Trying to output the x most common words in a text file

问题

I'm trying to write a program that will read in a text file and output a list of most common words (30 as the code is written now) along with their counts. so something like:

word1 count1
word2 count2
word3 count3
...   ...
...   ...
wordn countn

in order of count1 > count2 > count3 >... >countn. This is what I have so far but I cannot get the sorted function to perform what I want. The error I get now is:

TypeError: list indices must be integers, not tuple

I'm new to python. Any help would be appreciated. Thank you.

 def count_func(dictionary_list):
  return dictionary_list[1]

def print_top(filename):
  word_list = {}
  with open(filename, 'r') as input_file:

    count = 0

    #best
    for line in input_file:
      for word in line.split():
        word = word.lower()
        if word not in word_list:
          word_list[word] = 1
        else:
          word_list[word] += 1

#sorted_x = sorted(word_list.items(), key=operator.itemgetter(1))
#  items = sorted(word_count.items(), key=get_count, reverse=True)

  word_list = sorted(word_list.items(), key=lambda x: x[1])

  for word in word_list:
    if (count > 30):#19
      break
    print "%s: %s" % (word, word_list[word])
    count += 1


# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
  if len(sys.argv) != 3:
    print 'usage: ./wordcount.py {--count | --topcount} file'
    sys.exit(1)

  option = sys.argv[1]
  filename = sys.argv[2]
  if option == '--count':
    print_words(filename)
  elif option == '--topcount':
    print_top(filename)
  else:
    print 'unknown option: ' + option
    sys.exit(1)

if __name__ == '__main__':
  main()

回答1:

Use the collections.Counter class.

from collections import Counter

for word, count in Counter(words).most_common(30):
    print(word, count)

Some unsolicited advice: Don't make so many functions until everything is working as one big block of code. Refactor into functions after it works. You don't even need a main section for a script this small.

回答2:

Using itertools' groupby:

from itertools import groupby

words = sorted([w.lower() for w in open("/path/to/file").read().split()])
count = [[item[0], len(list(item[1]))] for item in groupby(words)]
count.sort(key=lambda x: x[1], reverse = True)
for item in count[:5]:
    print(*item)

This will list the file's words, sort them and list unique words and their occurrence. Subsequently, the found list is sorted by occurrence by:
```
count.sort(key=lambda x: x[1], reverse = True)
```
The reverse = True is to list the most common words first.
In the line:
```
for item in count[:5]:
```
[:5] defines the number of most occurring words to show.

回答3:

First method as others have suggested i.e. by using most_common(...) doesn't work according to your needs cause it returns the nth first most common words and not the words whose count is less than or equal to n:

Here's using most_common(...): note it just print the first nth most common words:

>>> import re
... from collections import Counter
... def print_top(filename, max_count):
...     words = re.findall(r'\w+', open(filename).read().lower())
...     for word, count in Counter(words).most_common(max_count):
...         print word, count
... print_top('n.sh', 1)
force 1

The correct way would be as follows, note it prints all the words whose count is less than equal to count:

>>> import re
... from collections import Counter
... def print_top(filename, max_count):
...     words = re.findall(r'\w+', open(filename).read().lower())
...     for word, count in filter(lambda x: x[1]<=max_count, sorted(Counter(words).items(), key=lambda x: x[1], reverse=True)):
...         print word, count
... print_top('n.sh', 1)
force 1
in 1
done 1
mysql 1
yes 1
egrep 1
for 1
1 1
print 1
bin 1
do 1
awk 1
reinstall 1
bash 1
mythtv 1
selections 1
install 1
v 1
y 1

回答4:

Here is my python3 solution. I was asked this question in an interview and the interviewer was happy this solution, albeit in a less time-constrained situation the other solutions provided above seem a lot nicer to me.

    dict_count = {}
    lines = []

    file = open("logdata.txt", "r")

    for line in file:# open("logdata.txt", "r"):
        lines.append(line.replace('\n', ''))

    for line in lines:
        if line not in dict_count:
            dict_count[line] = 1
        else:
            num = dict_count[line]
            dict_count[line] = (num + 1)

    def greatest(words):
        greatest = 0
        string = ''
        for key, val in words.items():
            if val > greatest:
                greatest = val
                string = key
        return [greatest, string]

    most_common = []
    def n_most_common_words(n, words):
        while len(most_common) < n:
            most_common.append(greatest(words))
            del words[(greatest(words)[1])]

    n_most_common_words(20, dict_count)

    print(most_common)

来源：https://stackoverflow.com/questions/39299600/trying-to-output-the-x-most-common-words-in-a-text-file

标签

python

sorting

dictionary

tuples

sorted