Given a 2D numpy array, I want to construct an array out of the column indices of the maximum value of each row. So far, arr.argmax(1) works well. However, for my specific case, for some rows, 2 or more columns may contain the maximum value. In that case, I want to select a column index randomly (not the first index as it is the case with .argmax(1)).
For example, for the following arr:
arr = np.array([
[0, 1, 0],
[1, 1, 0],
[2, 1, 3],
[3, 2, 2]
])
there can be two possible outcomes: array([1, 0, 2, 0]) and array([1, 1, 2, 0]) each chosen with 1/2 probability.
I have code that returns the expected output using a list comprehension:
idx = np.arange(arr.shape[1])
ans = [np.random.choice(idx[ix]) for ix in arr == arr.max(1, keepdims=True)]
but I'm looking for an optimized numpy solution. In other words, how do I replace the list comprehension with numpy methods to make the code feasible for bigger arrays?
Use scipy.stats.rankdata and apply_along_axis as follows.
import numpy as np
from scipy.stats import rankdata
ranks = rankdata(-arr, axis = 1, method = "min")
func = lambda x: np.random.choice(np.where(x==1)[0])
idx = np.apply_along_axis(func, 1, ranks)
print(idx)
It returns [1 0 2 0] or [1 1 2 0].
The main idea is rankdata calculates ranks of every value in each row, and the maximum value will have 1. func randomly choices one of index whose corresponding value is 1. Finally, apply_along_axis applies the func to every row of arr.
After some advice I got offline, it turns out that randomization of maximum values are possible when we multiply the boolean array that flags row-wise maximum values by a random array of the same shape. Then what remains is a simple argmax(1) call.
# boolean array that flags maximum values of each row
mxs = arr == arr.max(1, keepdims=True)
# random array where non-maximum values are zero and maximum values are random values
random_arr = np.random.rand(*arr.shape) * mxs
# row-wise maximum of the auxiliary array
ans = random_arr.argmax(1)
A timeit test shows that for data of shape (507_563, 12), this code runs in ~172 ms on my machine while the loop in the question runs for 11 sec, so this is about 63x faster.
I have two lists in python, say A and B.
List A is a list of lists with some integer indexes, for example [[2,3,1,3,2,3,3], [4,2,1,4],[5,4,3,3,3,4,]...] and so on.
List B has the same structure but instead of integers has numpy arrays.
[[array([0, 0]), array([0, 0]), array([0, 1]) ...][[array([0, 0]), array([0, 1])...]...]
These lists are correlated, so each numpy array corresponds to a integer, in other words, the sublists of A and the sublists of B has the same size. For example
[[2,3,3,3,2,4]...]
[[array([0, 0]), array([0, 1]), array([0, 1]), array([0, 1]), array([0, 0]), array([1, 0])]...]
The first integer in the sublist of A, "2" is linked to the first numpy array in the sublist of B.
As you can see there are repeated integers, and in consequence repeated numpy arrays. I want to recover the unique indexes without repeating, and thus their corresponding arrays.
Taking the example above the return of the procedure should be something like this:
[[2,3,4]...]
[[array([0, 0]), array([0, 1]), array([1, 0])]...]
How can I recover the unique elements from list A with their corresponded numpy array in list b?
My first attempt used the numpy.unique function so i can recover list A efficiently, but then I lose the information to recover the information from B. The line was
A = np.array([np.unique(a) for a in A])
TO RECAP
I have the following
import numpy as np
A = [] # PUT A REAL (short) LIST HERE
B = [] # PUT A REAL (short) LIST HERE
uniqueA = np.array([np.unique(a) for a in A])
print uniqueA #prints what I want/dont want
expectA = [1,4] # put what you would expect to get back
#ask additional questions here
First, get unique indices of A, then take them from A and B
import numpy as np
A = [[2,3,1,3,2,3,3], [4,2,1,4]]
B = [[np.zeros(2)]*len(A[0]), [np.zeros(2)]*len(A[1])]
indices = np.array([np.unique(a, True)[1] for a in A])
A = np.array([np.array(arr)[index] for arr, index in zip(A, indices)])
B = np.array([np.array(arr)[index] for arr, index in zip(B, indices)])
I am trying to use array slicing to reverse part of a NumPy array. If my array is, for example,
a = np.array([1,2,3,4,5,6])
then I can get a slice b
b = a[::-1]
Which is a view on the original array. What I would like is a view that is partially reversed, for example
1,4,3,2,5,6
I have encountered performance problems with NumPy if you don't play along exactly with how it is designed, so I would like to avoid "fancy" indexing if it is possible.
If you don't like the off by one indices
>>> a = np.array([1,2,3,4,5,6])
>>> a[1:4] = a[1:4][::-1]
>>> a
array([1, 4, 3, 2, 5, 6])
>>> a = np.array([1,2,3,4,5,6])
>>> a[1:4] = a[3:0:-1]
>>> a
array([1, 4, 3, 2, 5, 6])
You can use the permutation matrices (that's the numpiest way to partially reverse an array).
a = np.array([1,2,3,4,5,6])
new_order_for_index = [1,4,3,2,5,6] # Careful: index from 1 to n !
# Permutation matrix
m = np.zeros( (len(a),len(a)) )
for index , new_index in enumerate(new_order_for_index ):
m[index ,new_index -1] = 1
print np.dot(m,a)
# np.array([1,4,3,2,5,6])
I know there is a method for a Python list to return the first index of something:
>>> xs = [1, 2, 3]
>>> xs.index(2)
1
Is there something like that for NumPy arrays?
Yes, given an array, array, and a value, item to search for, you can use np.where as:
itemindex = numpy.where(array == item)
The result is a tuple with first all the row indices, then all the column indices.
For example, if an array is two dimensions and it contained your item at two locations then
array[itemindex[0][0]][itemindex[1][0]]
would be equal to your item and so would be:
array[itemindex[0][1]][itemindex[1][1]]
If you need the index of the first occurrence of only one value, you can use nonzero (or where, which amounts to the same thing in this case):
>>> t = array([1, 1, 1, 2, 2, 3, 8, 3, 8, 8])
>>> nonzero(t == 8)
(array([6, 8, 9]),)
>>> nonzero(t == 8)[0][0]
6
If you need the first index of each of many values, you could obviously do the same as above repeatedly, but there is a trick that may be faster. The following finds the indices of the first element of each subsequence:
>>> nonzero(r_[1, diff(t)[:-1]])
(array([0, 3, 5, 6, 7, 8]),)
Notice that it finds the beginning of both subsequence of 3s and both subsequences of 8s:
[1, 1, 1, 2, 2, 3, 8, 3, 8, 8]
So it's slightly different than finding the first occurrence of each value. In your program, you may be able to work with a sorted version of t to get what you want:
>>> st = sorted(t)
>>> nonzero(r_[1, diff(st)[:-1]])
(array([0, 3, 5, 7]),)
You can also convert a NumPy array to list in the air and get its index. For example,
l = [1,2,3,4,5] # Python list
a = numpy.array(l) # NumPy array
i = a.tolist().index(2) # i will return index of 2
print i
It will print 1.
Just to add a very performant and handy numba alternative based on np.ndenumerate to find the first index:
from numba import njit
import numpy as np
#njit
def index(array, item):
for idx, val in np.ndenumerate(array):
if val == item:
return idx
# If no item was found return None, other return types might be a problem due to
# numbas type inference.
This is pretty fast and deals naturally with multidimensional arrays:
>>> arr1 = np.ones((100, 100, 100))
>>> arr1[2, 2, 2] = 2
>>> index(arr1, 2)
(2, 2, 2)
>>> arr2 = np.ones(20)
>>> arr2[5] = 2
>>> index(arr2, 2)
(5,)
This can be much faster (because it's short-circuiting the operation) than any approach using np.where or np.nonzero.
However np.argwhere could also deal gracefully with multidimensional arrays (you would need to manually cast it to a tuple and it's not short-circuited) but it would fail if no match is found:
>>> tuple(np.argwhere(arr1 == 2)[0])
(2, 2, 2)
>>> tuple(np.argwhere(arr2 == 2)[0])
(5,)
l.index(x) returns the smallest i such that i is the index of the first occurrence of x in the list.
One can safely assume that the index() function in Python is implemented so that it stops after finding the first match, and this results in an optimal average performance.
For finding an element stopping after the first match in a NumPy array use an iterator (ndenumerate).
In [67]: l=range(100)
In [68]: l.index(2)
Out[68]: 2
NumPy array:
In [69]: a = np.arange(100)
In [70]: next((idx for idx, val in np.ndenumerate(a) if val==2))
Out[70]: (2L,)
Note that both methods index() and next return an error if the element is not found. With next, one can use a second argument to return a special value in case the element is not found, e.g.
In [77]: next((idx for idx, val in np.ndenumerate(a) if val==400),None)
There are other functions in NumPy (argmax, where, and nonzero) that can be used to find an element in an array, but they all have the drawback of going through the whole array looking for all occurrences, thus not being optimized for finding the first element. Note also that where and nonzero return arrays, so you need to select the first element to get the index.
In [71]: np.argmax(a==2)
Out[71]: 2
In [72]: np.where(a==2)
Out[72]: (array([2], dtype=int64),)
In [73]: np.nonzero(a==2)
Out[73]: (array([2], dtype=int64),)
Time comparison
Just checking that for large arrays the solution using an iterator is faster when the searched item is at the beginning of the array (using %timeit in the IPython shell):
In [285]: a = np.arange(100000)
In [286]: %timeit next((idx for idx, val in np.ndenumerate(a) if val==0))
100000 loops, best of 3: 17.6 µs per loop
In [287]: %timeit np.argmax(a==0)
1000 loops, best of 3: 254 µs per loop
In [288]: %timeit np.where(a==0)[0][0]
1000 loops, best of 3: 314 µs per loop
This is an open NumPy GitHub issue.
See also: Numpy: find first index of value fast
If you're going to use this as an index into something else, you can use boolean indices if the arrays are broadcastable; you don't need explicit indices. The absolute simplest way to do this is to simply index based on a truth value.
other_array[first_array == item]
Any boolean operation works:
a = numpy.arange(100)
other_array[first_array > 50]
The nonzero method takes booleans, too:
index = numpy.nonzero(first_array == item)[0][0]
The two zeros are for the tuple of indices (assuming first_array is 1D) and then the first item in the array of indices.
For one-dimensional sorted arrays, it would be much more simpler and efficient O(log(n)) to use numpy.searchsorted which returns a NumPy integer (position). For example,
arr = np.array([1, 1, 1, 2, 3, 3, 4])
i = np.searchsorted(arr, 3)
Just make sure the array is already sorted
Also check if returned index i actually contains the searched element, since searchsorted's main objective is to find indices where elements should be inserted to maintain order.
if arr[i] == 3:
print("present")
else:
print("not present")
For 1D arrays, I'd recommend np.flatnonzero(array == value)[0], which is equivalent to both np.nonzero(array == value)[0][0] and np.where(array == value)[0][0] but avoids the ugliness of unboxing a 1-element tuple.
To index on any criteria, you can so something like the following:
In [1]: from numpy import *
In [2]: x = arange(125).reshape((5,5,5))
In [3]: y = indices(x.shape)
In [4]: locs = y[:,x >= 120] # put whatever you want in place of x >= 120
In [5]: pts = hsplit(locs, len(locs[0]))
In [6]: for pt in pts:
.....: print(', '.join(str(p[0]) for p in pt))
4, 4, 0
4, 4, 1
4, 4, 2
4, 4, 3
4, 4, 4
And here's a quick function to do what list.index() does, except doesn't raise an exception if it's not found. Beware -- this is probably very slow on large arrays. You can probably monkey patch this on to arrays if you'd rather use it as a method.
def ndindex(ndarray, item):
if len(ndarray.shape) == 1:
try:
return [ndarray.tolist().index(item)]
except:
pass
else:
for i, subarray in enumerate(ndarray):
try:
return [i] + ndindex(subarray, item)
except:
pass
In [1]: ndindex(x, 103)
Out[1]: [4, 0, 3]
An alternative to selecting the first element from np.where() is to use a generator expression together with enumerate, such as:
>>> import numpy as np
>>> x = np.arange(100) # x = array([0, 1, 2, 3, ... 99])
>>> next(i for i, x_i in enumerate(x) if x_i == 2)
2
For a two dimensional array one would do:
>>> x = np.arange(100).reshape(10,10) # x = array([[0, 1, 2,... 9], [10,..19],])
>>> next((i,j) for i, x_i in enumerate(x)
... for j, x_ij in enumerate(x_i) if x_ij == 2)
(0, 2)
The advantage of this approach is that it stops checking the elements of the array after the first match is found, whereas np.where checks all elements for a match. A generator expression would be faster if there's match early in the array.
There are lots of operations in NumPy that could perhaps be put together to accomplish this. This will return indices of elements equal to item:
numpy.nonzero(array - item)
You could then take the first elements of the lists to get a single element.
Comparison of 8 methods
TL;DR:
(Note: applicable to 1d arrays under 100M elements.)
For maximum performance use index_of__v5 (numba + numpy.enumerate + for loop; see the code below).
If numba is not available:
Use index_of__v7 (for loop + enumerate) if the target value is expected to be found within the first 100k elements.
Else use index_of__v2/v3/v4 (numpy.argmax or numpy.flatnonzero based).
Powered by perfplot
import numpy as np
from numba import njit
# Based on: numpy.argmax()
# Proposed by: John Haberstroh (https://stackoverflow.com/a/67497472/7204581)
def index_of__v1(arr: np.array, v):
is_v = (arr == v)
return is_v.argmax() if is_v.any() else -1
# Based on: numpy.argmax()
def index_of__v2(arr: np.array, v):
return (arr == v).argmax() if v in arr else -1
# Based on: numpy.flatnonzero()
# Proposed by: 1'' (https://stackoverflow.com/a/42049655/7204581)
def index_of__v3(arr: np.array, v):
idxs = np.flatnonzero(arr == v)
return idxs[0] if len(idxs) > 0 else -1
# Based on: numpy.argmax()
def index_of__v4(arr: np.array, v):
return np.r_[False, (arr == v)].argmax() - 1
# Based on: numba, for loop
# Proposed by: MSeifert (https://stackoverflow.com/a/41578614/7204581)
#njit
def index_of__v5(arr: np.array, v):
for idx, val in np.ndenumerate(arr):
if val == v:
return idx[0]
return -1
# Based on: numpy.ndenumerate(), for loop
def index_of__v6(arr: np.array, v):
return next((idx[0] for idx, val in np.ndenumerate(arr) if val == v), -1)
# Based on: enumerate(), for loop
# Proposed by: Noyer282 (https://stackoverflow.com/a/40426159/7204581)
def index_of__v7(arr: np.array, v):
return next((idx for idx, val in enumerate(arr) if val == v), -1)
# Based on: list.index()
# Proposed by: Hima (https://stackoverflow.com/a/23994923/7204581)
def index_of__v8(arr: np.array, v):
l = list(arr)
try:
return l.index(v)
except ValueError:
return -1
Go to Colab
The numpy_indexed package (disclaimer, I am its author) contains a vectorized equivalent of list.index for numpy.ndarray; that is:
sequence_of_arrays = [[0, 1], [1, 2], [-5, 0]]
arrays_to_query = [[-5, 0], [1, 0]]
import numpy_indexed as npi
idx = npi.indices(sequence_of_arrays, arrays_to_query, missing=-1)
print(idx) # [2, -1]
This solution has vectorized performance, generalizes to ndarrays, and has various ways of dealing with missing values.
There is a fairly idiomatic and vectorized way to do this built into numpy. It uses a quirk of the np.argmax() function to accomplish this -- if many values match, it returns the index of the first match. The trick is that for booleans, there will only ever be two values: True (1) and False (0). Therefore, the returned index will be that of the first True.
For the simple example provided, you can see it work with the following
>>> np.argmax(np.array([1,2,3]) == 2)
1
A great example is computing buckets, e.g. for categorizing. Let's say you have an array of cut points, and you want the "bucket" that corresponds to each element of your array. The algorithm is to compute the first index of cuts where x < cuts (after padding cuts with np.Infitnity). I could use broadcasting to broadcast the comparisons, then apply argmax along the cuts-broadcasted axis.
>>> cuts = np.array([10, 50, 100])
>>> cuts_pad = np.array([*cuts, np.Infinity])
>>> x = np.array([7, 11, 80, 443])
>>> bins = np.argmax( x[:, np.newaxis] < cuts_pad[np.newaxis, :], axis = 1)
>>> print(bins)
[0, 1, 2, 3]
As expected, each value from x falls into one of the sequential bins, with well-defined and easy to specify edge case behavior.
Another option not previously mentioned is the bisect module, which also works on lists, but requires a pre-sorted list/array:
import bisect
import numpy as np
z = np.array([104,113,120,122,126,138])
bisect.bisect_left(z, 122)
yields
3
bisect also returns a result when the number you're looking for doesn't exist in the array, so that the number can be inserted in the correct place.
Note: this is for python 2.7 version
You can use a lambda function to deal with the problem, and it works both on NumPy array and list.
your_list = [11, 22, 23, 44, 55]
result = filter(lambda x:your_list[x]>30, range(len(your_list)))
#result: [3, 4]
import numpy as np
your_numpy_array = np.array([11, 22, 23, 44, 55])
result = filter(lambda x:your_numpy_array [x]>30, range(len(your_list)))
#result: [3, 4]
And you can use
result[0]
to get the first index of the filtered elements.
For python 3.6, use
list(result)
instead of
result
Use ndindex
Sample array
arr = np.array([[1,4],
[2,3]])
print(arr)
...[[1,4],
[2,3]]
create an empty list to store the index and the element tuples
index_elements = []
for i in np.ndindex(arr.shape):
index_elements.append((arr[i],i))
convert the list of tuples into dictionary
index_elements = dict(index_elements)
The keys are the elements and the values are their
indices - use keys to access the index
index_elements[4]
output
... (0,1)
For my use case, I could not sort the array ahead of time because the order of the elements is important. This is my all-NumPy implementation:
import numpy as np
# The array in question
arr = np.array([1,2,1,2,1,5,5,3,5,9])
# Find all of the present values
vals=np.unique(arr)
# Make all indices up-to and including the desired index positive
cum_sum=np.cumsum(arr==vals.reshape(-1,1),axis=1)
# Add zeros to account for the n-1 shape of diff and the all-positive array of the first index
bl_mask=np.concatenate([np.zeros((cum_sum.shape[0],1)),cum_sum],axis=1)>=1
# The desired indices
idx=np.where(np.diff(bl_mask))[1]
# Show results
print(list(zip(vals,idx)))
>>> [(1, 0), (2, 1), (3, 7), (5, 5), (9, 9)]
I believe it accounts for unsorted arrays with duplicate values.
Found another solution with loops:
new_array_of_indicies = []
for i in range(len(some_array)):
if some_array[i] == some_value:
new_array_of_indicies.append(i)
index_lst_form_numpy = pd.DataFrame(df).reset_index()["index"].tolist()