Where function in python returns nothing

Question

I have this array

     a = array([1,5,7])

I apply the where function

     where(a==8)

What is returned in this case is

    (array([], dtype=int64),)

However I would like the code to return the integer "0" whenever the where function returns an empty array. Is that possible?

What would you like it to return when there is a single match, and that match is at index zero? — NPE, Jan 25 '12 at 12:32
Why do you want that? What is good for to have to return types? Why don't make a check of the array size to know if you found it? And then use the array in case it's not emtpy? — ezdazuzena, Jan 25 '12 at 13:07
@aix: there is no possible "confusion" problem: if the match is at index 0, the result is `(array([0]),)`, not `0`. — Eric O. Lebigot, Jan 25 '12 at 16:21

score 5 · Accepted Answer · edited Jan 25 '12 at 16:19

5

def where0(vec):
    a = where(vec)
    return a if a[0] else 0
    # The return above is equivalent to:
    # if len(a[0]) == 0:
    #     return 0  # or whatever you like
    # else:
    #     return a

a = array([1,5,7])
print where0(a==8)

And consider also the comment from aix under your question. Instead of fixing where(), fix your algorithm

edited Jan 25 '12 at 16:19

Eric O. Lebigot

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answered Jan 25 '12 at 12:48

Jakub M.

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+1 for the general advice (it is best if functions only have a single return type), but your whole `if len(a[0]) == 0…` structure could be replaced by the much simpler and shorter `return a if a[0] else 0`. – Eric O. Lebigot Jan 25 '12 at 14:03
oh, I use python since long but never used this neat syntax... thanks! – Jakub M. Jan 25 '12 at 15:21

Brad123 · Answer 2 · 2019-05-17T15:39:57.223

0

a empty array will return 0 with .size

import numpy as np    
a = np.array([])    
a.size
>> 0

edited May 17 '19 at 15:39

answered May 16 '19 at 00:20

Brad123

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A brief explanation would improve your answer. – belwood May 16 '19 at 01:17

ezdazuzena · Answer 3 · 2012-01-25T16:22:43.130

0

Better use a function that has only one return type. You can check for the size of the array to know if it's empty or not, that should do the work:

 a = array([1,5,7])
 result = where(a==8)

 if result[0] != 0:
     doFancyStuff(result)
 else:
     print "bump"

edited Jan 25 '12 at 16:22

answered Jan 25 '12 at 13:20

ezdazuzena

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You probably meant `len(result[0])`? – Eric O. Lebigot Jan 25 '12 at 14:01
1

I wouldn't use 'is' on integers. For 0 it works because of the integer cache but a = 10000; a is 10000 is False. Using '==' avoids this problem. http://stackoverflow.com/questions/306313/python-is-operator-behaves-unexpectedly-with-integers – Free Monica Cellio Jan 25 '12 at 14:20
EOL, why would he want to take the len of result[0]? It's a one-dimensional array... – Free Monica Cellio Jan 25 '12 at 14:21
@ChadMiller: Thanks for the hint. Fixed it. – ezdazuzena Jan 25 '12 at 14:40
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@ChadMiller - "where" does not return an array, it returns a tuple of arrays (see the original question). Though I think this may have changed from earlier numpy versions. – JoeZuntz Jan 25 '12 at 14:48
@JoeZUntz: correct, see http://docs.scipy.org/doc/numpy/reference/generated/numpy.where.html?highlight=where#numpy.where – ezdazuzena Jan 25 '12 at 15:08
@JoeZuntz Thanks, I missed the ',' at the end of OP's output. – Free Monica Cellio Jan 25 '12 at 15:11
@ezdazuzena: your `if len(result[0]) != 0` could be more simply written as `if result[0]`. – Eric O. Lebigot Jan 25 '12 at 16:17

score -2 · Answer 4 · edited Jan 11 '18 at 23:31

-2

Try the below. This will handle the case where a test for equality to 0 will fail when index 0 is returned. (e.g. np.where(a==1) in the below case)

a = array([1,5,7])
ret = np.where(a==8)
ret = ret if ret[0].size else 0

edited Jan 11 '18 at 23:31

Machavity

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answered Jan 11 '18 at 14:28

surfertas

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Where function in python returns nothing

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