CODEWITHSUNDEEP
pythonnumpy
Essex
Essex

Reputation: 6138

Returned result is false with numpy conditions

I wrote a little script in order to return a list of 'characters' from a list of values :

List of values :

11
11
14
6
6
14

My script :

# -*- coding:utf-8 -*-

import numpy as np

data = np.loadtxt('/Users/valentinjungbluth/Desktop/produit.txt',
                  dtype = int)


taille = len(data)

for j in range (0, taille) :

    if data[j] == 1 or 2 or 3 or 11 or 12 or 13 :
        print 'cotisation'

    if data[j] == 14 :
        print 'donation'

    else :
        print 'part'

What I need to get :

'cotisation'
'cotisation'
'donation'
'part'
'part'
'donation'

What I get :

cotisation
part
cotisation
part
cotisation
donation

I don't see where I made an error .. If someone could read my script and maybe correct him ?

Thank you

Upvotes: 0

Views: 70

Answers (4)

Skycc
Skycc

Reputation: 3555

Scimonster point out your 1st issue in the code, you need elif for the second if.

The 2nd issue is the or part, your if condition is being intepret as 

 if (data[j] == 1) or 2 or 3 or 11 or 12 or 13

You could fixed them as below

if data[j] in [1, 2, 3, 11, 12, 13] :
    print 'cotisation'
elif data[j] == 14 :
    print 'donation'
else :
    print 'part'

Upvotes: 0

lemonhead
lemonhead

Reputation: 5518

There are a couple issues here:

if x == 1 or 2 or 3 or ... does not do what you are expecting.

Each clause is evaluated independently, so python does not know you are implying if x == 1 or x == 2 or x == 3, etc. It instead interprets this as if x == 1 or if 2 or if 3 ... if 2 and the like is actually a sensical if statement in python and is basically tantamount to asking if True (all non-zero integer values have a "truthiness" of True). So your statement is really equivalent to if data[j] == 1 or True or True or True or True, which reduces to if True. In other words, the condition is always satisfied regardless of the value. What you probably meant was:

if data[j] == 1 or data[j] == 2 or data[j] == 3 or data[j] == 11 ...

Your if-else construction has two blocks when one is intended

Currently boils down to something like the following in pseudocode:

if some condition:
    print 'something'

now regardless of the first condition, if some other disjoint condition:
    print 'something else'
otherwise:
    print 'does not match other condition'

In other words, by using two ifs, the second if block, of which the else is a part, is treated as completely independent of the first if. So cases that satisfy the first condition could also satisfy the else, which is sounds like is not what you wanted as each character should print exactly once. If you want it to be 3 disjoint cases, you need to use elif instead so it is all considered part of a single block:

E.g.

if condition:
    print 'something'
elif other condition:
    print 'something else'
else:
    print 'does not match EITHER of the two above condtions'

Upvotes: 1

JackTheCrab
JackTheCrab

Reputation: 132

Try:

for j in range(0, taille):
    if data[j] in {1, 2, 3, 11, 12, 13}:
        print ("cotisation")
    elif data[j] == 14:
        print ("donation")
    else:
        print ("part")

Upvotes: 1

Scimonster
Scimonster

Reputation: 33409

if data[j] == 1 or 2 or 3 or 11 or 12 or 13 :
    print 'cotisation'

if data[j] == 14 :
    print 'donation'

else :
    print 'part'

You have two separate ifs here, and one has an else. They should all flow together, so the second should be elif.

Upvotes: -1

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