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

Hello Guys, How are you all? Hope You all Are Fine. Today I get the following error ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() in python. So Here I am Explain to you all the possible solutions here.

Without wasting your time, Let’s start This Article to Solve This Error.

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

Today I get the following error ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() in python.

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

  1. How To Solve ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() Error ?

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

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

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

Solution 1

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)]

Solution 2

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 two boolean arrays.

The NumPy developers felt there was no one commonly understood way to evaluate an 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, the NumPy developers refused to guess and instead decided to raise a ValueError whenever one tries to evaluate an array in boolean context. Applying and to two numpy arrays causes the two arrays to be evaluated in boolean context (by calling __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().

Summery

It’s all About this issue. Hope all solution helped you a lot. Comment below Your thoughts and your queries. Also, Comment below which solution worked for you? Thank You.

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