Python error in output

Question

I wrote a code like below

t_p=0
f_p=0
f_n=0
t_n=0

while True:
    line=raw_input()
    if not line:
        break
    actual,predicted=line.split(',')
    print actual, predicted
    if (actual==1 and predicted==1):
        t_p+=50
        print "tp", t_p
    elif(actual==0 and predicted==1):
        f_p+=-25.1
        print "fp", f_p
    elif(actual==1 and predicted==0):
        f_n+=-50.0
        print "fn", f_n
    else:
        t_n=0.0
        print "tn", t_n

score=t_p+f_p+f_n+t_n
print score

Now when I pass the following:

0,0
0 0
tn 0.0
0,1
0 1
tn 0.0
1,0
1 0
tn 0.0
1,1
1 1
tn 0.0
1,1
1 1
tn 0.0

It seems to be taking tn value only always which shouldnt be since the values satisfy other conditions based on the values of those two variables.

Neo X · Accepted Answer · 2017-02-13 01:16:22Z

1

line=raw_input() and actual,predicted=line.split(',') will give you actual and predicted in string, which will never equal to int 1. A conversion of the type will fix the problem.

answered Feb 13, 2017 at 1:16

Neo X

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1 Comment

Baktaawar Over a year ago

Aah. I didn't realize. Let me check that

Chiheb Nexus · Accepted Answer · 2017-02-13 01:51:48Z

Without converting your input data into int, you can keep them like a strings and compare them to your validation data which are tuples like the example below:

t_p = 0
f_p = 0
f_n = 0
t_n = 0

while True:
    data = tuple(raw_input().split(','))

    if data[0] == 'q' or data[0] == '' :
        break
    else:
        print data


    if data == ('1','1'):
        t_p += 50 
        print 'tp', t_p

    if data == ('0','1'):
         f_p -= 25.1
         print 'fp', f_p

    if data == ('1', '0'):
        f_n -= 50.0
        print 'fn', f_n

    else:
        tn = 0.0
        print 'tn', t_n

score = t_n + f_n, + f_p, t_p
print score

Demo:

1,1
('1', '1')
tp 50
tn 0
1,0
('1', '0')
fn -50.0
0,1
('0', '1')
fp -25.1
tn 0
0,0
('0', '0')
tn 0

(-50.0, -25.1, 50)

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