CODEWITHSUNDEEP
pythonarrayscompare
mar
mar

Reputation: 333

How to compare two arrays in python based on a value

I am doing classification. I have two arrays, the first is 'Actual', and the second is 'Predicted'. I want to compare these two arrays. Suppose the first array is:

Actual = [1, 1, 2, 3, 1]

this tells us that the the first, second, and the last indexes are corresponding to class 1.

The 'Predicted' array is:

Predicted = [1, 1, 1, 3, 3]

this tells us that the first and second indexes have been predicted accurately.

I want the output the tells us just those indexes that accurately predicted as 1, like this:

output = [True, True, False, False, False]

Update I want to evaluate just based on value 1. If you see, the forth predicted value is accurately predicted by 3, but I do not want that, because I want evaluate 1 value.

Upvotes: 3

Views: 8426

Answers (4)

Ozgur Vatansever
Ozgur Vatansever

Reputation: 52223

Presuming length of two lists are same:

>>> [(x == y == 1) for x, y in zip(Actual, Predicted)]
[True, True, False, False, False]

To feel safe;

>>> from itertools import izip_longest
>>> [(x == y == 1) for x, y in izip_longest(Actual, Predicted, fillvalue=0)]
[True, True, False, False, False]

Upvotes: 6

Michael Aaron Safyan
Michael Aaron Safyan

Reputation: 95639

First, some array basics:

  1. To get the number of elements in an array, use len:

     x = ['a', 'b', c']
 y = len(x)  # y == 3

  2. To access the ith element of an array, use []:

     x = ['a','b', 'c']
 y = x[1]  # y == 'b'

  3. To obtain an iterator with values 0, 1, ..., n-1 ,use range(n):

     x = list(range(3))  # x = [0, 1, 2]

  4. To iterate through the values of an array, use for ... in:

     x = ['a', 'b', 'c']
 for value in x:
    process(value)   # called for 'a', 'b', and 'c'

  5. To compare items for equality, use == (or != for inequality).

Putting this altogether,now:

def ComputeArrayDifference(a, b):
   alen = len(a)
   blen = len(b)
   if alen != blen:
      raise DifferingSizesException('Inputs have different sizes', a, b)
   result = []
   for i in range(alen):
      result.append(a[i] == b[i])
   return result

Upvotes: 0

Anand S Kumar
Anand S Kumar

Reputation: 91009

If you do not mind using numpy library, then this can be done very easily -

In [10]: import numpy as np

In [11]: Actual=[1,1,2,3,1]

In [12]: ActualNp = np.array(Actual)

In [13]: Predicted=[1,1,1,3,3]

In [15]: PredictedNp = np.array(Predicted)

In [20]: (ActualNp == PredictedNp) & (PredictedNp == 1)
Out[20]: array([ True,  True, False, False, False], dtype=bool)

If not, assumming that you only want to check till the length of the smallest list (If they are of different lengths), you can use zip -

>>> Actual=[1,1,2,3,1]
>>> Predicted=[1,1,1,3,3]
>>> output = [a == b == 1 for a,b in zip(Actual,Predicted)]
>>> output
[True, True, False, False, False]

Upvotes: 2

helado
helado

Reputation: 895

A one liner simple approach would be 

def get_prediction_results(prediction, actual, predicted):
 return [a == predicted[i] == prediction for i, a in enumerate(actual)]

>>> get_prediction_results(1, [1,1,2], [1,1,2])
[True, True, False]

Upvotes: 0

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