 6:5}
{5:1, 2:2, 6:5}
{1:1, 3:7}
{5:1, 3:7}
{2:2, 4:6}
More Scenarios:
Input = [4,6,8,5,3]
target = 3
Output {4:3}
Input = [4,6,8,3,5,3]
target = 3
Output {5:3,3:3}
Input = [1,2,3,15]
target = 15
Output {3:15}
Below code covers for all above scenarios.
Scenarios to be handled, along with above.
Input =[1,6,7,1,3]
target=5
Output={0:1,3:1,4:3} , {0:1,0:1,4:3}, {3:1,3:1,4:3}
Input=[9,6,8,1,7]
target=5
Output={3:1,3:1,3:1,3:1,3:1}
As suggested by #Chris Doyle in previous question, will be using that code.
(How to find indices and combinations that adds upto given sum?)
Code:
from itertools import combinations
def find_sum_with_index(l1, target):
index_vals = [iv for iv in enumerate(l1) if iv[1] < target]
for r in range(1, len(index_vals) + 1):
for perm in combinations(index_vals, r):
if sum([p[1] for p in perm]) == target:
yield perm
l1 = [9, 1, 2, 7, 6, 1, 5]
target = 8
for match in find_sum_with_index(l1, target):
print(dict(match))
You can use dictionary comprehension
from itertools import combinations
l1 = [9,1, 2, 7, 6, 1, 5]
target = 8
for i in range(len(l1)):
for c in combinations(l1,i):
if sum(c) == target:
res = { i:x for i,x in enumerate(c)}
print(res)
I have a matrix (2d numpy ndarray, to be precise):
A = np.array([[4, 0, 0],
[1, 2, 3],
[0, 0, 5]])
And I want to roll each row of A independently, according to roll values in another array:
r = np.array([2, 0, -1])
That is, I want to do this:
print np.array([np.roll(row, x) for row,x in zip(A, r)])
[[0 0 4]
[1 2 3]
[0 5 0]]
Is there a way to do this efficiently? Perhaps using fancy indexing tricks?
Sure you can do it using advanced indexing, whether it is the fastest way probably depends on your array size (if your rows are large it may not be):
rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]
# Use always a negative shift, so that column_indices are valid.
# (could also use module operation)
r[r < 0] += A.shape[1]
column_indices = column_indices - r[:, np.newaxis]
result = A[rows, column_indices]
numpy.lib.stride_tricks.as_strided stricks (abbrev pun intended) again!
Speaking of fancy indexing tricks, there's the infamous - np.lib.stride_tricks.as_strided. The idea/trick would be to get a sliced portion starting from the first column until the second last one and concatenate at the end. This ensures that we can stride in the forward direction as needed to leverage np.lib.stride_tricks.as_strided and thus avoid the need of actually rolling back. That's the whole idea!
Now, in terms of actual implementation we would use scikit-image's view_as_windows to elegantly use np.lib.stride_tricks.as_strided under the hoods. Thus, the final implementation would be -
from skimage.util.shape import view_as_windows as viewW
def strided_indexing_roll(a, r):
# Concatenate with sliced to cover all rolls
a_ext = np.concatenate((a,a[:,:-1]),axis=1)
# Get sliding windows; use advanced-indexing to select appropriate ones
n = a.shape[1]
return viewW(a_ext,(1,n))[np.arange(len(r)), (n-r)%n,0]
Here's a sample run -
In [327]: A = np.array([[4, 0, 0],
...: [1, 2, 3],
...: [0, 0, 5]])
In [328]: r = np.array([2, 0, -1])
In [329]: strided_indexing_roll(A, r)
Out[329]:
array([[0, 0, 4],
[1, 2, 3],
[0, 5, 0]])
Benchmarking
# #seberg's solution
def advindexing_roll(A, r):
rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]
r[r < 0] += A.shape[1]
column_indices = column_indices - r[:,np.newaxis]
return A[rows, column_indices]
Let's do some benchmarking on an array with large number of rows and columns -
In [324]: np.random.seed(0)
...: a = np.random.rand(10000,1000)
...: r = np.random.randint(-1000,1000,(10000))
# #seberg's solution
In [325]: %timeit advindexing_roll(a, r)
10 loops, best of 3: 71.3 ms per loop
# Solution from this post
In [326]: %timeit strided_indexing_roll(a, r)
10 loops, best of 3: 44 ms per loop
In case you want more general solution (dealing with any shape and with any axis), I modified #seberg's solution:
def indep_roll(arr, shifts, axis=1):
"""Apply an independent roll for each dimensions of a single axis.
Parameters
----------
arr : np.ndarray
Array of any shape.
shifts : np.ndarray
How many shifting to use for each dimension. Shape: `(arr.shape[axis],)`.
axis : int
Axis along which elements are shifted.
"""
arr = np.swapaxes(arr,axis,-1)
all_idcs = np.ogrid[[slice(0,n) for n in arr.shape]]
# Convert to a positive shift
shifts[shifts < 0] += arr.shape[-1]
all_idcs[-1] = all_idcs[-1] - shifts[:, np.newaxis]
result = arr[tuple(all_idcs)]
arr = np.swapaxes(result,-1,axis)
return arr
I implement a pure numpy.lib.stride_tricks.as_strided solution as follows
from numpy.lib.stride_tricks import as_strided
def custom_roll(arr, r_tup):
m = np.asarray(r_tup)
arr_roll = arr[:, [*range(arr.shape[1]),*range(arr.shape[1]-1)]].copy() #need `copy`
strd_0, strd_1 = arr_roll.strides
n = arr.shape[1]
result = as_strided(arr_roll, (*arr.shape, n), (strd_0 ,strd_1, strd_1))
return result[np.arange(arr.shape[0]), (n-m)%n]
A = np.array([[4, 0, 0],
[1, 2, 3],
[0, 0, 5]])
r = np.array([2, 0, -1])
out = custom_roll(A, r)
Out[789]:
array([[0, 0, 4],
[1, 2, 3],
[0, 5, 0]])
By using a fast fourrier transform we can apply a transformation in the frequency domain and then use the inverse fast fourrier transform to obtain the row shift.
So this is a pure numpy solution that take only one line:
import numpy as np
from numpy.fft import fft, ifft
# The row shift function using the fast fourrier transform
# rshift(A,r) where A is a 2D array, r the row shift vector
def rshift(A,r):
return np.real(ifft(fft(A,axis=1)*np.exp(2*1j*np.pi/A.shape[1]*r[:,None]*np.r_[0:A.shape[1]][None,:]),axis=1).round())
This will apply a left shift, but we can simply negate the exponential exponant to turn the function into a right shift function:
ifft(fft(...)*np.exp(-2*1j...)
It can be used like that:
# Example:
A = np.array([[1,2,3,4],
[1,2,3,4],
[1,2,3,4]])
r = np.array([1,-1,3])
print(rshift(A,r))
Building on divakar's excellent answer, you can apply this logic to 3D array easily (which was the problematic that brought me here in the first place). Here's an example - basically flatten your data, roll it & reshape it after::
def applyroll_30(cube, threshold=25, offset=500):
flattened_cube = cube.copy().reshape(cube.shape[0]*cube.shape[1], cube.shape[2])
roll_matrix = calc_roll_matrix_flattened(flattened_cube, threshold, offset)
rolled_cube = strided_indexing_roll(flattened_cube, roll_matrix, cube_shape=cube.shape)
rolled_cube = triggered_cube.reshape(cube.shape[0], cube.shape[1], cube.shape[2])
return rolled_cube
def calc_roll_matrix_flattened(cube_flattened, threshold, offset):
""" Calculates the number of position along time axis we need to shift
elements in order to trig the data.
We return a 1D numpy array of shape (X*Y, time) elements
"""
# armax(...) finds the position in the cube (3d) where we are above threshold
roll_matrix = np.argmax(cube_flattened > threshold, axis=1) + offset
# ensure we don't have index out of bound
roll_matrix[roll_matrix>cube_flattened.shape[1]] = cube_flattened.shape[1]
return roll_matrix
def strided_indexing_roll(cube_flattened, roll_matrix_flattened, cube_shape):
# Concatenate with sliced to cover all rolls
# otherwise we shift in the wrong direction for my application
roll_matrix_flattened = -1 * roll_matrix_flattened
a_ext = np.concatenate((cube_flattened, cube_flattened[:, :-1]), axis=1)
# Get sliding windows; use advanced-indexing to select appropriate ones
n = cube_flattened.shape[1]
result = viewW(a_ext,(1,n))[np.arange(len(roll_matrix_flattened)), (n - roll_matrix_flattened) % n, 0]
result = result.reshape(cube_shape)
return result
Divakar's answer doesn't do justice to how much more efficient this is on large cube of data. I've timed it on a 400x400x2000 data formatted as int8. An equivalent for-loop does ~5.5seconds, Seberg's answer ~3.0seconds and strided_indexing.... ~0.5second.
I have a problem that involves taking a square matrix in list form and finding the absolute value of the difference between the primary diagonal and the secondary diagonal. For example, given the following list:
test_matrix = [[11, 2, 4], [4, 5, 6], [10, 8, -12]]
The correct solution returns
15
My solution which works is:
def diagonalDifference(test_matrix):
primary_diag = 0
secondary_diag = 0
for i, row in enumerate(test_matrix):
for j, val in enumerate(row):
if (i == j) and ((i + j) == len(test_matrix) - 1):
secondary_diag += val
primary_diag += val
elif i == j:
primary_diag += val
elif (i + j) == len(test_matrix) - 1:
print('secondary_diag:' + str(val))
secondary_diag += val
return abs(primary_diag - secondary_diag)
Which works fine but I was hoping if someone could tell me if there is a way to reproduce the solution using only one for loop. Any other improvements that can be made are also welcome.
Since you already have a core Python solution, let me offer a numpy-based solution that may be faster for large matrices:
import numpy as np
arr = np.array(test_matrix)
np.abs((np.diag(arr) - np.diag(np.fliplr(arr))).sum())
It is possible to solve the underlying problem in O(N), by simply iterating over possible row indices and using them to compute both row and column indices to index the (square) input data:
def diagonalDifference(test_matrix):
diff = 0
for i in range(len(test_matrix)):
diff += test_matrix[i][i]
diff -= test_matrix[i][-i-1]
return abs(diff)
test_matrix = [[11, 2, 4], [4, 5, 6], [10, 8, -12]]
print(diagonalDifference(test_matrix))
I am having list which as follows
input_list= [2, 3, 5, 2, 5, 1, 5]
I want to get all the indexes of maximum value. Need efficient solution. The output will be as follows.
output = [2,4,6] (The above list 5 is maximum value in a list)
I have tried by using below code
m = max(input_list)
output = [i for i, j in enumerate(a) if j == m]
I need to find any other optimum solution.
from collections import defaultdict
dic=defaultdict(list)
input_list=[]
for i in range(len(input_list)):
dic[input_list[i]]+=[i]
max_value = max(input_list)
Sol = dic[max_value]
You can use numpy (numpy arrays are very fast):
import numpy as np
input_list= np.array([2, 3, 5, 2, 5, 1, 5])
i, = np.where(input_list == np.max(input_list))
print(i)
Output:
[2 4 6]
Here's the approach which is described in comments. Even if you use some library, fundamentally you need to traverse at least once to solve this problem (considering input list is unsorted). So even lower bound for the algorithm would be Omega(size_of_list). If list is sorted we can leverage binary_search to solve the problem.
def max_indexes(l):
try:
assert l != []
max_element = l[0]
indexes = [0]
for index, element in enumerate(l[1:]):
if element > max_element:
max_element = element
indexes = [index + 1]
elif element == max_element:
indexes.append(index + 1)
return indexes
except AssertionError:
print ('input_list in empty')
Use a for loop for O(n) and iterating just once over the list resolution:
from itertools import islice
input_list= [2, 3, 5, 2, 5, 1, 5]
def max_indexes(l):
max_item = input_list[0]
indexes = [0]
for i, item in enumerate(islice(l, 1, None), 1):
if item < max_item:
continue
elif item > max_item:
max_item = item
indexes = [i]
elif item == max_item:
indexes.append(i)
return indexes
Here you have the live example
Think of it in this way, unless you iterate through the whole list once, which is O(n), n being the length of the list, you won't be able to compare the maximum with all values in the list, so the best you can do is O(n), which you already seems to be doing in your example.
So I am not sure you can do it faster than O(n) with the list approach.