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?
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.
To index on any criteria, you can so something like the following:
In : from numpy import * In : x = arange(125).reshape((5,5,5)) In : y = indices(x.shape) In : locs = y[:,x >= 120] # put whatever you want in place of x >= 120 In : pts = hsplit(locs, len(locs)) In : for pt in pts: .....: print(', '.join(str(p) 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 : ndindex(x, 103) Out: [4, 0, 3]
Yes, given an array,
array, and a value,
item to search for, you can use
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
would be equal to your item and so would be:
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)
The two zeros are for the tuple of indices (assuming first_array is 1D) and then the first item in the array of indices.
If you need the index of the first occurrence of only one value, you can use
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) 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.
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.
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 = 2 >>> index(arr2, 2) (5,)
This can be much faster (because it’s short-circuiting the operation) than any approach using
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)) (2, 2, 2) >>> tuple(np.argwhere(arr2 == 2)) (5,)
For 1D arrays, I’d recommend
np.flatnonzero(array == value), which is equivalent to both
np.nonzero(array == value) and
np.where(array == value) but avoids the ugliness of unboxing a 1-element tuple.
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 : l=range(100) In : l.index(2) Out: 2
In : a = np.arange(100) In : next((idx for idx, val in np.ndenumerate(a) if val==2)) Out: (2L,)
Note that both methods
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 : next((idx for idx, val in np.ndenumerate(a) if val==400),None)
There are other functions in NumPy (
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
nonzero return arrays, so you need to select the first element to get the index.
In : np.argmax(a==2) Out: 2 In : np.where(a==2) Out: (array(, dtype=int64),) In : np.nonzero(a==2) Out: (array(, dtype=int64),)
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 : a = np.arange(100000) In : %timeit next((idx for idx, val in np.ndenumerate(a) if val==0)) 100000 loops, best of 3: 17.6 µs per loop In : %timeit np.argmax(a==0) 1000 loops, best of 3: 254 µs per loop In : %timeit np.where(a==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
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.
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
to get the first index of the filtered elements.
For python 3.6, use
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")
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
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)
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
x < cuts (after padding
np.Infitnity). I could use broadcasting to broadcast the comparisons, then apply argmax along the
>>> 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)
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: applicable to 1d arrays under 100M elements.)
forloop; see the code below).
numbais not available:
enumerate) if the target value is expected to be found within the first 100k elements.
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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 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 return -1 # Based on: numpy.ndenumerate(), for loop def index_of__v6(arr: np.array, v): return next((idx 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
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,1)),cum_sum],axis=1)>=1 # The desired indices idx=np.where(np.diff(bl_mask)) # 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.
index_lst_form_numpy = pd.DataFrame(df).reset_index()["index"].tolist()