Where function in python returns nothing

问题

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?

回答1:

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

回答2:

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"

回答3:

a empty array will return 0 with .size

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

回答4:

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

来源：https://stackoverflow.com/questions/9002749/where-function-in-python-returns-nothing