显示Python字典中查询文件中每行单词的次数

Question

我有一个docs.txt文件，其中包含以下三行：

joyously urgently truthfully seemingly broadly urgently knowingly urgently steadily
joyously urgently truthfully seemingly rigidly broadly rigidly suddenly healthily commonly often
tremendously totally steadily sharply totally

我有一个querys.txt文件，其中包含：

urgently
rigidly suddenly
totally steadily

以及守则：

dictionary = {}
    angledictionary={}
    document= open('docs.txt', 'r')
    for line in document:
        lined=line.split()
        for word in lined:
            if word not in dictionary.keys():
                dictionary[word]=0
            dictionary[word]+=1
    dictionary=dict.fromkeys(dictionary,0)
    with open("queries.txt", "r") as open_queries:
        searchquery = open_queries.read().split("\n")
    with open('docs.txt', 'r') as openrelevancy:
        words = openrelevancy.read().split("\n")
    for query in searchquery:
        print('Query:', query)
        relevant = []
        line_number = 0
        for word in words:
            line_number += 1
            if query in word:
                relevant.append(line_number)
        print('Relevant Documents:', *relevant)

现在，每行的字典字数为0，我正在尝试这样做： 对于第1行：

{'joyously': 1, 'urgently': 3, 'truthfully': 1, 'seemingly': 1, 'broadly': 1, 'knowingly': 1, 'steadily': 1, 'rigidly': 0, 'suddenly': 0, 'healthily': 0, 'commonly': 0, 'often': 0, 'tremendously': 0, 'totally': 0, 'sharply': 0}

第2行：

{'joyously': 1, 'urgently': 1, 'truthfully': 1, 'seemingly': 1, 'broadly': 1, 'knowingly': 1, 'steadily': 1, 'rigidly': 2, 'suddenly': 1, 'healthily': 1, 'commonly': 1, 'often': 1, 'tremendously': 0, 'totally': 0, 'sharply': 0}

第3行：

{'joyously': 0, 'urgently': 0, 'truthfully': 0, 'seemingly': 0, 'broadly': 0, 'knowingly': 0, 'steadily': 0, 'rigidly': 0, 'suddenly': 0, 'healthily': 0, 'commonly': 0, 'often': 0, 'tremendously': 1, 'totally': 2, 'sharply': 1}

我怎样才能解决这个问题

Answer 1

下面的代码将计算给定文件中的每个单词

import os

def to_dict(file_path, delimiter=" "):
    dict_list = []  # Each index is a line
    if os.path.exists(file_path):
        with open(file_path, 'rb') as in_file:
            for line in in_file.readlines():  # Grab each line as a list
                line = line.split(delimiter)  # Split at our delim
                new_dict = {}
                for word in line:  # Code to count each index
                    word = word.rstrip()  # Remove formatting
                    if word in new_dict:
                        new_dict[word] += 1
                    else:
                        new_dict[word] = 1
                dict_list.append(new_dict)
        return dict_list
    else:
        print("{} Does not exist. Check the path and try again".format(file_path))

dict_count = to_dict(the_file_path, " ")
if dict_count:  # We found and converted file to a dict
    # function to query against words

Answer 2

一个简单的方法是使用collections.Counter：

from collections import Counter

with open("file.txt","r") as f:
    lineList = f.readlines()
for i,k in enumerate(map(lambda x:x.strip().split(" "),lineList)): # split each word
    print(f"Line {i+1}:{Counter(k)}")

结果:

Line 1:Counter({'urgently': 3, 'joyously': 1, 'truthfully': 1, 'seemingly': 1, 'broadly': 1, 'knowingly': 1, 'steadily': 1})
Line 2:Counter({'rigidly': 2, 'joyously': 1, 'urgently': 1, 'truthfully': 1, 'seemingly': 1, 'broadly': 1, 'suddenly': 1, 'healthily': 1, 'commonly': 1, 'often': 1})
Line 3:Counter({'totally': 2, 'tremendously': 1, 'steadily': 1, 'sharply': 1})

---

或者你可以用口述：

l = ['joyously', 'urgently', 'truthfully', 'seemingly', 'broadly', 'knowingly', 'steadily', 'rigidly', 'suddenly', 'healthily', 'commonly', 'often', 'tremendously', 'totally', 'sharply']
with open("file.txt","r") as f:
    lineList = f.readlines()

for line,i in enumerate(map(lambda x:x.strip().split(" "),lineList)):
    d = dict.fromkeys(l, 0) # load a dict from the list
    for j in i:
        d[j] += 1
    print(f"For line {line+1}:{d}")

结果:

For line 1:{'joyously': 1, 'urgently': 3, 'truthfully': 1, 'seemingly': 1, 'broadly': 1, 'knowingly': 1, 'steadily': 1, 'rigidly': 0, 'suddenly': 0, 'healthily': 0, 'commonly': 0, 'often': 0, 'tremendously': 0, 'totally': 0, 'sharply': 0}
For line 2:{'joyously': 1, 'urgently': 1, 'truthfully': 1, 'seemingly': 1, 'broadly': 1, 'knowingly': 0, 'steadily': 0, 'rigidly': 2, 'suddenly': 1, 'healthily': 1, 'commonly': 1, 'often': 1, 'tremendously': 0, 'totally': 0, 'sharply': 0}
For line 3:{'joyously': 0, 'urgently': 0, 'truthfully': 0, 'seemingly': 0, 'broadly': 0, 'knowingly': 0, 'steadily': 1, 'rigidly': 0, 'suddenly': 0, 'healthily': 0, 'commonly': 0, 'often': 0, 'tremendously': 1, 'totally': 2, 'sharply': 1}