Index of a numpy array within a sequence

Question

I want to find the index of a particular numpy array with in a sequence. For example, given:

import numpy as np

WHITE = np.array([255, 255, 255])
BLUE = np.array([0, 0, 255])
GRAY = np.array([192, 192, 192])
BLACK = np.array([0, 0, 0])
GREEN = np.array([0, 255, 0])
YELLOW = np.array([255, 255, 0])
COLORS = (WHITE, BLUE, GRAY, BLACK, GREEN)

I'd like to be able to ask for something like

print(COLORS.index(GRAY))

but this leads to the error:

ValueError: The truth value of an array with more than one element
is ambiguous. Use a.any() or a.all()

I do have a workaround, but it feels overly acrobatic:

def index_of(x, sequence):
    eq = list((item == x).all() for item in sequence)
    return eq.index(True)

print(index_of(GRAY, COLORS))

Is there are more elegant/Pythonic way to do this?

Can't you just use a dict? But essentially, no, since index uses == and == produces an array. — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Oct 19, 2017 at 21:22
next(i for (i, item) in enumerate(sequence) if (item == x).all()) would be slightly more elegant I think, although it is not very pretty either — tehforsch
– tehforsch, Commented Oct 19, 2017 at 21:30
Although, there are several ways to make your approach more elegant. Starting by not using list(<generator expression>) and just using a list-comprehension. Or better yet, just use a for-loop and return early.... — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Oct 19, 2017 at 21:30
@tehforsch you should just use a for-loop in that case. Wayyyy more elegant. — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Oct 19, 2017 at 21:30
@juanpa.arrivillaga and then have a break-statement? thats fine too, whether that is more elegant is a matter of taste i would say — tehforsch
– tehforsch, Commented Oct 19, 2017 at 21:31

Noel Díaz Rodríguez · Accepted Answer · 2017-10-19 22:46:40Z

2

May be a matter of taste but in this case a for loop could be more readable:

def index_of(search_for, arrays):
    for i, array in enumerate(arrays):
        if np.array_equal(search_for, array):
            return i
    raise ValueError('{} not in sequence'.format(search_for))

numpy function np.array_equal allows for comparison of arrays of different sizes in case you'll need it ((item == x).all() will raise an exception for different array sizes)

Also raise a ValueError exception to mimic the index function of the tuple.

answered Oct 19, 2017 at 22:46

Noel Díaz Rodríguez

211 bronze badge

Sign up to request clarification or add additional context in comments.

Comments

Ben Usman · Accepted Answer · 2017-10-20 17:55:03Z

1

Considering that you index not by values, but by actual objects, you should be able to use is:

def index_of(color, seq):
    return next(i for i, x in enumerate(seq) if x is color)

edited Oct 20, 2017 at 17:55

answered Oct 19, 2017 at 21:30

Ben Usman

8,4776 gold badges48 silver badges66 bronze badges

1 Comment

Peter Drake Over a year ago

I'm confused. Isn't that just going to return color, rather than the index where color appears?

hpaulj · Accepted Answer · 2017-10-19 21:38:01Z

If COLORS was a list of lists, then the index works:

In [39]: C = [c.tolist() for c in COLORS]
In [40]: C
Out[40]: [[255, 255, 255], [0, 0, 255], [192, 192, 192], [0, 0, 0], [0, 255, 0]]
In [41]: C.index(GRAY.tolist())
Out[41]: 2

With array elements, the index does something like:

In [44]: [c==GRAY for c in COLORS]
Out[44]: 
[array([False, False, False], dtype=bool),
 array([False, False, False], dtype=bool),
 array([ True,  True,  True], dtype=bool),
 array([False, False, False], dtype=bool),
 array([False, False, False], dtype=bool)]

Equality tests with array are done element by element, with the result being a boolean array. But using a boolean array in a context that expects a simple scalar boolean, results in the (all-too-common) ambiguity error. all can reduce those arrays to single values.

In [46]: [(c==GRAY).all() for c in COLORS]
Out[46]: [False, False, True, False, False]

In [47]: [(c==GRAY).all() for c in COLORS].index(True)
Out[47]: 2
In [48]: [(c==BLUE).all() for c in COLORS].index(True)
Out[48]: 1

If COLORS is turned into a 2d array, we can do test like:

In [49]: CA = np.array(COLORS)
In [50]: CA
Out[50]: 
array([[255, 255, 255],
       [  0,   0, 255],
       [192, 192, 192],
       [  0,   0,   0],
       [  0, 255,   0]])
In [51]: CA==GRAY
Out[51]: 
array([[False, False, False],
       [False, False, False],
       [ True,  True,  True],
       [False, False, False],
       [False, False, False]], dtype=bool)
In [52]: (CA==GRAY).all(axis=1)
Out[52]: 
array([False, False,  True, False, False], dtype=bool)

Collectives™ on Stack Overflow

Index of a numpy array within a sequence

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related