To figure that last part out I just did a simple if statement.
for k in sorted(word_count, key=lambda x:word_count[x], reverse=True):
if word_count[k] >=300:
print (k, word_count[k])
And the output was correct.
I did have one more question though.
import os
from wordtools import extract_words
source_filepath=input("Enter the path to the source file:")
dest_filepath =input("Enter the path to the destination file:")
sourcef=open(source_filepath, 'r')
destf=open(dest_filepath, 'w')
for line in sourcef:
destf.write(line)
file=input ("Would you like to process another file?(Y/N):")
if file== "Y":
source_filepath=input("Enter the path to the source file:")
dest_filepath =input("Enter the path to the destination file:")
else:
This code asks the user for a source/dest_filepath.
I'm wondering how I can make it so the program can tell if the
source/dest_filepath the user entered is actually a program on the computer.
Also i have to ask the user if they would like to "process another
file(Y/N)?" and I'm not sure where to put that.
On Sun, May 4, 2014 at 10:38 PM, Jake Blank <jakenbl...@gmail.com> wrote:
> Hi,
>
> I finally got it.
> This was the code:
> for k in sorted(word_count, key=lambda x:word_count[x], reverse=True):
> print (k, word_count[k])
>
> The only question i have now is how to limit the amount of returns the
> program runs to the first 15 results.
>
>
>
>
> On Sun, May 4, 2014 at 10:19 PM, Brian van den Broek <
> brian.van.den.br...@gmail.com> wrote:
>
>> Hi Jake,
>>
>> Please do be sure to use Reply All rather than just Reply. I'm sending
>> my reply and and quotes from yours to the list; that way, others can
>> follow along, learn and help.
>>
>> Also, in general, reply under the messages to which you respond,
>> ideally trimming what isn't needed away. (You will see that is what I
>> have done below.) Yes, that's not how email is used outside of
>> technical circles. I'd maintain the technical circles's preference for
>> not top posting is right. But, right or wrong, it is what those whom
>> you are asking for free help prefer, so it is prudent to do it,
>> gritting your teeth if you must :-)
>>
>> On 4 May 2014 21:36, Jake Blank <jakenbl...@gmail.com> wrote:
>> > Hey Thanks for responding.
>> >
>> > So now my code looks like this:
>> > from wordtools import extract_words
>> >
>> > source_filepath=input("Enter the path to the source file:")
>> > dest_filepath =input("Enter the path to the destination file:")
>> >
>> > sourcef=open(source_filepath, 'r')
>> > destf=open(dest_filepath, 'w')
>> > for line in sourcef:
>> > destf.write(line)
>> > file=input ("Would you like to process another file?(Y/N):")
>> > if file== "Y":
>> > source_filepath=input("Enter the path to the source file:")
>> > dest_filepath =input("Enter the path to the destination file:")
>> > else:
>> > word_count = {}
>> > file = open (source_filepath, 'r')
>> > full_text = file.read().replace('--',' ')
>> > full_text_words = full_text.split()
>> >
>> > for words in full_text_words:
>> > stripped_words = words.strip(".,!?'`\"- ();:")
>> > try:
>> > word_count[stripped_words] += 1
>> > except KeyError:
>> > word_count[stripped_words] = 1
>> >
>> > ordered_keys = word_count.keys()
>> > sorted(ordered_keys)
>> > print ('This is the output file for Alice in Wonderland')
>> > for k in sorted(ordered_keys):
>> > print (k, word_count[k])
>> >
>> > The first part about the user specifying the file is a little off but
>> > besides that I am able to return all of the words in the story with the
>> > number of times they occur alphabetically. In order to return the
>> sorted
>> > list by number of times that each word occurs I am a little confused if
>> i
>> > have to change something in my print statement? I understand how i
>> have to
>> > sort the words by their associated values i'm confused where in my code
>> i
>> > would do that.
>> >
>> > Thanks, Jake
>>
>> > On Sun, May 4, 2014 at 9:16 PM, Brian van den Broek
>> > <brian.van.den.br...@gmail.com> wrote:
>> >>
>> >> On May 4, 2014 8:31 PM, "Jake Blank" <jakenbl...@gmail.com> wrote:
>>
>>
>> <snip Jake's original question>
>>
>>
>> >> Hi Jake,
>> >>
>> >> You are sorting the dictionary keys by the keys themselves, whereas
>> >> what you want is the keys sorted by their associated values.
>> >>
>> >> Look at the key parameter in
>> >> https://docs.python.org/3.4/library/functions.html#sorted.
>> >>
>> >> To get you started, here is an example in the vicinity:
>> >>
>> >> >>> data = ['abiab', 'cdocd', 'efaef', 'ghbgh']
>> >> >>> sorted(data)
>> >> ['abiab', 'cdocd', 'efaef', 'ghbgh']
>> >> >>> sorted(data, key=lambda x:x[2])
>> >> ['efaef', 'ghbgh', 'abiab', 'cdocd']
>> >> >>> def get_third(x): return x[2]
>> >> ...
>> >> >>> sorted(data, key=get_third)
>> >> ['efaef', 'ghbgh', 'abiab', 'cdocd']
>> >> >>>
>> >>
>> >> In case the lambda version is confusing, it is simply a way of doing
>> >> the get_third version without having to create a function outside of
>> >> the context of the sorted expression.
>> >>
>> >> If that sorts you, great. If not, please do ask a follow-up. (I was
>> >> trying not to do it for you, but also not to frustrate by giving you
>> >> too little of a push.)
>>
>>
>> So, the code in your second message didn't seem to reflect any changes
>> in light of the hint I gave. Did you not see how to apply it? If so,
>> that's fine. But, rather than leave me guessing, it would be better to
>> say; that way I, or others in the thread, can better direct efforts to
>> help.
>>
>> Does this help more?
>>
>> >>> data = {'a':2, 'b':4, 'c':3, 'd':1}
>> >>> sorted(data,key=lambda x:data[x])
>> ['d', 'a', 'c', 'b']
>> >>> sorted(data,key=lambda x:data[x])
>> ['d', 'a', 'c', 'b']
>> >>> data = {'a':2, 'b':4, 'c':3, 'd':1}
>> >>> sorted(data)
>> ['a', 'b', 'c', 'd']
>> >>> sorted(data,key=lambda x:data[x])
>> ['d', 'a', 'c', 'b']
>> >>> for letter in sorted(data,key=lambda x:data[x]):
>> ... print(letter, data[letter])
>> ...
>> d 1
>> a 2
>> c 3
>> b 4
>> >>> for letter in sorted(data,key=lambda x:data[x],reverse=True):
>> ... print(letter, data[letter])
>> ...
>> b 4
>> c 3
>> a 2
>> d 1
>> >>>
>>
>> Can you see how key=lambda x:data[x] is forcing the sort to consider
>> not the keys of the dictionary, but their associated values?
>>
>> Best,
>>
>> Brian vdB
>>
>
>
_______________________________________________
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
https://mail.python.org/mailman/listinfo/tutor
