NumPy 数组中元素的索引

Use np.where to get the indices where a given condition is True.

Examples:

For a 2D np.ndarray called a:

i, j = np.where(a == value) # when comparing arrays of integers


i, j = np.where(np.isclose(a, value)) # when comparing floating-point arrays

For a 1D array:

i, = np.where(a == value) # integers


i, = np.where(np.isclose(a, value)) # floating-point

Note that this also works for conditions like >=, <=, != and so forth...

You can also create a subclass of np.ndarray with an index() method:

class myarray(np.ndarray):
def __new__(cls, *args, **kwargs):
return np.array(*args, **kwargs).view(myarray)
def index(self, value):
return np.where(self == value)

Testing:

a = myarray([1,2,3,4,4,4,5,6,4,4,4])
a.index(4)
#(array([ 3,  4,  5,  8,  9, 10]),)

I'm torn between these two ways of implementing an index of a NumPy array:

idx = list(classes).index(var)
idx = np.where(classes == var)

Both take the same number of characters, but the first method returns an int instead of a numpy.ndarray.

You can convert a numpy array to list and get its index .

for example:

tmp = [1,2,3,4,5] #python list
a = numpy.array(tmp) #numpy array
i = list(a).index(2) # i will return index of 2, which is 1

this is just what you wanted.

You can use the function numpy.nonzero(), or the nonzero() method of an array

import numpy as np


A = np.array([[2,4],
[6,2]])
index= np.nonzero(A>1)
OR
(A>1).nonzero()

Output:

(array([0, 1]), array([1, 0]))

First array in output depicts the row index and second array depicts the corresponding column index.

This problem can be solved efficiently using the numpy_indexed library (disclaimer: I am its author); which was created to address problems of this type. npi.indices can be viewed as an n-dimensional generalisation of list.index. It will act on nd-arrays (along a specified axis); and also will look up multiple entries in a vectorized manner as opposed to a single item at a time.

a = np.random.rand(50, 60, 70)
i = np.random.randint(0, len(a), 40)
b = a[i]


import numpy_indexed as npi
assert all(i == npi.indices(a, b))

This solution has better time complexity (n log n at worst) than any of the previously posted answers, and is fully vectorized.

If you are interested in the indexes, the best choice is np.argsort(a)

a = np.random.randint(0, 100, 10)
sorted_idx = np.argsort(a)