python: how to identify if a variable is an array or a scalar
I have a function that takes the argument NBins . I want to make a call to this function with a scalar 50 or an array [0, 10, 20, 30] . How can I identify within the function, what the length of NBins is? or said differently, if it is a scalar or a vector?
I tried this:
>>> N=[2,3,5]
>>> P = 5
>>> len(N)
3
>>> len(P)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: object of type 'int' has no len()
>>>
As you see, I can't apply len to P , since it's not an array.... Is there something like isarray or isscalar in python?
thanks
>>> isinstance([0, 10, 20, 30], list)
True
>>> isinstance(50, list)
False
To support any type of sequence, check collections.Sequence instead of list .
note : isinstance also supports a tuple of classes, check type(x) in (..., ...) should be avoided and is unnecessary.
You may also wanna check not isinstance(x, (str, unicode))
Previous answers assume that the array is a python standard list. As someone who uses numpy often, I'd recommend a very pythonic test of:
if hasattr(N, "__len__")
Combining @jamylak and @jpaddison3's answers together, if you need to be robust against numpy arrays as the input and handle them in the same way as lists, you should use
import numpy as np
isinstance(P, (list, tuple, np.ndarray))
This is robust against subclasses of list, tuple and numpy arrays.
And if you want to be robust against all other subclasses of sequence as well (not just list and tuple), use
import collections
import numpy as np
isinstance(P, (collections.Sequence, np.ndarray))
Why should you do things this way with isinstance and not compare type(P) with a target value? Here is an example, where we make and study the behaviour of NewList , a trivial subclass of list.
>>> class NewList(list):
... isThisAList = '???'
...
>>> x = NewList([0,1])
>>> y = list([0,1])
>>> print x
[0, 1]
>>> print y
[0, 1]
>>> x==y
True
>>> type(x)
<class '__main__.NewList'>
>>> type(x) is list
False
>>> type(y) is list
True
>>> type(x).__name__
'NewList'
>>> isinstance(x, list)
True
Despite x and y comparing as equal, handling them by type would result in different behaviour. However, since x is an instance of a subclass of list , using isinstance(x,list) gives the desired behaviour and treats x and y in the same manner.
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