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

I had the same problem (i.e. indexing with multi-conditions, here it’s finding data in a certain date range). The (a-b).any() or (a-b).all() seem not working, at least for me.

Alternatively I found another solution which works perfectly for my desired functionality (The truth value of an array with more than one element is ambigous when trying to index an array).

Instead of using suggested code above, use:

numpy.logical_and(a, b)

The reason for the exception is that and implicitly calls bool. First on the left operand and (if the left operand is True) then on the right operand. So x and y is equivalent to bool(x) and bool(y).

However the bool on a numpy.ndarray (if it contains more than one element) will throw the exception you have seen:

>>> import numpy as np
>>> arr = np.array([1, 2, 3])
>>> bool(arr)
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

The bool() call is implicit in and, but also in if, while, or, so any of the following examples will also fail:

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

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

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

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

There are more functions and statements in Python that hide bool calls, for example 2 < x < 10 is just another way of writing 2 < x and x < 10. And the and will call bool: bool(2 < x) and bool(x < 10).

The element-wise equivalent for and would be the np.logical_and function, similarly you could use np.logical_or as equivalent for or.

For boolean arrays – and comparisons like <, <=, ==, !=, >= and > on NumPy arrays return boolean NumPy arrays – you can also use the element-wise bitwise functions (and operators): np.bitwise_and (& operator)

>>> np.logical_and(arr > 1, arr < 3)
array([False,  True, False], dtype=bool)

>>> np.bitwise_and(arr > 1, arr < 3)
array([False,  True, False], dtype=bool)

>>> (arr > 1) & (arr < 3)
array([False,  True, False], dtype=bool)

and bitwise_or (| operator):

>>> np.logical_or(arr <= 1, arr >= 3)
array([ True, False,  True], dtype=bool)

>>> np.bitwise_or(arr <= 1, arr >= 3)
array([ True, False,  True], dtype=bool)

>>> (arr <= 1) | (arr >= 3)
array([ True, False,  True], dtype=bool)

A complete list of logical and binary functions can be found in the NumPy documentation:

• “Logic functions”
• “Binary operations”

if you work with pandas what solved the issue for me was that i was trying to do calculations when I had NA values, the solution was to run:

df = df.dropna()

And after that the calculation that failed.

This typed error-message also shows while an if-statement comparison is done where there is an array and for example a bool or int. See for example:

... code snippet ...

if dataset == bool:
    ....

... code snippet ...

This clause has dataset as array and bool is euhm the "open door"… True or False.

In case the function is wrapped within a try-statement you will receive with except Exception as error: the message without its error-type:

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

Taking up @ZF007’s answer, this is not answering your question as a whole, but can be the solution for the same error. I post it here since I have not found a direct solution as an answer to this error message elsewhere on Stack Overflow.

The error, among others, appears when you check whether an array was empty or not.

• if np.array([1,2]): print(1) –> ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all().

• if np.array([1,2])[0]: print(1) –> no ValueError, but: if np.array([])[0]: print(1) –> IndexError: index 0 is out of bounds for axis 0 with size 0.

• if np.array([1]): print(1) –> no ValueError, but again will not help at an array with many elements.

• if np.array([]): print(1) –> DeprecationWarning: The truth value of an empty array is ambiguous. Returning False, but in future this will result in an error. Use 'array.size > 0' to check that an array is not empty.

• if np.array([]).size is not None: print(1): Taking up a comment by this user, this does not work either. This is since no np.array can ever be the same object as None – that object is unique – and thus will always match is not None (i.e. never match is None) whether or not it’s empty.

Doing so:

• if np.array([]).size: print(1) solved the error.

For me, this error occurred on testing, code with error below:

pixels = []
self.pixels = numpy.arange(1, 10)
self.assertEqual(self.pixels, pixels)

This code returned:

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

Because i cannot assert with a list the object returned by method arrange of numpy.

Solution as transform the arrange object of numpy to list, my choice was using the method toList(), as following:

pixels = []
self.pixels = numpy.arange(1, 10).toList()
self.assertEqual(self.pixels, pixels)

Normally, when you compare two single digits the Python regular codes work correctly, but inside an array there are some digits (more than one number) that should be processed in parallel.

For example, let us assume the following:

a = np.array([1, 2, 3])
b = np.array([2, 3, 4])

And you want to check if b >= a: ?

Because, a and b are not single digits and you actually mean if every element of b is greater than the similar number in a, then you should use the following command:

if (b >= a).all():
 print("b is greater than a!")

Simplest answer is use "&" instead of "and".

>>> import numpy as np
>>> arr = np.array([1, 4, 2, 7, 5])
>>> arr[(arr > 3) and (arr < 6)]   # this will fail
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
>>> arr[(arr > 3) & (arr < 6)]   # this will succeed
array([4, 5])

Cause

This error occurs any time that the code attempts to convert a Numpy array to boolean (i.e., to check its truth value, as described in the error message). For a given array a, this can occur:

• Explicitly, by using bool(a).

• Implicitly with boolean logical operators: a and a, a or a, not a.

• Implicitly using the built-in any and all functions. (These can accept a single array, regardless of how many dimensions it has; but cannot accept a list, tuple, set etc. of arrays.)

• Implicitly in an if statement, using if a:. While it’s normally possible to use any Python object in an if statement, Numpy arrays deliberately break this feature, because it could easily be used to write incorrect code by mistake otherwise.

Numpy arrays and comparisons (==, !=, <, >, <=, >=)

Comparisons have a special meaning for Numpy arrays. We will consider the == operator here; the rest behave analogously. Suppose we have

import numpy as np
>>> a = np.arange(9)
>>> b = a % 3
>>> a
array([0, 1, 2, 3, 4, 5, 6, 7, 8])
>>> b
array([0, 1, 2, 0, 1, 2, 0, 1, 2])

Then, a == b does not mean "give a True or False answer: is a equal to b?", like it would usually mean. Instead, it will compare the values element by element, and evaluate to an array of boolean results for those comparisons:

>>> a == b
array([ True,  True,  True, False, False, False, False, False, False])

In other words, it does the same kind of broadcasting that mathematical operators (like b = a % 3) do.

It does not make sense to use this result for an if statement, because it is not clear what to do: should we enter the if block, because some of the values matched? Or should we enter the else block, because some of the values didn’t match? Here, Numpy applies an important principle from the Zen of Python: "In the face of ambiguity, refuse the temptation to guess."

Thus, Numpy will only allow the array to be converted to bool, if it contains exactly one element. (In some older versions, it will also convert to False for an empty array; but there are good logical reasons why this should also be treated as ambiguous.)

Similarly, comparing a == 4 will not check whether the array is equal to the integer (of course, no array can ever be equal to any integer). Instead, it will broadcast the comparison across the array, giving a similar array of results:

>>> a == 4
array([False, False, False, False,  True, False, False, False, False])

Fixing expressions

• If the code is explicitly converting to bool, choose between applying .any or .all to the result, as appropriate. As the names suggest, .any will collapse the array to a single boolean, indicating whether any value was truthy; .all will check whether all values were truthy.

  >>> (a == 4).all() # `a == 4` contains some `False` values
  False
  >>> (a == 4).any() # and also some `True` values
  True
  >>> a.all() # We can check `a` directly as well: `0` is not truthy,
  False
  >>> a.any() # but other values in `a` are.
  True
  

  If the goal is to convert a to boolean element-wise, use a.astype(bool), or (only for numeric inputs) a != 0.

• If the code is using boolean logic (and/or/not), use bitwise operators (&/|/~, respectively) instead:

  >>> ((a % 2) != 0) & ((a % 3) != 0) # N.B. `&`, not `and`
  array([False,  True, False, False, False,  True, False,  True, False])
  

  Note that bitwise operators also offer access to ^ for an exclusive-or of the boolean inputs; this is not supported by logical operators (there is no xor).

• For a list (or other sequence) of arrays that need to be combined in the same way (i.e., what the built-ins all and any do), instead build the corresponding (N+1)-dimensional array, and use np.all or np.any along axis 0:

  >>> a = np.arange(100) # a larger array for a more complex calculation
  >>> sieves = [a % p for p in (2, 3, 5, 7)]
  >>> all(sieves) # won't work
  Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
  ValueError: The truth value of an array with more than one element is ambiguous.
   Use a.any() or a.all()
  >>> np.all(np.array(sieves), axis=0) # instead:
  array([False,  True, False, False, False, False, False, False, False,
         False, False,  True, False,  True, False, False, False,  True,
         False,  True, False, False, False,  True, False, False, False,
         False, False,  True, False,  True, False, False, False, False,
         False,  True, False, False, False,  True, False,  True, False,
         False, False,  True, False, False, False, False, False,  True,
         False, False, False, False, False,  True, False,  True, False,
         False, False, False, False,  True, False, False, False,  True,
         False,  True, False, False, False, False, False,  True, False,
         False, False,  True, False, False, False, False, False,  True,
         False, False, False, False, False, False, False,  True, False,
         False])
  

Fixing if statements

First, keep in mind that if the code has an if statement that uses a broken expression (like if (a % 3 == 0) or (a % 5 == 0):), then both things will need to be fixed.

Generally, an explicit conversion to bool (using .all() or .any() as above) will avoid an exception:

>>> a = np.arange(20) # enough to illustrate this
>>> if ((a % 3 == 0) | (a % 5 == 0)).any():
...     print('there are fizzbuzz values')
... 
there are fizzbuzz values

but it might not do what is wanted:

>>> a = np.arange(20) # enough to illustrate this
>>> if ((a % 3 == 0) | (a % 5 == 0)).any():
...     a = -1
... 
>>> a
-1

If the goal is to operate on each value where the condition is true, then the natural way to do that is to use the mask as a mask. For example, to assign a new value everywhere the condition is true, simply index into the original array with the computed mask, and assign:

>>> a = np.arange(20)
>>> a[(a % 3 == 0) | (a % 5 == 0)] = -1
>>> a
array([-1,  1,  2, -1,  4, -1, -1,  7,  8, -1, -1, 11, -1, 13, 14, -1, 16,
       17, -1, 19])

This indexing technique is also useful for finding values that meet a condition. Building on the previous sieves example:

>>> a = np.arange(100)
>>> sieves = [a % p for p in (2, 3, 5, 7)]
>>> a[np.all(np.array(sieves), axis=0)]
array([ 1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71,
       73, 79, 83, 89, 97])

(Exercise: study the code and understand why this result isn’t quite a list of primes under 100; then fix it.)

Using Pandas

The Pandas library has Numpy as a dependency, and implements its DataFrame type on top of Numpy’s array type. All the same reasoning applies, such that Pandas Series (and DataFrame) objects cannot be used as boolean: see Truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

The Pandas interface for working around the problem is a bit more complicated – and best understood by reading that Q&A. The question specifically covers Series, but the logic generally applies to DataFrames as well. If you need more specific guidance, though, see If condition with a dataframe.