Type Erorr nr2

[Mackjan]

Hi
I have got the TypeError. I use lists at the same way in my other programs it works excellent, but just in this program I have got problem don’t know how to solve it?
Traceback (most recent call last):
File "E:\extra_b.py", line 42, in <module>
ordna(pile)
File "E:\extra_b.py", line 22, in ordna
new_num = num-1
TypeError: unsupported operand type(s) for -: 'list' and 'int'

import random
def lasin ():
     c=int(input ("Enter an integer "))
     return c                     
def dela(c):
     sum=0
     pile = []
     while sum < c:
         b = random.randrange (1, 10 - sum + 1)
         pile.append(b)
         sum += 1
     print pile,sum
     return sum, pile
def ordna(x):
    b = 0
    y = []
    for num in x:
        new_num = num-1
        if new_num<0:
            del new_num
        else:
            y.append(new_num)
            b+=1
    y.append (b)
    y.sort()
    y.reverse ()
    print " ".join(map(str,y))
    return y
c=lasin ()
pile=dela(c)
ordna(pile)

Thanks a lot for your help.

[jrcagle]

Instead of debugging, if it's OK, I'll talk about *how* to debug.

Since the error is coming from line 42 (line 31 in the code above), insert a print statement just prior to line 31:

...
pile = dela(c)
print pile
ordna(pile)

That way, you can check your assumption that "pile" is a list of numbers. (Hint: it isn't!)

Jeff

[Mackjan]

I see I want to work with a list of number and my first function return a list of number. I don’t want to mix a deferent type int and list.
But when I did as you said, I got a deferent list which looks like that:
First print --> [2, 5, 7, 7] 4
Second print --> (4, [2, 5, 7, 7])
I don’t know why they are deferent?
May be same where my program has mixed a deferent kind of type?? I don’t see it !

[woooee]

Obviously this is not a list of numbers, so you want to take a look at the function that returns this. Add print statements to see what is going on throughout the program until you understand what is happening

[Freaky_Chris]

First print --> [2, 5, 7, 7] 4
Second print --> (4, [2, 5, 7, 7])

the first print is showing 2 different objects.
The second print is showing 1 object.

Read about lists & tuples, embedding in particular, this may help you understand

Chris

[Mackjan]

Hi
Thanks guys I find the reason of error but I still don’t understand fallowing.
My first function dela ( ) return pile and sum, pile is a list of numbers and sum is a digit but I used only pile which is a list and sent it to my other function. Sum which is a digit and returns by function but didn´t used.
If a function return a, b can I use only a?? And how b can influence result of a??

I have changed my cod and it works now.

def dela(c):
     sum=0
     pile = []
     while sum < c:
         b = random.randrange (1, 10 - sum + 1)
         pile.append(b)
         sum += 1
     return pile

[Freaky_Chris]

def dela(c):
     sum=0
     pile = []
     while sum < c:
         b = random.randrange (1, 10 - sum + 1)
         pile.append(b)
         sum += 1
     print pile,sum
     return sum, pile
def ordna(x):
    b = 0
    y = []
    for num in x:
        new_num = num-1
        if new_num<0:
            del new_num
        else:
            y.append(new_num)
            b+=1
    y.append (b)
    y.sort()
    y.reverse ()
    print " ".join(map(str,y))
    return y
c=lasin ()
pile=dela(c)
ordna(pile)

This is fine you can use both a and b that are returned, when you return them in the way you have the result is a tuple.

just means that you have to access each bit indervidual.

so to use the list of numbers that is returned in this case you would have to put

ordna(pile[1])

If you print them you will see that you can access both

print pile[0]
print pile[1]

Notice that the first line will print the interger that you returned and the second line will print the list of numbers you returned.

Chris

[woooee]

To work the way you want, your original code would have been
c=lasin ()
sum, pile=dela(c)
ordna(pile)
If the function returns two variables, but the calling line only receives one variable, then python returns one tuple, because you only have one block of memory allocated for the return.