Median: 'list' object has no attribute 'len' . . . HELP!

#1



https://www.codecademy.com/courses/python-intermediate-en-rCQKw/2/5?curriculum_id=4f89dab3d788890003000096#


Error: (with [1] supplied)
Oops, try again.
median([1]) resulted in an error: 'list' object has no attribute 'len'


I need help here obviously. It keeps erroring out. With median[4.5.5.4], it kept returning 4 instead of 4.5. And with median[1], I keep getting the one I mentioned above: median([1]) resulted in an error: 'list' object has no attribute 'len'

Any ideas?


def median(x):
    new_lst = x
    s = new_lst.sort()  #sortiing list
    leng = new_lst.len(s)  #getting length
        #Below: was getting an error when the list was just a [1] so made the 'if' and 'elif' to check if length is greater than 1 and then the other half of the argument.
    if leng > 1 and leng % 2 != 0:  
        med = (leng / 2.0)
        return med
    elif leng > 1 and leng % 2 == 0:
        a = (leng / 2.0)
        b = a - 1
        c = s.index[a]
        d = s.index[b]
        med = s[c] + s[d] / 2.0
        return med
    else:
        return s

Thanks in advance for your help!!!

#2

this:

new_lst = x

will not make a copy of the list. use list() to make a copy:

new_list = list(x)

this doesn't work either:

s = new_lst.sort()

.sort() is a built in function, it returns the sorted list in the same variable:

new_list.sort()

then you use len really weird, if you want the length of something you just use len:

x = len([1,2,3,4])

no need for the dot notation.

Good, now you fiddle with it, if you need more help post an updated version of your code

#3

@stetim94

Thank you!

Ugh I know len() for length, duh. I can't believe I did that. I think it's because I had new_lst.count() in there, realized that's not what I needed and just replaced the dot count with a dot len. Stupid LOL!

So with new_lst.sort() will just sort the new_lst and I don't need an argument?

And for the copy: I thought that that is how you make a working copy so as to not screw up the original list...

I'll work on these and might be back. Thanks for your help!!!

#4

yes, exactly.

Nope, that is not the case. You can make variable which refer to the same list

#5

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