# the truth value of an array with more than one element is ambiguous. use a.any() or a.all()

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## Problem :

I got following rather cryptic error message:
the truth value of an array with more than one element is ambiguous. use a.any() or a.all()

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## Solution :

I had the same problem with indexing with multiple-conditions while finding the data in a certain date range. The (a-b).any() OR (a-b).all() was not working for me at all.

I found the solution which works perfectly for my functionality.

Simply used the numpy.logical_and(a,b) and it worked for me. Following is the way to rewrite the code :

`selected  = r[numpy.logical_and(r["dt"] >= startdate, r["dt"] <= enddate)]`
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The error message explains itself very well.

``````ValueError: The truth value of an array with more than one element is ambiguous.

Use a.any() or a.all().``````

# Solutions:

## Array comparison return boolean:

Make sure your array comparison returns a boolean array. Methods any() and all() reduce values over the array either the value is logical_or or logical_and. Probably you also don’t need to check for equality.

## What should bool return?

The main question is what should the bool(np.array([False, False, True])) have to return. There are many arguments.

• True, because bool(np.array(x)) should have to return same as bool(list(x)), and non empty lists.
• True, Because at least one element has to be true.
• False, because all the elements are not false value.

Since, the truth value of an array having more than one element is ambiguous so you have to use .any() or .all().

### Example:

``````>>> v = np.array([1, 2, 3]) == np.array([1, 2, 4])

>>> v

Array([True, True, False], dtype=bool)

>>> v.any()

True

>>> v.all()

False``````

## Float values:

You have to use np.allclose, if you are making comparisons of array for float data type.

### Example:

``````>>> np.allclose(np.array([1, 2, 3_1e-8]), np.array([1, 2, 3]))

True.``````

I hope you can understand and this answer will help to solve your problem.

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