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pythonnumpydecimalnan
Juyun Lee
Juyun Lee

Reputation: 51

how do I check decimal.is_nan() for all values in array?

Suppose I have my array like this:

from decimal import Decimal
array = [Decimal(np.nan), Decimal(np.nan), Decimal(0.231411)]

I know that if the types are float, I can check if all the values are nan or not , as:

np.isnan(array).all()

Is there a way for type Decimal?

The solution would be better without iteration.

Upvotes: 1

Views: 996

Answers (2)

Tonechas
Tonechas

Reputation: 13743

You could use NumPy's vectorize to avoid iteration.

In [40]: from decimal import Decimal

In [41]: import numpy as np

In [42]: nums = [Decimal(np.nan), Decimal(np.nan), Decimal(0.231411)]

In [43]: nums
Out[43]: 
[Decimal('NaN'),
 Decimal('NaN'),
 Decimal('0.2314110000000000055830895462349872104823589324951171875')]

In [44]: np.all(np.vectorize(lambda x: x.is_nan())(np.asarray(nums)))
Out[44]: False

In [45]: np.all(np.vectorize(lambda x: x.is_nan())(np.asarray(nums[:-1])))
Out[45]: True

In the snippet above nums is a list of instances of class Decimal. Notice that you need to convert that list into a NumPy array.

Upvotes: 1

s3dev
s3dev

Reputation: 9721

From my comment above, I realise it’s an iteration. The reason is that np.isnan does not support Decimal as an input type; therefore, I don’t believe this can be done via broadcasting, without converting the datatype - which means a potential precision loss, which is a reason to use a Decimal type.

Additionally, as commented by @juanpa.arrivillaga, as the Decimal objects are in a list, iteration is the only way. Numpy is not necessary in this operation.

One method is:

all([i.is_nan() for i in array])

Upvotes: 0

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