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Date: Tue, 7 Feb 2012 21:37:20 +0000
Subject: Re: iterating over list with one mising value
From: Arnaud Delobelle <arnodel@gmail.com>
To: Sammy Danso <samdansobe@yahoo.com>
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On 7 February 2012 20:23, Sammy Danso <samdansobe@yahoo.com> wrote:
>
> Hi Expert,
> Thanks for your responses and help. thought I should provide more informa=
tion for clarity.

Please don't top-post.

> Please find the error message below for more information
>
> =C2=A0=C2=A0=C2=A0for (key, value) in wordFreq2:
> ValueError: need more than 1 value to unpack
>
> this is a sample of my data
>
> ['with', 3, 'which', 1, [...] , 3, 'other']
>
> What I would like to do is to pad the last value 'other' with a default s=
o I can iterate sucessfully

* It would be better if you provided the code that produces the error,
rather than just the error.  This would allow others to diagnose your
problem without having to guess what you're really doing (see my guess
below)

* Also, don't truncate the traceback.

My guess: you're running a loop like this, where each item is unpacked:

    for (key, value) in wordFreq2:
        print key, value

on data like that:

    wordFreq2 =3D ['with', 3, 'which', 1, 'were', 2, 'well', 1]

Your list is flat so the unpacking fails.  For it to work, you need
your list to be of the form:

    wordFreq2 =3D [('with', 3), ('which', 1), ('were', 2), ('well', 1)]

Then it will work.  The quickest way to transform your list to the
required form is something like this:

    def pairs(seq, fillvalue):
        it =3D iter(seq)
        return list(izip_longest(it, it, fillvalue=3Dfillvalue)

so you can do:

    word_freq_pairs =3D pairs(wordFreq2)

--=20
Arnaud