CODEWITHSUNDEEP
pythonnumpy
azal
azal

Reputation: 1260

ValueError in simple Python calculation

This is my function and the variable tracks is a list and each element of the list is a n x 3 array:

temp = np.array(np.zeros((n, n)))
for j in range(n-1):
    for w in range(j + 1, n):  
        mindistance = np.zeros(len(tracks[j]))
        for i in range(len(tracks[j])):   
            mindistance[i] = np.linalg.norm(min(np.fabs(np.array(tracks[w]) - tracks[j][i])))
        temp[j][w]=np.sum(mindistance)/len(tracks[j])

I'm trying to calculate the minimum distances between the arrays of the list which represent 3d lines in space but I am getting the error:

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

The error is probably related to the call to min() but I can't solve it. Following is the error traceback:

Traceback (most recent call last):

  File "<ipython-input-14-7fb640816626>", line 1, in <module>
    runfile('/Users/G_Laza/Desktop/functions/Main.py', wdir='/Users/G_Laza/Desktop/functions')

  File "/Applications/Spyder.app/Contents/Resources/lib/python2.7/spyderlib/widgets/externalshell/sitecustomize.py", line 580, in runfile
    execfile(filename, namespace)

  File "/Users/G_Laza/Desktop/functions/Main.py", line 42, in <module>
    tempA = distance_calc.dist_calc(len(subset_A), subset_A)  # distance matrix calculation

  File "distance_calc.py", line 23, in dist_calc
    mindistance[i] = np.linalg.norm(min(np.fabs(np.array(tracks[w]) - tracks[j][i])))

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

Upvotes: 0

Views: 1119

Answers (1)

wwii
wwii

Reputation: 23753

The error occurs because you cannot determine whether a complete array is True or False. What would be the boolean state of an array where all elements are True but one?

min takes an iterable for an argument and compares each element to the other, each comparison results in a boolean value. Iterating over a 1-d numpy array produces individual elements - min works for a 1-d numpy array.

>>> a
array([-4, -3, -2, -1,  0,  1,  2,  3,  4])
>>> for thing in a:
     print thing,


-4 -3 -2 -1 0 1 2 3 4
>>> min(a)
-4
>>>

Iterating over a 2-d numpy array produces rows.

>>> b
array([[-4, -3, -2],
       [-1,  0,  1],
       [ 2,  3,  4]])
>>> for thing in b:
     print thing

[-4 -3 -2]
[-1  0  1]
[2 3 4]
>>>

min won't won't work for 2-d arrays because it is comparing arrays and - The truth value of an array with more than one element is ambiguous.

>>> c
array([0, 1, 2, 3])
>>> c < 2
array([ True,  True, False, False], dtype=bool)
>>> bool(c < 2)

Traceback (most recent call last):
  File "<pyshell#74>", line 1, in <module>
    bool(c < 2)
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
>>> 
>>> bool(np.array((True, True)))

Traceback (most recent call last):
  File "<pyshell#75>", line 1, in <module>
    bool(np.array((True, True)))
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
>>> 
>>> bool(np.array((True, False)))

Traceback (most recent call last):
  File "<pyshell#76>", line 1, in <module>
    bool(np.array((True, False)))
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
>>>

If you need to find the element with the minimum value, use numpy.amin or the ndarray.min method.

>>> 
>>> np.amin(b)
-4
>>> b.min()
-4
>>>

Upvotes: 2

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