如何在每个单词出现之前写出文本文件的名称？

Question

How can i write the text file name in each word frequency so that it first shows the fileno and then frequency of word in that file. for example: { like:['file1',2,'file2,'4'] } here like is the word that both file contains, i want to write file1 and file2 before their frequencies. It should be general for any number of files.

Here is my code

file_list = [open(file, 'r') for file in files] 
    num_files = len(file_list) 
    wordFreq = {}  
    for i, f in enumerate(file_list): 
        for line in f: 
            for word in line.lower().split():
                if not word in wordFreq:
                    wordFreq[word] = [0 for _ in range(num_files)]
                wordFreq[word][i] += 1

Answer 1

I know that my code is not pretty and not exactly what you want, but it is a solution. I would prefer using dictionary instead of a list structure like ['file1',2,'file2,'4']

Let's define 2 files as an example:

file1.txt:

this is an example

file2.txt:

this is an example
but multi line example

Here is the solution:

from collections import Counter

filenames = ["file1.txt", "file2.txt"]

# First, find word frequencies in files
file_dict = {}
for filename in filenames:
    with open(filename) as f:
        text = f.read()
    words = text.split()

    cnt = Counter()
    for word in words:
        cnt[word] += 1
    file_dict[filename] = dict(cnt)

print("file_dict: ", file_dict)

#Then, calculate frequencies in files for each word 
word_dict = {}
for filename, words in file_dict.items():
    for word, count in words.items():
        if word not in word_dict.keys():
            word_dict[word] = {filename: count}
        else:
            if filename not in word_dict[word].keys():
                word_dict[word][filename] = count    
            else:
                word_dict[word][filename] += count


print("word_dict: ", word_dict)

Output:

file_dict:  {'file1.txt': {'this': 1, 'is': 1, 'an': 1, 'example': 1}, 'file2.txt': {'this': 1, 'is': 1, 'an': 1, 'example': 2, 'but': 1, 'multi': 1, 'line': 1}}
word_dict:  {'this': {'file1.txt': 1, 'file2.txt': 1}, 'is': {'file1.txt': 1, 'file2.txt': 1}, 'an': {'file1.txt': 1, 'file2.txt': 1}, 'example': {'file1.txt': 1, 'file2.txt': 2}, 'but': {'file2.txt': 1}, 'multi': {'file2.txt': 1}, 'line': {'file2.txt': 1}}

Answer 2

This is a good use case for collections.Counter ; I suggest making a counter for each file.

from collections import Counter

def make_counter(filename):
    cnt = Counter()

    with open(filename) as f:
        for line in f:                # read line by line, is more performant for big files
            cnt.update(line.split())  # split line by whitespaces and updated word counts

    print(filename, cnt)
    return cnt

This function can be used for each file, making a dict that holds all the counters:

filename_list = ['f1.txt', 'f2.txt', 'f3.txt']
counter_dict = {                      # this will hold a counter for each file
    fn: make_counter(fn)
    for fn in filename_list}

Now a set can be used to get all the different words that appear in the files:

all_words = set(                      # this will hold all different words that appear
    word                              # in any of the files
    for cnt in counter_dict.values()
    for word in cnt.keys())

And these lines print each word and the count that word has in each file:

for word in sorted(all_words):
    print(word)
    for fn in filename_list:
        print('  {}: {}'.format(fn, counter_dict[fn][word]))

Obviously, you can adjust the printing to your specific needs, but this approach should allow you the flexibility you need.

---

If you rather have one dict with all the words as keys and their counts as values, you could try something like this:

all_words = {}

for fn, cnt in counter_dict.items():
    for word, n in cnt.items():
        all_words.setdefault(word, {}).setdefault(fn, 0)
        all_words[word][fn] += 0