Python: "Use a.any() or a.all()" while traversing a numpy.ndarray

Question

In a numpy.ndarray like this one:

myarray=
array([[ 0.47174344,  0.45314669,  0.46395022,  0.47440382,  0.50709627,
         0.53350065,  0.5233444 ,  0.49974663,  0.48721607,  0.46239652,
         0.4693633 ,  0.47263569,  0.47591957,  0.436558  ,  0.43335574,
         0.44053621,  0.42814804,  0.43201894,  0.43973886,  0.44125302,
         0.41176999],
       [ 0.46509004,  0.46221505,  0.48824086,  0.50088744,  0.53040384,
         0.53592231,  0.49710228,  0.49821022,  0.47720381,  0.49096272,
         0.50438366,  0.47173162,  0.48813669,  0.45032002,  0.44776794,
         0.43910269,  0.43326132,  0.42064458,  0.43472954,  0.45577299,
         0.43604956]])

I want to count how many cells exceed a given value, let's say 0.5, and set those that don't to 0.0. This is what I do:

count=0 
value=0.5
for i in range(myarray.shape[0]):
   for j in range(myarray.shape[1]):
       if myarray[i][j]<value:
          myarray[i][j]=0 
       elif myarray[i][j]>=value:     
          count=count+1 
percentage=round(100*count/(myarray.shape[0]*myarray.shape[1]),2)

However, I get this error: ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all(), pointing at the line where I check if myarray[i][j]<value.

Why does this happen and how to fix it? What is the truth value?

No errors. Same script as yours. Got the percentage value as 16. — gora
– gora, Commented Aug 4, 2016 at 21:57

Noam · Accepted Answer · 2016-08-04 22:11:45Z

6

Normally, you can compare two numbers to get a truth value. For example:

elem = 5
if elem < 6:
    # do something

is equivalent to:

if true:
    # do something

However, you can't compare an array to a value. For example:

elem = [5,7]
if elem < 6:
    # this doesn't make sense

Instead, you can get the truth value of whether any or all elements satisfy the condition. For example:

elem = np.array([5,7])
if np.any(elem<6):
    # this is true, because 5 < 6
if np.all(elem<6):
    # this isn't true, because 7 > 6

I ran your example code above and found no error, so I'm not sure what the issue is. But this is what you should look out for. Consider printing the element you are comparing to see if it is an array.

Also, this is a shorter way of doing what you want to do:

myarray = np.array( putarrayhere )
count = sum(myarray >= value)

edited Aug 4, 2016 at 22:11

answered Aug 4, 2016 at 21:56

Noam

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Comments

Sen Qiao · Accepted Answer · 2016-08-04 22:07:14Z

Yeah I think your numpy.array has an extra bracket or it is encompassing another array.

Tried Manually set array as

myarray=np.array([[ 0.47174344,  0.45314669,  0.46395022,  0.47440382,  0.50709627,0.53350065,  0.5233444 ,  0.49974663,  0.48721607,  0.46239652, 0.4693633 ,  0.47263569,  0.47591957,  0.436558  ,  0.43335574,0.44053621,  0.42814804,  0.43201894,  0.43973886, 0.44125302, 0.41176999],[ 0.46509004,  0.46221505,  0.48824086,  0.50088744,  0.53040384,0.53592231,  0.49710228,  0.49821022,  0.47720381,  0.49096272,0.50438366,  0.47173162,  0.48813669,  0.45032002,  0.44776794,0.43910269,  0.43326132,  0.42064458,  0.43472954,  0.45577299,0.43604956]])

and the code works

but setting:

myarray=np.array([[[ 0.47174344,  0.45314669,  0.46395022,  0.47440382,  0.50709627,0.53350065,  0.5233444 ,  0.49974663,  0.48721607,  0.46239652, 0.4693633 ,  0.47263569,  0.47591957,  0.436558  ,  0.43335574,0.44053621,  0.42814804,  0.43201894,  0.43973886, 0.44125302, 0.41176999],[ 0.46509004,  0.46221505,  0.48824086,  0.50088744,  0.53040384,0.53592231,  0.49710228,  0.49821022,  0.47720381,  0.49096272,0.50438366,  0.47173162,  0.48813669,  0.45032002,  0.44776794,0.43910269,  0.43326132,  0.42064458,  0.43472954,  0.45577299,0.43604956]]])

yielded similar errors

Ohad Eytan · Accepted Answer · 2016-08-04 22:12:34Z

1

Regardless the error you can simply do:

myarray[myarray<value]=0
np.count_nonzero(myarray)

to get your desired result

answered Aug 4, 2016 at 22:12

Ohad Eytan

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