How detect length of a numpy array with only one element?

Question

I am reading in a file using numpy.genfromtxt which brings in columns of both strings and numeric values. One thing I need to do is detect the length of the input. This is all fine provided there are more than one value read into each array.

But...if there is only one element in the resulting array, the logic fails. I can recreate an example here:

import numpy as np
a = np.array(2.3)

len(a) returns an error saying:

TypeError: len() of unsized object

however, If a has 2 or more elements, len() behaves as one would expect.

import numpy as np
a = np.array([2.3,3.6])

len(a) returns 2

My concern here is, if I use some strange exception handling, I can't distinguish between a being empty and a having length = 1.

EDIT: @noskio suggested setting a = np.array([2.3]). The problem is, the actual genesis of a is by using numpy.genfromtxt. The code looks like this:

import numpy as np
indata = np.genfromtxt(some_filename, names=True,dtype=None)
a = indata['one_col_headername']

As a result, if indata is only one row in the file, a is a 0-d array.

array([2]) is an array with one element and 1 dimension. array(2) is an array with zero rank or zero dimensions. — endolith
– endolith, Commented Dec 16, 2012 at 5:12

leoschet · Accepted Answer · 2018-05-22 13:52:16Z

41

If you need a one-liner (assuming the answer you are expecting is 1):

In [1]: import numpy as np

In [2]: a = np.array(2.3)

In [3]: len(np.atleast_1d(a))
Out[3]: 1

This page explains why it was decided to implement 0-dimensional arrays in numpy.

edited May 22, 2018 at 13:52

leoschet

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answered Dec 30, 2010 at 20:12

pberkes

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1 Comment

pberkes Over a year ago

@wim it looks like the link has been removed, and I cannot find a replacement for it.

unutbu · Accepted Answer · 2010-12-30 20:26:20Z

6

import numpy as np

tests=[np.array(2.3),np.array([]),np.array([2.3]),np.array([2.3,3.6])]

print('{a:30}{s:<10}{l:<10}{sl:<10}'.format(a='repr',s='shape',sl='len(shape)',l='length'))
for a in tests:
    s=a.shape
    l=len(a) if a.shape else 0
    sl=len(s)
    print('{a!r:30}{s:<10}{l:<10}{sl:<10}'.format(a=a,l=l,s=s,sl=sl))

yields

repr                          shape     length    len(shape)
array(2.2999999999999998)     ()        0         0         
array([], dtype=float64)      (0,)      0         1         
array([ 2.3])                 (1,)      1         1         
array([ 2.3,  3.6])           (2,)      2         1

You can distinguish between an "empty" array (e.g. np.array([])) and a numpy scalar (e.g. np.array(2.3)) by looking at the length of the shape.

edited Dec 30, 2010 at 20:26

answered Dec 30, 2010 at 20:06

unutbu

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1 Comment

mishaF Over a year ago

Thanks @unubtu. That's cool, but how can I distinguish between a with length of 1 and an empty a?

Brandon · Accepted Answer · 2014-03-04 22:50:21Z

3

It looks like the size property of ndarrays will work in this case if you know that the array is one-dimensional. In my opinion, a.size is a lot more readable than len(np.atleast_1d(a)). However, note that the size property will return the total number of elements in the array if it has more than one dimension:

In [1]: import numpy as np

In [2]: np.array(2.3).size
Out[2]: 1

In [3]: np.array([1, 2]).size
Out[3]: 2

In [4]: np.array([[1,2], [3,4]]).size
Out[4]: 4

answered Mar 4, 2014 at 22:50

Brandon

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Comments

nosklo · Accepted Answer · 2010-12-30 20:03:36Z

0

a = np.array([2.3])
print len(a)

answered Dec 30, 2010 at 20:03

nosklo

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1 Comment

ecoe Over a year ago

this works for a constant as you show, but not for an array input

Aaryan Bhatia · Accepted Answer · 2023-08-26 20:37:12Z

0

Had this same issue, I found a simple way to distinguish the two use cases mentioned by mishaF and leoschet:

    import numpy as np
    array_1=np.array(2.3) #not in a list
    array_2=np.array([2.3]) #in a list of len 1
    print(len(array_1)) #gives the type error mentioned in the question
    print(len(array_2)) #returns 1 as desired

edited Aug 26, 2023 at 20:37

answered Aug 26, 2023 at 20:36

Aaryan Bhatia

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Community Over a year ago

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