count method in 'if' doesn't work - python

Question

I don't get it, I'm trying to count the 2 in this list and when it is like this:

hand=['D2', 'H5', 'S2', 'SK', 'CJ', 'H7', 'CQ', 'H9', 'D10', 'CK']
f=''.join(hand)
count2=f.count('2')
print count2

it works perfectly and it prints me 2 as the number of times the 2 is in the list. But when I'm putting it in an if it doesn't work:

def same_rank(hand, n):
    if hand.count('2')>n:
        print hand.count('2')
    else:
        print 'bite me'



hand=['D2', 'H5', 'S2', 'SK', 'CJ', 'H7', 'CQ', 'H9', 'D10', 'CK']
f=''.join(hand)
n=raw_input('Give n ')
print same_rank(hand,n)

If the user gives the n=1 then it is supposed to print 2 because the number 2 is twice in the list and I want it to be more than one that it is! So why it doesn't return that?

Martijn Pieters · Accepted Answer · 2014-05-16 14:00:57Z

5

raw_input() returns a string; strings are always sorted after numbers, so 2 > '1' is always False:

>>> 2 > '1'
False

Convert your input to an integer first:

n = int(raw_input('Give n '))

Had you used Python 3, you'd have gotten an exception instead:

>>> 2 > '1'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: int() > str()

Because Python 3 has done away with giving arbitrary types a relative ordering.

Next, you don't pass in f, you are passing in hand, the list:

>>> hand.count('2')
0
>>> f
'D2H5S2SKCJH7CQH9D10CK'
>>> f.count('2')
2

You probably wanted to pass in the latter, your function doesn't work otherwise.

edited May 16, 2014 at 14:00

answered May 16, 2014 at 13:44

Martijn Pieters

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4 Comments

RedRose23 Over a year ago

Nope just a total newbie tried to pass my university exam :P I tried it and it still returns me: bite me None I don't get it even if I put a value on n at first place I got this error

Martijn Pieters Over a year ago

@RedRose23: Ah, that's because you are passing in the wrong value. :-)

RedRose23 Over a year ago

Oh look it's finally working after 3 hours! Even though it still prints None with 2 but who cares? Thanks a lot :D Ok now it doesn't even print None cause I change print to return in def. Oh yeah! You're awesome! Thanks :D

Martijn Pieters Over a year ago

@RedRose23: print prints whatever your function returns. It returns None (the default). You don't need that print call, remove it.

chepner · Accepted Answer · 2014-05-16 13:45:12Z

0

n is a string, not an integer, when you pass it to same_rank. Either use

n = int(raw_input('Give n '))

or convert n when passing the value:

print same_rank(hand, int(n))

or have same_rank handle the conversion:

def same_rank(hand, n):
    n = int(n)
    if hand.count('2')>n:
        print hand.count('2')
    else:
        print 'bite me'

answered May 16, 2014 at 13:45

chepner

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1 Comment

RedRose23 Over a year ago

I tried all of them but the it prints: bite me None and not 2 as it supposed to print. I don't get what I'm doing wrong.

MONTYHS · Accepted Answer · 2014-05-16 13:49:09Z

0

n is string type please change into int Working Code:

def same_rank(hand, n):
    f=''.join(hand)
    if f.count('2')>int(n):
        print f.count('2')
    else:
        print 'bite me'



hand=['D2', 'H5', 'S2', 'SK', 'CJ', 'H7', 'CQ', 'H9', 'D10', 'CK']
f=''.join(hand)
n=raw_input('Give n ')
print same_rank(hand,n)

answered May 16, 2014 at 13:49

MONTYHS

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