In a given numpy array X:
X = array([1,2,3,4,5,6,7,8,9,10])
I would like to replace indices (2, 3) and (7, 8) with a single element -1 respectively, like:
X = array([1,2,-1,5,6,7,-1,10])
In other words, I replaced values at indices (2, 3) and (7,8) of the original array with a singular value.
Question is: Is there a numpy-ish way (i.e. without for loops and usage of python lists) around it? Thanks.
Note: This is NOT equivalent of replacing a single element in-place with another. Its about replacing multiple values with a "singular" value. Thanks.
A solution using numpy.delete, similar to #pault, but more efficient as it uses pure numpy indexing. However, because of this efficient indexing, it means that you cannot pass jagged arrays as indices
Setup
a = np.array([1,2,3,4,5,6,7,8,9,10])
idx = np.stack([[2, 3], [7, 8]])
a[idx] = -1
np.delete(a, idx[:, 1:])
array([ 1, 2, -1, 5, 6, 7, -1, 10])
I'm not sure if this can be done in one step, but here's a way using np.delete:
import numpy as np
from operator import itemgetter
X = np.array(range(1,11))
to_replace = [[2,3], [7,8]]
X[list(map(itemgetter(0), to_replace))] = -1
X = np.delete(X, list(map(lambda x: x[1:], to_replace)))
print(X)
#[ 1 2 -1 5 6 7 -1 10]
First we replace the first element of each pair with -1. Then we delete the remaining elements.
Try np.put:
np.put(X, [2,3,7,8], [-1,0]) # `0` can be changed to anything that's not in the array
print(X[X!=0]) # whatever You put as an number in `put`
So basically use put to do the values for the indexes, then drop the zero-values.
Or as #khan says, can do something that's out of range:
np.put(X, [2,3,7,8], [-1,np.max(X)+1])
print(X[X!=X.max()])
All Output:
[ 1 2 -1 5 6 7 -1 10]
Say that I have 4 numpy arrays
[1,2,3]
[2,3,1]
[3,2,1]
[1,3,2]
In this case, I've determined [1,2,3] is the "minimum array" for my purposes, as it is one of two arrays with lowest value at index 0, and of those two arrays it has the the lowest index 1. If there were more arrays with similar values, I would need to compare the next index values, and so on.
How can I extract the array [1,2,3] in that same order from the pile?
How can I extend that to x arrays of size n?
Thanks
Using the python non-numpy .sort() or sorted() on a list of lists (not numpy arrays) automatically does this e.g.
a = [[1,2,3],[2,3,1],[3,2,1],[1,3,2]]
a.sort()
gives
[[1,2,3],[1,3,2],[2,3,1],[3,2,1]]
The numpy sort seems to only sort the subarrays recursively so it seems the best way would be to convert it to a python list first. Assuming you have an array of arrays you want to pick the minimum of you could get the minimum as
sorted(a.tolist())[0]
As someone pointed out you could also do min(a.tolist()) which uses the same type of comparisons as sort, and would be faster for large arrays (linear vs n log n asymptotic run time).
Here's an idea using numpy:
import numpy
a = numpy.array([[1,2,3],[2,3,1],[3,2,1],[1,3,2]])
col = 0
while a.shape[0] > 1:
b = numpy.argmin(a[:,col:], axis=1)
a = a[b == numpy.min(b)]
col += 1
print a
This checks column by column until only one row is left.
numpy's lexsort is close to what you want. It sorts on the last key first, but that's easy to get around:
>>> a = np.array([[1,2,3],[2,3,1],[3,2,1],[1,3,2]])
>>> order = np.lexsort(a[:, ::-1].T)
>>> order
array([0, 3, 1, 2])
>>> a[order]
array([[1, 2, 3],
[1, 3, 2],
[2, 3, 1],
[3, 2, 1]])
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()