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

`r`

is a numpy (rec)array. So `r["dt"] >= startdate`

is also a (boolean)array. For numpy arrays the `&`

operation returns the elementwise-and of the twoboolean arrays.

The NumPy developers felt there was no one commonly understood way to evaluatean array in boolean context: it could mean `True`

if *any* element is`True`

, or it could mean `True`

if *all* elements are `True`

, or `True`

if the array has non-zero length, just to name three possibilities.

Since different users might have different needs and different assumptions, theNumPy developers refused to guess and instead decided to raise a ValueErrorwhenever one tries to evaluate an array in boolean context. Applying `and`

totwo numpy arrays causes the two arrays to be evaluated in boolean context (bycalling `__bool__`

in Python3 or `__nonzero__`

in Python2).

Your original code

`mask = ((r["dt"] >= startdate) & (r["dt"] <= enddate))selected = r[mask]`

looks correct. However, if you do want `and`

, then instead of `a and b`

use `(a-b).any()`

or `(a-b).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, simply using a `numpy.logical_and(a,b)`

would work. Here you may want to rewrite the code as

`selected = r[numpy.logical_and(r["dt"] >= startdate, r["dt"] <= enddate)]`

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