CODEWITHSUNDEEP
pythonnumpyfor-loopambiguity
FaCoffee
FaCoffee

Reputation: 7929

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

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?

Upvotes: 2

Views: 17312

Answers (3)

Ohad Eytan
Ohad Eytan

Reputation: 8464

Regardless the error you can simply do:

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

to get your desired result

Upvotes: 1

Noam
Noam

Reputation: 550

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)

Upvotes: 6

Sen Qiao
Sen Qiao

Reputation: 126

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

Upvotes: 1

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