"Josep M. Fontana" wrote
: return sorted(
: word_table.items(), key=lambda item: item[1], reverse=True
: )
By the way, I know what a lambda function is and I read about the
key
parameter in sorted() but I don't understand very well what
"key=lambda item: item[1]" does.
...
What I don
"col speed" wrote
Just my twopenneth, I'm a noob and I'm not going to try such a big
file on
my old machine, but:
1. Maybe create a *set* from the wordlist, loop through that, so you
call
"count" on wordlist only once. OR
This would be an improvement but still involves traversing the en
Alan Gauld wrote:
> The loop is a bit clunky. it would be clearer just to iterate over
> a_list:
>
> for item in a_list:
>words[item] = a_list.count(item)
This is a very inefficient approach because you repeat counting the number
of occurrences of a word that appears N times N times:
>>> w
"Josep M. Fontana" wrote
The code I started writing to achieve this result can be seen below.
You will see that first I'm trying to create a dictionary that
contains the word as the key with the frequency as its value. Later
on
I will transform the dictionary into a text file with the desire
